Trigonometry Booster with Problems and Solutions for IIT JEE Main and Advan

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Trigonometry Booster with Problems & Solutions for

JEE Main and Advanced

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About the Author REJAUL MAKSHUD (RM) Post Graduated from Calcutta University in PURE MATHEMATICS. Presently, he trains IIT Aspirants at RACE IIT Academy, Jamshedpur.

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Trigonometry Booster with Problems & Solutions for

JEE Main and Advanced

Rejaul Makshud M. Sc. (Calcutta University, Kolkata)

McGraw Hill Education (India) Private Limited CHENNAI McGraw Hill Education Offices Chennai New York St Louis San Francisco Auckland Bogotá Caracas Kuala Lumpur Lisbon London Madrid Mexico City Milan Montreal San Juan Santiago Singapore Sydney Tokyo Toronto

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McGraw Hill Education (India) Private Limited Published by McGraw Hill Education (India) Private Limited 444/1, Sri Ekambara Naicker Industrial Estate, Alapakkam, Porur, Chennai - 600 116

Coordinate Geometry Booster Copyright © 2017, McGraw Hill Education (India) Private Limited. No part of this publication may be reproduced or distributed in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise or stored in a database or retrieval system without the prior written permission of the publishers. The program listings (if any) may be entered, stored and executed in a computer system, but they may not be reproduced for publication. This edition can be exported from India only by the publishers, McGraw Hill Education (India) Private Limited ISBN (13): 978-93-5260-248-3 ISBN (10): 93-5260-248-X Information contained in this work has been obtained by McGraw Hill Education (India), from sources believed to be reliable. However, neither McGraw Hill Education (India) nor its authors guarantee the accuracy or completeness of any information published herein, and neither McGraw Hill Education (India) nor its authors shall be responsible for any errors, omissions, or damages arising out of use of this information. This work is published with the understanding that McGraw Hill Education (India) and its authors are supplying information but are not attempting to render engineering or other professional services. If such services are required, the assistance of an appropriate professional should be sought. Typeset at The Composers, 260, C.A. Apt., Paschim Vihar, New Delhi 110 063 and text and cover printed at

Cover Designer: Creative Designer visit us at: www.mheducation.co.in

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Dedicated to My Parents

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Preface

TRIGONOMETRY BOOSTER with Problems & Solutions for JEE Main and Advanced is meant for aspirants preparing for the entrance examination of different technical institutions, especially NIT/IIT/BITSAT/IISc. In writing this book, I have drawn heavily from my long teaching experience at National Level Institutes. After many years of teaching I have realised the need of designing a book that will help the readers to build their base, improve their level of mathematical concepts and enjoy the subject. This book is designed keeping in view the new pattern of questions asked in JEE Main and Advanced Exams. It has five chapters. Each chapter has the concept booster followed by a large number of exercises with the exact solutions to the problems as given below: Level - I Level - II Level - III Level - IV (0.......9) Passages Matching Reasoning Previous years papers

: : : : : : : : :

Problems based on Fundamentals Mixed Problems (Objective Type Questions) Problems for JEE Advanced Exam Tougher Problems for JEE Advanced Exams Integer type Questions Comprehensive Link Passages Matrix Match Assertion and Reason Questions asked in Previous Years’ IIT-JEE Exams

Remember friends, no problem in mathematics is difficult. Once you understand the concept, they will become easy. So please don’t jump to exercise problems before you go through the Concept Booster and the objectives. Once you are confident in the theory part, attempt the exercises. The exercise problems are arranged in a manner that they gradually require advanced thinking. I hope this book will help you to build your base, enjoy the subject and improve your confidence to tackle any type of problem easily and skilfully. My special thanks goes to Mr. M.P. Singh (IISc. Bangalore), Mr. Manoj Kumar (IIT, Delhi), Mr. Nazre Hussain (B. Tech.), Dr. Syed Kashan Ali (MBBS) and Mr. Shahid Iqbal, who have helped, inspired and motivated me to accomplish this task. As a matter of fact, teaching being the best learning process, I must thank all my students who inspired me most for writing this book. I would like to convey my affectionate thanks to my wife, who helped me immensely and my children who bore with patience my neglect during the period I remained devoted to this book. I also convey my sincere thanks to Mr Biswajit Das of McGraw Hill Education for publishing this book in such a beautiful format.

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viii

Preface

I owe a special debt of gratitude to my father and elder brother, who taught me the first lesson of Mathematics and to all my learned teachers— Mr. Swapan Halder, Mr. Jadunandan Mishra, Mr. Mahadev Roy and Mr. Dilip Bhattacharya, who instilled the value of quality teaching in me. I have tried my best to keep this book error-free. I shall be grateful to the readers for their constructive suggestions toward the improvement of the book.

Rejaul Makshud M. Sc. (Calcutta University, Kolkata)

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Contents

Preface

Chapter 1

vii

The Ratios and Identities

1.1–1.87

Introduction 1.1 Measurement of Angles 1.1 Trigonometrical Ratios 1.2 Limits of the values of Trigonometrical Functions 1.2 Sign of Trigonometric Ratios 1.2 T-ratios of the Angle (–q ), in Terms of q, for All Values of q 1.3 T-ratios of the Different Angles in Terms of q, for All Values of q 1.3 Graph of Trigonometric Functions 1.3 T-ratios of Compound Angles 1.4 Some Important Deductions 1.4 Transformation Formulae 1.5 Multiple Angles 1.6 Some Important Deductions 1.6 The Maximum and Minimum Values of 1.7 Sub–Multiple Angles 1.8 Conditional Trigonometrical Identities 1.11 Trigonometrical Series 1.12 Exercises 1.12 Answers 1.30 Hints and Solutions 1.30

Chapter 2

Trigonometric Equations

2.1–2.53

Definition 2.1 Solution of a Trigonometric Equation 2.1 General solution of Trigonometric Equations 2.1 A Trigonometric Equation is of the Form 2.1 Principal Value 2.1 Method to Find Out the Principal Value 2.2 Solutions in Case of Two Equations are Given 2.2 Some Important Remarks to Keep in Mind while Solving a Trigonometric Equation 2.2 Types of Trigonometric Equations: 2.2 Exercises 2.3 Answers 2.17 Hints and Solutions 2.18

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x

Chapter 3

Contents

Trigonometric In-Equation

3.1–3.15

Trigonometric Inequalities 3.1 Exercises 3.2 Answers 3.4 Hints and Solutions 3.5

Chapter 4

Inverse Trigonometric Functions

4.1–4.84

Inverse Function 4.1 Introduction to Inverse Function 4.1 Inverse Trigonometric Functions 4.1 Graphs of Inverse Trigonometric Functions 4.2 Constant Property 4.4 Conversion of Inverse Trigonometric Functions 4.4 Composition of Trigonometric Functions and its Inverse 4.5 Composition of Inverse Trigonometric Functions and Trigonometric Functions 4.5 Sum of Angles 4.6 Multiple Angles 4.7 More Multiple Angles 4.7 Exercises 4.8 Answers 4.24 Hints and Solutions 4.25

Chapter 5

Properties of Triangles

5.1–5.92

Properties of Triangles 5.1 Introduction 5.1 Sine Rule 5.1 Cosine Rule 5.1 Projection Formulae 5.1 Napier’s Analogy (Law of Tangents) 5.1 Half-Angled Formulae 5.1 Area of Triangle 5.2 m-n Theorem 5.2 Radii of Circle Connected with a Triangle 5.3 Inscribed Circle and its Radius 5.3 Escribed Circle of a Triangle and Their Radii 5.3 Regular Polygon 5.4 Orthocentre and Pedal Triangle of Any Triangle 5.4 Distance between the Circumcentre and Orthocentre 5.5 Distance Between the Circumcentre and the Incentre 5.6 Distance between the Circumcentre and Centroid 5.6 Distance Between the Incentre and Orthocentre 5.6 Excentral Triangle 5.7 Quadrilateral 5.9 Exercises 5.12 Answers 5.30 Hints and Solutions 5.30

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1.1

The Ratios and Identities

C H A P T E R

1

The Ratios and Identities

CONCEPT BOOSTER 1.1

C

q

INTRODUCTION

Trigonometry (from Greek trigonon “triangle” + metron “measure”) is a branch of mathematics that studies triangles and the relationships of the lengths of their sides and the angles between those sides. Trigonometry defines the trigonometric functions, which describe those relationships and have applicability to cyclical phenomena, such as waves. This field, evolved during the third century BC as a branch of geometry, was used extensively for astronomical studies. It is also the foundation of the practical art of surveying. Trigonometry basics are often taught in school either as a separate course or as part of a pre-calculus course. The trigonometric functions are pervasive in parts of pure mathematics and applied mathematics such as Fourier analysis and the wave equation, which are in turn essential to many branches of science and technology.

1.2

–ve angle O

MEASUREMENT OF ANGLES

1. Angle: The measurement of an angle is the amount of rotation from the initial side to the terminal side. 2. Sense of an Angle: The sense of an angle is +ve or –ve based on whether the initial side rotates in the anticlock-wise or clockwise direction to get the terminal side.

D

3. System of measuring angles There are three systems of measuring angles such as (i) Sexagesimal system (ii) Centisimal system (iii) Circular system 4. In sexagesimal system, we have 1 right angle = 90o 1o = 60¢ 1¢ = 60≤ 5. In centasimal system, we have 1 right angle = 100g 1g = 100¢ 1¢ = 100≤ 6. In circular system, the unit of measurement is radian Radian: One radian is the measure of an angle subtended at the centre of a circle by an arc of length equal to the radius of the circle. Here, –AOB = 1 radian = 1e. B

O

1°

A

B

O

A Positive angle

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Notes (i) When an angle is expressed in radians, the word radian is omitted.

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1.2

Trigonometry Booster

Ê 22 ˆ (ii) Since 180 = p radian = Á radian = 0.01746 Ë 7 ¥ 180 ˜¯ radian 180∞ Ê 180 ˆ (iii) 1 radian = =Á ¥ 7˜ = 57o16¢22¢ Ë ¯ p 22 (iv) The angle between two consecutive digits is 30° Êp ˆ ÁË radians˜¯ 6

1.3.3

o

(v) The hour hand rotates through an angle of 30° in one Ê 1ˆ hour (i.e. Á ˜ in one minute) Ë 2¯ (vi) The minute hand rotates through an angle of 6° in one minute. (vii) The relation amongst three systems of measurement of an angle is D G 2R = = 90∞ 100 p (viii) The number of radians in an angle subtended by an Arc s arc of a circle at the centre is , i.e., q = Radius r

1.3 1.3.1

TRIGONOMETRICAL RATIOS Definitions of Trigonometric Ratios C h

A

q

p

B

b

1. sin q =

p h

2. cosec q =

3. cos q =

b h

4. sec q =

h b

5. tan q =

p b

6. cot q =

b p

1.3.2

h p

Signs of Trigonometrical Ratios

The signs of the trigonometrical ratios in different quadrants are remembered by the following chart. sin and cosec are +ve and rest are –ve tan and cot are +ve and rest are –ve

(i) sin q ◊ cosec q (ii) cos q ◊ sec q (iii) tan q ◊ cot q sin q Step-II: (i) tan q = cos q cos q (ii) cot q = sin q Step-I:

=1 =1 =1

sin q ◊ cosec q = 1 cos q ◊ sec q = 1 tan q ◊ cot q = 1 sin2 q + cos2 q = 1 sec2 q = 1 + tan2 q cosec2 q = 1 + cot2 q

Step-III: (i) (ii) (iii) Step-IV: (i) (ii) (iii)

Step-V: Ranges of odd power t-ratios (i) –1 £ sin2n + 1 q, cos2n + 1 q £ 1 (ii) – < tan2n + 1 q, cot2n + 1 q < (iii) cosec2n + 1 q, sec2n + 1 q ≥ 1 cosec2n + 1 q, sec2n + 1 q £ –1 where n Œ W Step-VI: Ranges of even power t-ratios (i) 0 £ sin2n q, cos2n q £ 1 (ii) 0 £ tan2n q, cot2n q < (iii) 1 £ cosec2n q, sec2n q < where n Œ N

1.4 1. 2. 3. 4. 5. 6.

1.5

LIMITS OF THE VALUES OF TRIGONOMETRICAL FUNCTIONS –1 £ sin q £ 1 –1 £ cos q £ 1 cosec q ≥ 1 and cosec q £ –1 sec q ≥ 1 and sec q £ –1 – < tan q < – < cot q <

SIGN OF TRIGONOMETRIC RATIOS

(E) Rotation 90

All t-ratios are +ve cos and sec are +ve and rest are –ve

It is also known as all, sin, tan, cos formula.

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Relation between the Trigonometrical Ratios of an Angle

0, 360

180

270

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1.3

The Ratios and Identities –270

0, –360

–180

–90

1.6

T-RATIOS OF THE ANGLE (–q ), IN TERMS OF q, FOR ALL VALUES OF q

1. (i) (ii) (iii) (iv) (v) (vi)

1.7

sin (–q) = –sin q cos (–q) = cos q tan (–q) = –tan q cosec (–q) = –cosec q sec (–q) = sec q cot (–q) = –cot q

T-RATIOS OF THE DIFFERENT ANGLES IN TERMS OF q, FOR ALL VALUES OF q

2. (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) 3 (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) 4. (i) (ii) (iii) (iv) (v) (vi) (vii) (viii)

sin (90 – q) = sin (90° ¥ 1 – q) = cos q sin (90 + q) = sin (90° ¥ 1 + q) = cos q sin (180 – q) = sin (90° ¥ 2 – q) = sin q sin (180 + q) = sin (90° ¥ 2 + q) = –sin q sin (270 – q) = sin (90° ¥ 3 – q) = –cos q sin (270 + q) = sin (90° ¥ 3 + q) = –cos q sin (360 – q) = sin (90° ¥ 4 – q) = –sin q sin (360 + q) = sin (90° ¥ 4 + q) = sin q cos (90–q) = cos (90° ¥ 1 – q) = sin q cos (90 + q) = cos (90° ¥ 1 + q) = –sin q cos (180 – q) = cos (90° ¥ 2 – q) = –cos q cos (180 + q) = cos (90° ¥ 2 + q) = –cos q cos (270 – q) = cos (90° ¥ 3 – q) = –sin q cos (270 + q) = cos (90° ¥ 3 + q) = –sin q cos (360 – q) = cos (90° ¥ 4 – q) = cos q cos (360 + q) = cos (90° ¥ 4 + q) = cos q. tan (90 –q) = tan (90° ¥ 1 – q) = cot q tan (90 + q) = tan (90° ¥ 1 + q) = –cot q tan (180 – q) = tan (90° ¥ 2 – q) = –tan q tan (180 + q) = tan (90° ¥ 2 + q) = tan q tan (270 – q) = tan (90° ¥ 3 – q) = cot q tan (270 + q) = tan (90° ¥ 3 + q) = –cot q tan (360 – q) = tan (90° ¥ 4 – q) = –tan q tan (360 + q) = tan (90° ¥ 4 + q) = tan q

Note: All the above results can be remembered by the following simple rule. 1. If q be measured with an even multiple of 90° by + or – sign, then the T-ratios remains unaltered (i.e., sine remains sine and cosine remains cosine, etc.) and treating q as an acute angle, the quadrant in which the associated angle lies, is determined and then the sign

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of the T-ratio is determined by the All – Sin – Tan – Cos formula. 2. If q be associated with an odd multiple of 90 by +ve or –ve sign, then the T-ratios is altered in form (i.e., sine becomes cosine and cosine becomes sine, tangent becomes cotangent and conversely, etc.) and the sign of the ratio is determined as in the previous paragraph. 3. If the multiple of 90 is more than 4, then divide it by 4 and find out remainder. If remainder is 0, then the degree lies on right of x-axis, if remainder is 1, then the degree lies on the +ve y-axis, if remainder is 2, then the degree lies on –ve of x-axis and if the remainder is 3, then the degree lies on the –ve of y-axis respectively.

1.8

GRAPH OF TRIGONOMETRIC FUNCTIONS

1. Graph of f(x) = sin x Y y=1 X¢

0

p

X y=1

Y¢

Characteristics of Sine Function 1. It is an odd function, since sin (–x) = –sin x 2. It is a periodic function with period 2p p 3. sin x = 1 fi x = (4n + 1) , n Œ I 2 4. sin x = 0 fi x = np, n Œ I p 5. sin x = –1 fi x = (4n - 1) , n Œ I 2 Graph of f(x) = cos x Y

X¢

O

X

Y¢

Characteristics of cosine function 1. It is an even function, since cos (–x) = cos x 2. It is a periodic function with period 2p. 3. cos x = 1 fi x = 2np, n Œ I p 4. cos x = 0 fi x = (2n + 1) , n Œ I 2 5. cos x = –1 fi x = (2n + 1)p, n Œ I

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1.4

Trigonometry Booster

3. Graph of f(x) = tan x

Graph of f(x) = cosec x Y

Y Y¢

X¢

O

X

y=1 X¢

O

Y¢

Characteristics of tangent function 1. It is an odd function, since tan (–x) = –tan x 2. It is a periodic function with period p p 3 tan x = 1 fi x = (4n + 1) , n Œ I 4 4. tan x = 0 fi x = np, n Œ I p 5. tan x = –1 fi x = (4n - 1) , n Œ I 4

Y¢

Y

O

X

Characteristics of cotangent function 1. It is an odd function, since cot (–x) = –cot x 2. It is a periodic function with period 2p p 3. cot x = 1 fi x = (4n + 1) , n Œ I 4 p 4. cot x = 0 fi x = (2n + 1) , n Œ I 2 p 5. cot x = –1 fi x = (4n - 1) , n Œ I 4 Graph of f(x) = sec x

Y¢

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1.11

Definition

The Addition Formula

1. sin (A + B) = sin A cos B + cos A sin B 2. cos (A + B) = cos A cos B – sin A sin B tan A + tan B 3. tan ( A + B) = 1 - tan A ◊ tan B

1.9.2

y=1

Characteristics of secant function 1. It is an even function, sec(–x) = sec x 2. It is a periodic function with period 2p 3. sec x can never be zero. 4. sec x = 1 fi x = 2np, n Œ I 5. sec x = –1 fi x = (2n + 1)p, n Œ I

T-RATIOS OF COMPOUND ANGLES

1.9.1

Y

O

1.9

The algebraic sum or difference of two or more angles is called a compound angle such as A + B, A – B, A + B + C, A + B – C, etc.

Y¢

X¢

y = –1

Characteristics of co-secant function 1. It is an odd function, since cosec (–x) = –cosec x 2. It is a periodic function with period 2p p 3. cosec x = 1 fi x = (4n + 1) , n Œ I 2 4. cosec x can never be zero. p 5. sec x = –1 fi x = (4n - 1) , n Œ I 2

4. Graph of f(x) = cot x

X¢

X

X y = –1

Subtraction Formulae

1. sin (A – B) = sin A cos B – cos A sin B 2. cos (A – B) = cos A cos B + sin A sin B tan A - tan B 3. tan (A – B) = 1 + tan A tan B

1.10

SOME IMPORTANT DEDUCTIONS

Deduction 1 sin (A + B) sin (A – B) = sin2 A – sin2 B = cos2 B – cos2 A Proof: We have sin (A + B) sin (A – B) = {sin A cos B + cos A sin B} ¥ {sin A cos B – cos A sin B} = {sin2 A cos2 B – cos2 A sin2 B} = {sin2 A (1 – sin2 B) – (1 – sin2 A) sin2 B} = {sin2 A – sin2 A sin2 B – sin2 B – sin2 A sin2 B = sin2 A – sin2 B

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1.5

The Ratios and Identities

= (1 – cos2 A) – (1 – cos2 B) = cos2 B – cos2 A Deduction 2 cos (A + B) cos (A – B) = cos2 A – sin2 B = cos2 B – sin2 A Proof: We have, cos (A + B) cos (A – B) = {cos A cos B + sin A sin B} ¥ {cos A cos B – sin A sin B} = {cos2 A cos2 B – sin2 A sin2 B} = {cos2 A (1 – sin2 B) – (1 – cos2 A) sin2 B} = {cos2 A – cos2 A sin2 B – sin2 B + cos2 A sin2 B} = cos2 A – sin2 B Deduction-3 cot (A + B) =

cot A cot B - 1 cot B + cot A

Proof: We have, cos (A + B) cos ( A + B) = sin ( A + B ) cos A cos B - sin A sin B = sin A cos B + cos A sin B cos A cos B sin A sin B sin A sin B sin A sin B = sin A cos B cos A sin B + sin A sin B sin A sin B cot A cot B - 1 = cot B + cot A Deduction 4 cot (A – B) =

cot A cot B + 1 cot B - cot A

Proof: We have, cot (A – B) cos ( A - B) = sin ( A - B ) cos A cos B + sin A sin B = sin A cos B - cos A sin B cos A cos B sin A sin B + sin A sin B sin A sin B = sin A cos B cos A sin B sin A sin B sin A sin B cot A cot B + 1 = cot B - cot A Deduction 5 sin (A + B + C) = cos A cos B cos C (tan A + tan B + tan C – tan A tan B tan C) Proof: We have sin (A + B + C) = sin (A + B) cos C + cos (A + B) sin C = {sin A ◊ cos B + cos A ◊ sin B} cos C + {cos A cos B – sin A sin B}sin C

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= sin A ◊ cos B ◊ cos C + sin B ◊ cos A ◊ cos C + sin C ◊ cos A cos B – sin A ◊ sin B ◊ sin C = cos A ◊ cos B ◊ cos C [tan A + tan B + tan C] –tan A ◊ tan B ◊ tan C] Deduction 6 cos (A + B + C) = cos A cos B cos C ¥ [1 – tan A tan B – tan B tan C – tan C tan A] Proof: We have, cos (A + B + C) = cos (A + B) cos C – sin (A + B) sin C = {cos A cos B – sin A sin B}cos C –{sin A cos B + cos A sin B}sin C = cos A cos B cos C – sin A sin B cos C {–sin A sin C cos B – cos A sin B sin C} = cos A cos B cos C ¥ [1 – tan A tan B – tan B tan C – tan C tan A] Deduction 7 tan (A + B + C) sin (A + B + C ) = cos (A + B + C ) cos A cos B cos C (tan A + tan B + tan C - tan A tan B tan C ) = cos A cos B cos C (1 - tan A tan B - tan B tan C - tan C tan A) Proof: We have, tan (A + B + C) sin ( A + B + C ) = cos ( A + B + C ) cos A cos B cos C {tan A + tan B + tan C - tan A tan B tan C} = cos A cos B cos C {1 - tan A tan B - tan B tan C - tan C tan A}

1.11

TRANSFORMATION FORMULAE

1.11.1 1. 2. 3. 4.

Transformation of Products into Sums or Differences

2 sin A cos B = sin (A + B) + sin (A – B) 2 cos A sin B = sin (A + B) – sin (A – B) 2 cosA cos B = cos (A + B) + cos (A – B) 2 sin A sin B = cos (A – B) – cos (A + B)

1.11.2

Transformations of Sums or Differences into Products

C+D C-D cos 2 2 C+D C-D sin 2. sin C – sin B = 2 cos 2 2 C+D C-D cos 3. cos C + cos D = 2 cos 2 2 C+D C-D sin 4. cos C – tan D = – 2 sin 2 2

1. sin C + sin D = 2 sin

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1.6

Trigonometry Booster

1.12

MULTIPLE ANGLES

1.12.1

Definition

An angle is of the form nA, n ΠZ, is called a multiple angle of A. Such as 2A, 3A, 4A, etc. are each multiple angles of A.

1.12.2

Trigonometrical Ratios of 2A in Terms of t-ratio of A

1. sin 2A = 2 sin A cos A 2. cos 2A = cos2 A – sin2 A = 2 cos2 A – 1 = 1 – 2 sin 2 A 2 tan A 3. tan 2A = 1 - tan 2 A

1.12.3

T-ratios of Angle 2A

2 tan A , 1 + tan 2 A 1 - tan 2 A cos 2A = 1 + tan 2 A 1 – cos 2A = 2 sin2 A, 1 + cos 2A = 2 cos2 A sin 2A tan A = , 1 + cos 2A 1 - cos 2A tan A = sin 2A

4. sin 2A = 5. 6. 7. 8. 9.

1.12.4

Trigonometrical Ratios of 3A in Terms of t-ratio of A

10. sin 3A = 3 sin A – 4 sin3 A 11. cos 3A = 4 cos3 A – 3 cos A 3 tan A - tan 3 A 12. tan 3A = 1 - 3 tan 2 A

1.13

SOME IMPORTANT DEDUCTIONS

Deduction 1 1 sin2 A = (1 – cos 2A) 2 Proof: We have, 1 1 sin2 A = (2 sin 2 A) = (1 - cos 2A) 2 2 Deduction 2 1 cos2 A = (1 + cos 2A) 2 Proof: We have, cos2 A 1 1 = (2 cos 2 A) = (1 + cos 2A) 2 2 Deduction-3 1 sin3 A = (3 sin A – sin 3A) 4

TR_01.indd 6

Proof: We have, sin 3A = 3 sin A – 4 sin3 A fi 4 sin3 A = 3 sin A – sin 3A 1 fi sin 3 A = (3 sin A - sin 3A) 4 Deduction-4 1 cos3 A = (cos 3A + 3 cos A) 4 Proof: We have, cos 3A = 4 cos3 A – 3 cos A fi 4 cos3A = cos 3A + 3 cos A 1 cos3 A = (cos 3A + 3 cos A) fi 4 Deduction 5 sin A sin (60 – A) ◊ sin (60 + A) =

1 sin 3A 4

Proof: We have, sin A ◊ sin (60° – A) ◊ sin (60° + A) = sin A ◊ (sin2 60° – sin2 A) Ê3 ˆ = sin A ◊ Á - sin 2 A˜ Ë4 ¯

(

)

sin A ◊ 3 – 4 sin 2 A 4 1 = (3 sin A – 4 sin 3 A) 4 1 = ¥ sin 3A 4 =

Deduction 6 cos A ◊ cos (60 – A) . cos (60 + A) =

1 cos 3A 4

Proof: We have, cos A ◊ cos (60° – A) ◊ cos (60° + A) = cos A ◊ (cos2 60° – sin2 A) Ê1 ˆ = cos A ◊ Á - 1 + cos 2 A˜ Ë4 ¯ Ê 3 ˆ = cos A ◊ Á - + cos 2 A˜ Ë 4 ¯ cos A ◊ (- 3 + 4 cos 2 A) 4 1 = ◊ (- 3 cos A + 4 cos3 A) 4 1 = ◊ (4 cos3 A – 3 cos A) 4 1 = ¥ cos 3A 4 =

Deduction 7 tan A ◊ tan (60 – A) . tan (60 + A) = tan 3A Proof: We have, tan A ◊ tan (60° – A) ◊ tan (60° + A)

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1.7

The Ratios and Identities

= (3 sin A – 4 sin3 A)(1 – 2 sin2 A) + 2(4 cos3 A – 3 cos A) sin A cos A = (3 sin A – 4 sin3 A)(1 – 2 sin2 A) + 2(4 cos2 A – 3) sin A cos2 A = (3 sin A – 4 sin3 A)(1 – 2 sin2 A) + 2(1 – 4 sin2 A)(sin A – sin3 A) = (3 sin A – 4 sin3 A – 6 sin3 A + 8 sin5 A) + 2(sin A – 4 sin3 A – sin3 A + 4 sin5 A) = 5 sin A – 20 sin3 A + 16 sin5 A = 16 sin 5 A – 20 sin3 A + 5 sin A

sin A ◊ sin (60° - A) ◊ sin (60° + A) cos A ◊ cos (60° - A) ◊ cos (60° + A) 1 sin 3A = 4 1 cos 3A 4 sin 3A = cos 3A = tan 3 A =

Deduction 8 sin 4A = 4 sin cos A – 8 cos A sin3 A

Deduction 12 cos 5A = 16 cos5 A – 20 cos3 A + 5 cos A

Proof: We have, sin 4A = 2 sin 2A ◊ cos 2A = 2(2 sin A ◊ cos A)(1 – 2 sin2 A) = 4 sin A ◊ cos A(1 – 2 sin2 A) = 4 sin A ◊ cos A – 8 sin3 A ◊ cos A

Proof: We have, cos 5A = cos (3A + 2A) = cos 3A cos 2A – sin 3A sin 2A = (4 cos3 A – 3 cos A)(2 cos2 A – 1) – (3 sin A – 4 sin3 A)(2 sin A cos A) = 8 cos5 A – 6 cos3 A – 4 cos3 A + 3 cos A – (3 – 4 sin2 A) 2 cos A (1 – cos2 A) = 8 cos5 A – 10 cos3 A + 3 cos A – (4 cos2 A – 1) (2 cos A – 2 cos3 A) = 8 cos5 A – 10 cos3 A + 3 cos A – 8 cos3 A + 2 cos A + 8 cos5 A – 2 cos3 A = 16 cos5 A – 20 cos3 A + 5 cos A

Deduction 9 cos 4A = 1 – 8 sin2 A + 8 sin4 A Proof: We have, cos 4A = cos 2 (2A) = 1– 2 sin2 (2A) = 1 – 2 (2 sin A ◊ cos A)2 = 1 – 8 sin2 A ◊ cos2 A = 1 – 8 sin2 A (1 – sin2 A) = 1 – 8 sin2 A + 8 sin4 A Deduction 10 4 tan A - 4 tan 3 A tan 4 A = 1 - 6 tan 2 A + tan 4 A Proof: We have, tan 4A = tan 2 ◊ (2 A) 2 tan 2A = 1 + tan 2 2A 4 tan A 1 - tan 2 A = 2 Ê 2 tan A ˆ 1- Á Ë 1 - tan 2 A ˜¯ =

4 tan A (1 - tan 2 A) (1 – tan 2 A) 2 - 4 tan 2 A

=

4 tan A - 4 tan 3 A 1 – 6 tan 2 A + tan 4 A

Deduction 11 sin 5A = 16 sin5 A – 20 sin3 A + 5 sin A Proof: We have, sin 5A = sin (3A + 2A) = sin 3A cos 2A + cos 3A ◊ sin 2A

TR_01.indd 7

Deduction 13 sin 6A = (6 sin A – 32 sin3 A + 32 sin5 A) cos A Proof: We have, sin 6A = sin 2 (3A) = 2 sin 3A ◊ cos 3A = 2(3 sin A – 4 sin3 A)(4 cos3 A – 3 cos A) = 2(3 sin A – 4 sin3 A)(1 – 4 sin2 A) cos A = 2(3 sin A – 4 sin3 A – 12 sin3 A + 16 sin 5A) cos A = 2(3 sin A – 16 sin3 A + 16 sin5 A) cos A = (6 sin A – 32 sin3 A + 32 sin5 A) cos A Deduction 14 cos 6 A = 32 cos6 A – 48 cos4 A + 18 cos2 A – 1 Proof: We have, cos 6A = cos 2(3A) = 2 cos2 (3A) – 1 = 2(4 cos3 A – 3 cos A)2 – 1 = 2(16 cos6 A – 24 cos4 A + 9 cos2 A) – 1 = 32 cos6 A – 48 cos4 A + 18 cos2 A – 1

1.14

THE MAXIMUM AND MINIMUM VALUES OF

f (x) = a cos x + b sin x + c We have, f (x) = a cos x + b sin x + c Let a = r sin q and b = r cos q a Then r = a 2 + b 2 and tan (q ) = b

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1.8

Trigonometry Booster

f (x) = a cos x + b sin x + c = r(sin q cos x + cos q sin x) = r sin (q + x) As we know that, –1 £ sin (q + x) £ 1 fi –r + c £ r sin (q + x) + c £ r + c fi –r + c £ f (x) £ r + c

1.15.3

Now,

fi

- a 2 + b 2 + c £ f (x) £ a 2 + b 2 + c

Thus, the maximum value of f (x) is

a 2 + b2 + c

and the minimum values of f (x) is - a 2 + b 2 + c .

1.15 1.15.1

SUB–MULTIPLE ANGLES Definition

A , n Œ Z (π 0), is called a subn A A A A multiple angle of A. Thus , , , , etc. are each a sub2 3 4 5 multiple angle of A. An angle is of the form

1.15.2

Ê Aˆ

Ê Aˆ

T-ratios of angle ÁË 2 ˜¯ and ÁË ˜¯ 3

Ê Aˆ Ê Aˆ 1. sin A = 2 sin Á ˜ cos Á ˜ = Ë 2¯ Ë 2¯

Ê Aˆ 2 tan Á ˜ Ë 2¯ Ê Aˆ 1 + tan 2 Á ˜ Ë 2¯

2Ê

Aˆ Ê Aˆ 2. cos A = cos Á ˜ - sin 2 Á ˜ Ë 2¯ Ë 2¯

Aˆ = 1 - 2 sin Á ˜ = Ë 2¯ Ê Aˆ 2 tan Á ˜ Ë 2¯ 3. tan A = Ê Aˆ 1 - tan 2 Á ˜ Ë 2¯

Ê Aˆ 1 - tan 2 Á ˜ Ë 2¯ Ê Aˆ 1 + tan 2 Á ˜ Ë 2¯

Ê Aˆ Ê Aˆ 4. sin A = 3 sin Á ˜ - 4 sin 3 Á ˜ Ë 3¯ Ë 3¯ Ê Aˆ 3 Ê Aˆ 5. cos A = 4 cos Á ˜ - 3 cos Á ˜ Ë 3¯ Ë 3¯ Ê Aˆ Ê Aˆ 3 tan Á ˜ - tan 3 Á ˜ Ë 3¯ Ë 3¯ 6. tan A = A Ê ˆ 1 - 3 tan 2 Á ˜ Ë 3¯

TR_01.indd 8

Ê 5 - 1ˆ 1. sin (18°) = Á Ë 4 ˜¯ Proof : Let A = 18° fi 5A = 90° fi 2A = 90° – 3A fi sin 2A = sin (90° – 3A) = cos 3A fi 2 sin A cos A = 4 cos3 A – 3 cos A fi 2 sin A = 4 cos2 A – 3 fi 2 sin A = 4 – 4 sin2A – 3 = 1 – 4 sin2A fi 4 sin2 A + 2 sin A – 1 = 0 fi

sin A =

- 2 ± 20 - 2 ± 2 5 = 8 8

fi

sin A =

-1 ± 5 4

fi

sin A =

fi sin (18°) = rant. 2. cos 18° =

5 -1 - 5 -1 , 4 4 5 -1 , since 18° lies on the first quad4

1 10 + 2 5 4

Proof: We have, cos (18°) = 1 - sin 2 (18°) Ê 5 - 1ˆ = 1- Á Ë 4 ˜¯

2

Ê 5 +1- 2 5ˆ = 1- Á ˜¯ Ë 16

Ê Aˆ = 2 cos 2 Á ˜ - 1 Ë 2¯ 2Ê

Values of sin 18°, cos 18° and tan 18°

Ê 16 - 5 - 1 + 2 5 ˆ = Á ˜¯ Ë 16 =

1 10 + 2 5 4

3. tan 18° =

5 -1

10 + 2 5 Proof: We have, tan (18°) sin (18°) = cos (18°) Ê 5 - 1ˆ ÁË ˜ 4 ¯ = 10 + 2 5 4 Ê 5 -1 ˆ =Á ˜ Ë 10 + 2 5 ¯

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1.9

The Ratios and Identities

Notes: (i) sin 72° = cos 18° =

1 10 + 2 5 4

=

Values of sin 36°, cos 36° and tan 36°

1. cos 36° =

sin (36∞) cos (36∞)

10 - 2 5 4 = 5 +1 4

5 -1 4

(ii) cos 72° = sin 18° =

1.15.4

=

5 +1 4

=

10 - 2 5 ( 5 + 1) ( 5 - 1) ¥ ( 10 - 2 5 ) 4

Proof: We have, cos (36°) Notes:

= cos 2 (18°) = 1 – 2 sin2 (18°) Ê 5 - 1ˆ = 1 - 2Á Ë 4 ˜¯

2

(ii) cos 54° = sin 36° =

Ê 5 +1- 2 5ˆ = 1 - 2Á ˜¯ Ë 16 Ê8 - 5 -1+ 2 5ˆ =Á ˜¯ Ë 8 Ê 2 + 2 5ˆ =Á ˜ Ë 8 ¯

1.15.5

Some Important Deductions

Deduction 1 Ê 1 ˆ tan Á 7 ∞˜ = 6 - 4 - 3 + 2 Ë 2 ¯ 1 - cos (2q ) sin (2q )

Ê 1 ˆ 1 - cos (15°) tan Á 7 ∞˜ = Ë 2 ¯ sin (15°)

Proof: We have, sin (36°) = 1 - cos 2 (36°) 2

1=

3 +1 2 2 3 -1 2 2

Ê 5 +1+ 2 5ˆ = 1- Á ˜¯ Ë 16

=

Ê 16 - 5 - 1 - 2 5 ˆ = Á ˜¯ Ë 16

=

Ê 10 - 2 5 ˆ = Á Ë 16 ˜¯

=

(2 2 - 3 - 1)( 3 + 1) 2

1 10 - 2 5 4

=

2 6 - 3 - 3 + 2 2 - 3 -1 2

=

2( 6 - 4 - 3 + 2) 2

=

3. tan 36° =

1 ¥ ( 5 - 1) ¥ 10 - 2 5. 4

Proof: We have tan (36°)

TR_01.indd 9

1 10 - 2 5 4

1 Put q = 7 ∞ , then 2

1 10 - 2 5 4

Ê 5 + 1ˆ = 1- Á Ë 4 ˜¯

5 +1 4

Proof: As we know that, tan q =

Ê 5 + 1ˆ =Á Ë 4 ˜¯ 2. sin 36° =

(i) sin 54° = cos 36° =

2 2 - 3 -1 3 -1 (2 2 - 3 - 1)( 3 + 1) ( 3 - 1)( 3 + 1)

= ( 6 - 4 - 3 + 2)

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1.10

Trigonometry Booster

Deduction 2 Ê 1 ˆ cot Á 7 °˜ = 6 + 4 + 3 + 2 Ë 2 ¯ 1 + cos (2q ) Proof: As we know that, cot (q ) = sin (2q ) 1 Put q = 7 ∞ , 2 Now,

Ê 1 ˆ cot Á 7 ∞˜ Ë 2 ¯ =

=

1 + cos (15∞) sin (15∞)

Ê 1 ˆ 2 cos 2 Á 22 ∞˜ = 1 + cos (45∞) Ë 2 ¯ Ê 1 ˆ =1+ Á Ë 2 ˜¯ =

2 2 + 3 +1 3 -1 (2 2 + 3 + 1)( 3 + 1)

fi

Ê 1 ˆ cos Á 22 ∞˜ = ± Ë 2 ¯

fi

Ê 1 ˆ cos Á 22 ∞˜ = Ë 2 ¯

( 3 - 1)( 3 + 1)

2( 6 + 4 + 3 + 2) 2

= ( 6 + 4 + 3 + 2)

2 +1 2 2 -1 2 2 2 +1 2 2

1 since, 22 ∞ lies in the first quadrant. 2 Ê 1 ˆ 1 Thus, cos Á 22 ∞˜ = 2+ 2 Ë 2 ¯ 2 Deduction 5 Ê 1 ˆ tan Á 22 ∞˜ = 2 - 1 Ë 2 ¯

Deduction 3 Ê 1 ˆ 1 sin Á 22 ∞˜ = 2- 2 Ë 2 ¯ 2

Proof: As we know that, tan q =

Proof: As we know that, 2 sin2 (q) = 1 – cos 2q 1 Put, q = 22 ∞ , 2 Ê 1 ˆ 2 sin 2 Á 22 ∞˜ = 1 - cos (45∞) Ë 2 ¯ Ê 1 ˆ =1- Á Ë 2 ˜¯ =

TR_01.indd 10

Ê 1 ˆ 1 sin Á 22 ∞˜ = 2- 2 Ë 2 ¯ 2

Deduction 4 Ê 1 ˆ 1 cos Á 22 ∞˜ = 2+ 2 Ë 2 ¯ 2

fi

(2 6 + 3 + 3 + 2 2 + 3 + 1) = 2 =

fi

Proof: As we know that, 2 cos2 (q) = 1 – cos 2q 1 q = 22 ∞ , Put, 2

3 +1 1+ 2 2 = 3 -1 2 2 =

1 since, 22 ∞ lies in the first quadrant. 2

fi

Ê 1 ˆ sin Á 22 ∞˜ = ± Ë 2 ¯

fi

Ê 1 ˆ sin Á 22 ∞˜ = Ë 2 ¯

2 -1 2 2 -1 2 2 2 -1 2 2

1 - cos (2q ) sin (2q )

1 Ê 1 ˆ 1 - cos (45∞) Put q = 22 ∞, tan Á 22 ∞˜ = Ë 2 ¯ 2 sin (45∞) 1 12 = 2 -1 = 1 2 Deduction 6 Ê 1 ˆ cot Á 22 ∞˜ = 2 + 1 Ë 2 ¯ Proof: As we know that, 1 + cos (2q ) cot (q ) = sin (2q ) 1 Put q = 22 ∞ , 2

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1.11

The Ratios and Identities

Ê 1 ˆ 1 + cos (45∞) cot Á 22 ∞˜ = Ë 2 ¯ sin (45∞)

fi

=

1+

1 2 = 2 +1

1 2

Deduction 7 1 ˆ 1 Ê sin Á112 ∞˜ = 2+ 2 Ë 2 ¯ 2 1 ˆ Ê Proof: We have, sin Á112 ∞˜ Ë 2 ¯ 1 ˆ Ê = sin Á 90° ¥ 1 + 22 °˜ Ë 2 ¯ Ê 1 ˆ = cos Á 22 ∞˜ Ë 2 ¯ =

1 2+ 2 2

Deduction 8 1 ˆ 1 Ê cos Á112 ∞˜ = 2- 2 Ë 2 ¯ 2 1 Proof: We have, cos ÊÁ112 ∞ˆ˜ Ë 2 ¯ 1 ˆ Ê = cos Á 90° ¥ 1 + 22 ∞˜ Ë 2 ¯ Ê 1 ˆ = – sin Á 22 ∞˜ Ë 2 ¯ 1 =2+ 2 2 Deduction 9 1 ˆ Ê tan Á112 ∞˜ = - ( 2 + 1) Ë 2 ¯ 1 ˆ Ê Proof: We have, tan Á112 ∞˜ Ë 2 ¯ 1 ˆ Ê = tan Á 90° ¥ 1 + 22 °˜ Ë 2 ¯ Ê 1 ˆ = – cot Á 22 °˜ Ë 2 ¯ = - ( 2 + 1)

1.16

CONDITIONAL TRIGONOMETRICAL IDENTITIES

Here, we shall deal with trigonometrical identities involving two or more angles. In establishing such identities we will be frequently using properties of supplementary and comple-

TR_01.indd 11

mentary angles and hence students are advised to go through all the above formulae, starting from the Ist topic. We have certain trigonometrical identities like, sin2 q + cos2 q = 1 and sec2 q = 1 + tan2 q, etc. Such identities are identities in the sense that they hold for all values of the angles which satisfy the given condition amongst them and they are called Conditional Identities. If A, B, C denote the angles of a triangle ABC, then the relation A + B + C = p enables us to establish many important identities involving trigonometric ratios of these angles. (i) If A + B + C = p, then A + B = p – C, B + C = p – A and C + A = p – B (ii) If A + B + C = p, then sin (A + B) = sin (p – C) = sin C similarly, sin (B – C) = sin (p – A) = sin A and sin (C + A) = sin (p – B) = sin B (iii) If A + B + C = p, then cos (A + B) = cos (p – C) = –cos C Similarly, cos (B + C) = cos (p – A) = – cos A and cos (C + A) = cos (p – B) = – cos B (iv) If A + B + C = p, then tan (A + B) = tan (p – C) = –tan C Similarly, tan (B + C) = tan (p – A) = – tan A and tan (C + A) = tan (p – B) = – tan B A+ B p C B+C p A (v) If A + B + C = p, then = - , = 2 2 2 2 2 2 C+A p B and = 2 2 2 Therefore, Ê A + Bˆ Êp Cˆ Ê Cˆ sin Á = sin Á - ˜ = cos Á ˜ Ë 2¯ Ë 2 ˜¯ Ë 2 2¯ Ê A + Bˆ Êp Cˆ Ê Cˆ cos Á = cos Á - ˜ = sin Á ˜ Ë 2¯ Ë 2 ˜¯ Ë 2 2¯ Ê A + Bˆ Êp Cˆ Ê Cˆ tan Á = tan Á - ˜ = cot Á ˜ Ë 2 ˜¯ Ë 2¯ Ë 2 2¯ Note: Dear students, please recollect the following formulae from basic trigonometry Step I: (i) sin 2A = 2 sin A cos A (ii) cos 2A = 2 cos2 A – 1 (iii) cos 2A = 1 – 2 sin2 A (iv) cos 2A = cos2 A – sin2 A Step II: (i) 1 + cos 2A = 2 cos2 A (ii) 1 – cos 2A = 2 sin 2 A (iii) 1 + cos A = 2 cos2 (A/2) (iv) 1 – cos A = 2 sin2 (A/2) Step III: (i) cos (A + B) + cos (A – B) = 2 cos A cos B (ii) cos (A – B) – cos (A + B) = 2 sin A sin B

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1.12

Trigonometry Booster

(i) to express each term as a difference of the two terms directly or by manipulation and then addition, or (ii) to arrange the series in such a way that it follows some standard trigonometrical expansion.

Step IV: Ê C + Dˆ Ê C - Dˆ cos Á (i) cos C + cos D = 2 cos Á Ë 2 ˜¯ Ë 2 ˜¯ Ê C + Dˆ Ê D - Cˆ (ii) cos C – cos D = 2 sin Á sin Á Ë 2 ˜¯ Ë 2 ˜¯

1.17.2

Ê C + Dˆ Ê C - Dˆ (iii) sin C + sin D = 2 sin Á cos Á Ë 2 ˜¯ Ë 2 ˜¯

1. A trigonometrical series involved with the terms of sines or cosines. Rule: Whenever angles are in AP and the trigonometrical terms involved with sines or cosines having power 1. 1. We must multiply each term by Ê common diffrence of angles ˆ 2 sin Á ˜¯ Ë 2

Ê C + Dˆ Ê C - Dˆ (iv) sin C – sin D = 2 cos Á sin Á Ë 2 ˜¯ Ë 2 ˜¯

1.17

TRIGONOMETRICAL SERIES

1.17.1

Different Types of the Summation of a Trigonometrical Series

2. and then express each term as a difference of two terms, 3. And finally add them.

Introduction

In this section, we are mainly concerned with different procedures to find out the summation of trigonometrical series. To find out the sum of different trigonometrical series, first we observe the nature of the angles of the trigonometrical terms. We must observe whether the angles form any sequence or not. If they form any sequences, then we must check, what kind of sequence it is. We also observe the sequence formed (if any) by the coefficients of terms of the series. So, our main attempt will be

1.17.3

A Trigonomeytrical Series Based on Method of Difference

Rules: 1. Express each term of the series as a difference of two expressions. 2. Finally adding them and we shall get the required result.

E XERCISES

LEVEL I

(Problems Based on Fundamentals)

MEASUREMENT OF ANGLES

1. If the radius of the earth 4900 km, what is the length of its circumference? 2. The angles of a triangle are in the ratio 3 : 4 : 5. Find the smallest angle in degrees and the greatest angle in radians. 3. The angles of a triangle are in AP and the number of degrees in the least is to the number of radians in the greatest as 60 to p, find the angles in degrees. 4. The number of sides in two regular polygons are 5 : 4 and the difference between their angles is 9. Find the number of sides of the polygon. 5. The angles of a quadrilateral are in AP and the greatest is double the least. Express the least angles in radians. 6. Find the angle between the hour hand and the minute hand in circular measure at half past 4. 7. Find the length of an arc of a circle of radius 10 cm subtending an angle of 30° at the centre.

TR_01.indd 12

8. The minute hand of a watch is 35 cm long. How far does its tip move in 18 minutes? 9. At what distance does a man, whose height is 2 m subtend an angle of 10°? 1 10. Find the distance at which a globe 5 cm in diameter, 2 will subtend an angle of 6°. 11. The radius of the earth being taken to 6400 km and the distance of the moon from the earth being 60 times the radius of the earth. Find the radius of the moon which subtends an angle of 16° at the earth. 12. The difference between the the acute angles of a right 2p angled triangle is radians. Express the angles in 3 degrees. 13. The angles of a quadrilateral are in AP and the greatest angle is 120°. Find the angles in radians. 1 14. At what distance does a man 5 ft in height, subtend 2 an angle of 15≤? 15. Find the angle between the hour hand and minute-hand in circular measure at 4 o’clock.

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1.13

The Ratios and Identities TRIGONOMETRICAL RATIOS AND IDENTITIES

16. If sec q + tan q = 3,where q lies in the first quadrant, then find the value of cos q. 1 17. If cosec q – cot q = , then find the value of sin q 5 18. If a = c cos q + d sin q and b = c sin q – d cos q such that am + bn = cp + dq, where m, n, p, q Œ N then find the value of m + n + p + q + 42. 19. If 3 sin q + 4 cos q = 5, then find the value of 3 cos q – 4 sin q. 20. If x = r cos q sin j, y = r cos q cos j and z = r sin q such that xm + yn + zp = r2, where m, n, p Œ N, then find the value of (m + n + p – 4)m + n + p + 4. 2 sin a 21. If x = , then find the value of 1 + cos a + 3 sin a sin a - 3 cos a + 3 . 2 - 2 cos a 22. If P = sec6 q – tan6 q – 3 sec2 q tan2 q, Q = cosec6 q – cot6 q – 3 cosec2 q cot2 q and R = sin6 q + cos6 q + 3 sin2 q cos2 q, then find the value of (P + Q + R)(P + Q + R). Ê pˆ 23. If 3 sin x + 4 cos x = 5, for all x in Á 0, ˜ , Ë 2¯ then find the value of 2 sin x + cos x + 4 tan x 24. If sin A + sin B + sin C + 3 = 0, then find the value of cos A + cos B + cos C + 10. 5 25. If (1 + sin q )(1 + cos q ) = , 4 then find the value of (1 – sin q)(1 – cos q). 26. Find the minimum value of the expression 9x 2 sin 2 x + 4 f (x) = , for all x in (0, p). x sin x 27. If cos q + sin q = 2 cos q , then prove that cos q - sin q = 2 sin q 28. If tan2 q = 1 – e2 then prove that sec q + tan3 q ◊ cosec q = (2 – e2)3/2 29. If sin q + sin2 q + sin3 q = 1, then prove that, cos6 q – 4 cos4 q + 8 cos2 q = 4 30. If x = that

2 sin q , then prove 1 + cos q + sin q

1 - cos q + sin q =x 1 + sin q

31. Prove that 3(sin x – cos x)4 + 6(sin x + cos x)2 + 4(sin6x + cos6x) = 13. 32. If sin x + sin2 x = 1, then find the value of cos8 x + 2 cos6 x + cos4 x

TR_01.indd 13

2

2

33. If 0 £ q £ 180° and 81sin q + 81cos q = 30 , then find the value of q 34. Let fk(q) = sink(q) + cosk(q). 1 1 Then find the value of f 6 (q ) - f 4 (q ) 6 4 3 3 35. If x sin a + y cos a = sin a cos a and x sin a = y cos a then prove that x2 + y2 = 1. 36. If tan q + sin q = m, tan q – sin q = n, then prove that m2 – n2 = 4 mn . cos 4 x sin 4 x + = 1 , then prove that 37. If cos 2 y sin 2 y cos 4 y sin 4 y + =1 cos 2 x sin 2 x 38. If fn(q) = sinn q + cosn q, prove that 2f6(q) – 3f4(q) + 1 = 0 39. If

40. If

sin A cos A = p, = q , prove that sin B cos B p Ê q2 - 1 ˆ tan A ◊ tan B = Á . q Ë 1 - p 2 ˜¯ sin 4 a cos 4 a 1 , then + = a b a+b

prove that

sin 8 a cos8 a 1 + = a3 b3 ( a + b )3

MEASUREMENT OF DIFFERENT T-RATIOS

41. Find the value of (i) sin 120° (ii) sin 150° (iii) sin 210° (iv) sin 225° (v) sin 300° (vi) sin 330° (vii) sin 405° (viii) sin 650° (ix) sin 1500° (x) sin 2013° 42. Find the value of cos (1°) ◊ cos (2°) ◊ cos (3°)…cos (189°) 43. Find the value of tan (1°) ◊ tan (2°) ◊ tan (3°)…tan (89°) 44. Find the value of tan 35° ◊ tan 40° ◊ tan 45° ◊ tan 50° ◊ tan 55° 45. Find the value of sin (10°) + sin (20°) + sin (30°) + sin (40°) + … + sin (360°) 46. Find the value of cos (10°) + cos (20°) + cos (30°) + cos (40°) + … + cos (360°) 47. Find the value of sin2 5° + sin2 10° + sin2 15° + … + sin2 90°

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1.14

Trigonometry Booster

48. Find the value of Êpˆ Êpˆ Ê 4p ˆ Ê 7p ˆ sin 2 Á ˜ + sin 2 Á ˜ + sin 2 Á ˜ + sin 2 Á ˜ Ë 18 ¯ Ë 9¯ Ë 9 ¯ Ë 18 ¯ 49. Find the value of tan (20°) tan (25°) tan (45°) tan (65°) tan (70°) 50. Find the value of cos (q1) + cos (q2) + cos (q3) if sin (q1) + sin (q2) + sin (q3) = 3 51. Find the value of sin2 6° + sin2 12° + … + sin2 90° 52. Find the value of sin2 10° + sin2 20° + … + sin2 90° 53. Find the value of sin2 9° + sin2 18° + … + sin2 90° 54. Find the value of tan 1° ◊ tan 2° ◊ tan 3° … tan 89° 55. Find the value of cos 1° ◊ cos 2° ◊ cos 3° … cos 189° 56. Solve for q; 2 sin2q + 3 cos q = 0 where 0 < q < 360°. 57. Solve for q; cos q + 3 sin q = 2 , where 0 < q < 360°. 58. If 4na = p, then prove that tan a tan 2a tan 3a... tan (2n –1)a = 1. 59. Find the value of cos (18°) + cos (234°) + cos (162°) + cos (306°). 60. Find the value of cos (20°) + cos (40°) + cos (60°) +… + cos (180°) 61. Find the value of sin (20°) + sin (40°) + sin (60°) + … + sin (360°) 62. Draw the graphs of (i) f (x) = sin x + 1 (ii) f (x) = sin x – 1 (iii) f (x) = –sin x (iv) f (x) = 1 – sin x (v) f (x) = –1 – sin x (vi) f (x) = sin 2x, sin 3x (vii) f (x) = sin2x (viii) f (x) = cos2x (ix) f (x) = max(sin x, cos x} (x) f (x) = min{sin x, cos x} 1 (xi) f (x) = min sin x, , cos x 2 (xii) f (x) = max{tan x, cot x} (xiii) f (x) = min{tan x, cot x} 63. Find the number of solutions of 1 (i) sin x = , " x Œ [0, 2p] 2 3 (ii) cos x = , " x Œ[0, 3p] 2 (iii) 4 sin2x – 1 = 0, " x Œ[0, 3p] (iv) sin2x – 3 sin x + 2 = 0, " x Œ [0, 3p] (v) cos2x – cos x – 2 = 0, " x Œ [0, 3p]

{

TR_01.indd 14

}

COMPOUND ANGLES

64. Find the values of (i) sin (15°), (ii) cos (15°), (iii) tan (15°) 65. Find the value of tan (75°) + cot (75°) 66. Prove that cos (9°) + sin (9°) = 2 sin (54°) 67. Prove that tan (70°) = 2 tan (50°) + tan (20°) cos 20∞ - sin 20∞ 68. Prove that = tan 25° cos 20∞ + sin 20∞ cos 7∞ + sin 7∞ 69. Prove that = tan 52°. cos 7∞ - sin 7∞ 70. Prove that tan 20° + tan 25° + tan 20° tan 25° = 1. 71. If A + B = 45°, then find the value of (1 + tan A)(1 + tan B) 72. Find the value of (1 + tan 245°)(1 + tan 250°)(1 + tan 260°) (1 – tan 200°)(1 – tan 205°)(1 – tan 215°) 73. Prove that tan 13A – tan 9A – tan 4A = tan 4A ◊ tan 9A ◊ tan 13A 74. Prove that tan 9A – tan 7A – tan 2A = tan 2A ◊ tan 7A ◊ tan 9A 1 m 75. If tan a = , , tan b = 2m + 1 m +1 p then prove that a + b = . 4 76. Prove that sin2 B = sin2A + sin2 (A – B) – 2 sin A cos B sin (A – B) 77. Prove that cos (2x + 2y) = cos 2x cos 2y + cos2(x + y) – cos2(x – y) x- y 78. If sin q = , x+ y x Êp qˆ then prove that tan Á + ˜ = ± Ë 4 2¯ y 79. If tan a =

Q sin b , P + Q cos b

prove that tan (b - a ) =

P sin a Q + P cos a

80. If cos (a – b) + cos (b – g) + cos (g - a ) = prove that cos a + cos b + cos g = 0 and sin a + sin b + sin g = 0 81. If tan (a + q) = n tan (a – q), show that

3 2

sin 2q n - 1 = sin 2a n + 1

82. If sin a + sin b = a and cos a + cos b = b, then show that b2 - a 2 (i) cos (a + b ) = 2 b + a2 (ii) sin (a + b ) =

2ab b + a2 2

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1.15

The Ratios and Identities

83. If a and b are the roots of a cos q + b sin q = c, then prove that a 2 - b2 (i) cos (a + b ) = 2 a + b2 (ii) cos (a - b ) =

2c - (a 2 + b 2 ) a 2 + b2

Êa - bˆ = 4 cos 2 Á Ë 2 ˜¯

84. If a and b are the roots of a tan q + b sec q = c, then show that 2ac tan (a + b ) = 2 2 a -c 85. If tan (p cos q) = cot (p sin q), prove that pˆ 1 Ê cos Á q - ˜ = ± . Ë ¯ 4 2 2 86. If tan q =

then prove that tan (a – b) = (1 – n) tan a 89. If x + y + z = 0, then prove that cot (x + y – z) ◊ cot (y + z – x) + cot (y + z – x) ◊ cot (z + x – y) + cot (z + x – y) ◊ cot (x + y – z) = 1 90. If 2 tan a = 3 tan b, then show sin 2b tan (a - b ) = 5 - cos 2b TRANSFORMATION FORMULAE

91. Prove that: sin 5 A - sin 3 A (i) = tan A cos 5 A + cos 3 A sin A + sin 3 A = tan 2 A (ii) cos A + cos 3 A sin A + sin B Ê A + Bˆ = tan Á (iii) Ë 2 ˜¯ cos A + cos B 92. Prove that: (i) sin 38° + sin 22° = sin 82° (ii) sin 105° + cos 105° = cos 45° (iii) cos 55° + cos 65° + cos 175° = 0 (iv) cos 20° + cos 100° + cos 140° = 0 (v) sin 50° – sin 70 + sin 10° = 0 93. Prove that sin (47°) + cos (77°) = cos (17°) 94. Prove that cos (80°) + cos (40°) – cos (20°) = 0

97. Prove that (cos a – cos b)2 + (sin a – sin b)2 Êa - bˆ = 4 sin 2 Á Ë 2 ˜¯ 98. Prove that (sin a – sin b)2 + (cos a – cos b)2 Êa - bˆ = 4 sin 2 Á Ë 2 ˜¯ 99. Prove that cos 20 ◊ cos 40 ◊ cos 80 =

x sin j y sin q and tan j = , 1 - x cos j 1 - y cos q

then prove that, x sin j = y sin q. 87. If tan a + tan b = a, cot a + cot b = b and tan (a + b) = c then find a relation in a, b and c. n sin a cos a 88. If tan b = , 1 - n sin 2 a

TR_01.indd 15

95. Prove that sin (10°) + sin (20°) + sin (40°) + sin (50°) –sin (70°) – sin (80°) = 0 96. Prove that (cos a + cos b)2 + (sin a + sin b)2

that

100. Prove that sin 20 ◊ sin 40 ◊ sin 80 =

1 8 3 8

3 16 3 112. Prove that cos 10 ◊ cos 30. cos 50 ◊ cos 70° = 16 113. Prove that: sin A + sin 3 A + sin 5 A (i) = tan 3 A cos A + cos 3 A + cos 5 A

101. Prove that sin 10 ◊ sin 50 ◊ sin 60 ◊ sin 70 =

(ii)

cos 4x + cos 3x + cos 12x = cot 3x sin 4x + sin 3x + sin 2x

(iii)

sin A + sin 3A + sin 5A + sin 7A = tan 4A cos A + cos 3A + cos 5A + cos 7A

(iv)

sin A + sin 2A + sin 4A + sin 5A = tan 3 A cos A + cos 2A + cos 4A + cos 5A

114. If sin A - sin B = Ê A + Bˆ tan Á Ë 2 ˜¯ 115. If cos A + cos B =

1 1 and cos A - cos B = , then find 2 3

1 1 , sin A + sin B = , 2 4

Ê A + Bˆ 1 then prove that tan Á = Ë 2 ˜¯ 2 116. If cosec A + sec A = cosec B + sec B, then Ê A + Bˆ prove that tan A tan B = cot Á Ë 2 ˜¯ 117. If sin 2A = l sin 2B, then prove that tan (A + B) l + 1 = tan (A - B) l - 1 118. Find the value of

3 cot (20°) - 4 cos (20°)

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1.16

Trigonometry Booster

119. If sin A + sin B = a and cos A + cos B = b, then find cos (A + B) 120. If

2 cos A = x +

1 , x

2 cos B = y +

cos (A – B)

1 , y

Ans.

137. Prove that 4 cos (q) ◊ cos (60° – q) ◊ cos 60° + q = cos (3q)

then

find

1Ê 1ˆ xy + ˜ Á 2Ë xy ¯

121. Prove that sin (47°) + sin (61°) – sin (11°) – sin (25°) = cos (7°). m 1 , and tan b = 122. If tan a = , then find m +1 2m + 1 tan (a + b) 123. Find the number of integral values of k for which 7 cos x + 5 sin x = 2k + 1 has a solution. 124. Prove that cos a + cos b + cos g + cos (a + b + g) Êa + bˆ Êb +g ˆ Êg +aˆ = 4 cos Á cos Á cos Á ˜ ˜ Ë 2 ¯ Ë 2 ¯ Ë 2 ˜¯

3 8

139. Prove that tan (q) + tan (60° + q) – tan (60° – q) = 3 tan (3q) 140. Prove that cos (q) cos (2q) ◊ cos (22q) ◊ cos (23q) … cos (2n–1q) =

sin (2nq ) 2n sin q

Ê 2p ˆ Ê 4p ˆ Ê 8p ˆ 1 141. Prove that cos Á ˜ cos Á ˜ cos Á ˜ = Ë 7 ¯ Ë 7 ¯ Ë 7¯ 8

143. If M = a 2 cos 2 q + b 2 sin 2 q + a 2 sin 2 q + b 2 cos 2 q

Ê 1 - cos 2q ˆ 125. Prove that Á = tan q Ë sin 2q ˜¯

such that maximum (M2) = m1 and minimum (M2) = m2, then find the value of m1 – m2.

Ê 1 + cos 2q ˆ 126. Prove that Á = cot q Ë sin 2q ˜¯

144. Prove that tan 4q =

127. Prove that cot q – tan q = 2 cot (2q) 128. Prove that tan q + 2 tan (2q) + 4 tan (4q) + 8 cot 8q = cot q b 129. If tan q = , a prove that a cos (2q) + b sin (2q) = a 130. Prove that 3 cosec (20°) - sec (20°) = 4 131. Prove that tan (9°) – tan (27°) – tan (63°) + tan 81°) = 4 Ê sec 8A - 1ˆ tan 8A 132. Prove that Á = Ë sec 4A - 1˜¯ tan 2A 133. Prove that ˆ Ê 2p ˆ 3 2 2 Ê 2p - q ˜ + cos 2 Á + q˜ = (i) cos (q ) + cos Á Ë 3 ¯ Ë 3 ¯ 2 Êp ˆ Êp ˆ 3 (ii) cos 2 q + cos 2 Á - q ˜ + cos 2 Á + q ˜ = Ë3 ¯ Ë3 ¯ 2 134 Prove that 3 2

135. Prove that 4 sin (q) sin (60° + q) sin (60° – q) = sin 3q 3 136. Prove that sin (20∞) ◊ sin (40∞) ◊ sin (80∞) = 8

TR_01.indd 16

cos (10°) ◊ cos (50°) ◊ cos (70°)=

1 Ê 2p ˆ Ê 4p ˆ Ê 6p ˆ 142. Prove that cos Á ˜ + cos Á ˜ + cos Á ˜ = - . Ë 7 ¯ Ë 7 ¯ Ë 7 ¯ 2

MULTIPLE ANGLES

sin 2 q + sin 2 (120° + q ) + sin 2 (240° + q ) =

138. Prove that

4 tan q - 4 tan 3q 1 - 6 tan 2q + tan 4q

145. Prove that Ê sin x sin 3x sin 9x ˆ 1 ÁË cos 3x + cos 9x + cos 27x ˜¯ = 2 (tan 27x - tan x) 146. Prove that Êqˆ tan Á ˜ (1 + sec q )(1 + sec 2q )(1 + sec 22q ) Ë 2¯ …(1 + sec(2nq)) = tan (2nq) Maximum or minimum values of f (x) = q cos + b sin c) 147. Find the maximum and minimum values of (i) f (x) = 3 sin x + 4 cos x + 10 (ii) f (x) = 3 sin (100) x + 4 cos (100) x + 10 (iii) f (x) = 3 sin x + 4 (iv) f (x) = 2 cos x + 5 (v) f (x) = sin x + cos x (vi) f (x) = sin x – cos x (vii) f (x) = sin (sin x) (viii) f (x) = cos (cos x) (ix) f (x) = sin (sin x) + cos (sinx) (x) f (x) = cos (sin x) + sin (cos x) 148. Find the range of f (x) = sin x + cos x + 3 149. Find the greatest and the least values of 2 sin2 q + 3 cos2 q 150. Prove that p –4 < 5 cos q + 3 cos ÊÁ q + ˆ˜ + 3 < 10 Ë 3¯

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1.17

The Ratios and Identities

151. Find the maximum and minimum values of cos2q – 6 sin q cos q + 3 sin2q + 2 152. Find the least value of cosec2 x + 25 sec2 x 153. Find the ratio of the greatest value of 2 – cos x + sin2 x to its least value. 154. If y = 4 sin2 q – cos 2q, then y lies in the interval............. 155. If m is the minimum value of f (x) = 3 sin x + 5 and n is the maximum value of g(x) = 3 – 2 sin x then find the value of (m + n + 2). 156. Find the maximum and the minimum values of f (x) = sin2 x + cos4 x. 157. Find the maximum and the minimum values of f (x) = cos2 x + sin4 x. 158. Find the maximum and minimum values of f (x) = sin4 x + cos4 x. 159. Find the maximum and minimum values of f (x) = sin6 x + cos6 x. 160. If A = cos2 q + sin4 q and B = cos4 q + sin2 q such that m1 = Maximum of A and m2 = Minimum of B then find the value of m12 + m22 + m1m2 161. Find the maximum and minimum values of f (x) = (sin x + cot + cosec 2x)3 Ê pˆ where x ŒÁ 0, ˜ Ë 2¯ 162. Find the maximum and minimum values of 5 f (x) = sin 2 q - 6 sin q cos q + 3 cos 2 q 163. Find the minimum value of f (x) =

a2 b2 Ê pˆ , x ŒÁ 0, ˜ + 2 Ë 2¯ cos x sin 2 x

164. Find the minimum value of x 2 sin 2 x + 4 f (x) = , x sin x Ê pˆ where x ŒÁ 0, ˜ Ë 2¯ 165. Find the maximum and minimum values of f (x) = logxy + logyx, where x > 1, y > 1 166. Find the minimum values of f (x) = 2 log10 x - log x (0.01), x > 1 167. Find the minimum value of ( x 2 + 1)( y 2 + 1)( z 2 + 1) f ( x, y , z ) = , x, y , z > 0 xyz 168. Find the minimum value of ( x3 + 2)( y 3 + 2)( z 3 + 2) f ( x, y , z ) = , x, y , z > 0 xyz

TR_01.indd 17

169. Find the minimum value of (a 2 + 1)(b 2 + 1)(c 2 + 1)(d 2 + 1) f (a, b, c, d ) = abcd where a, b, c, d > 0 SUB-MULTIPLE ANGLES

170. Prove that q q (i) tan + cot = 2 cosec q 2 2 q (ii) cosec q – cot q = tan 2 1 + sin q - cos q q (iii) = tan 1 + sin q + cos q 2 Êp qˆ (iv) sec q + tan q = tan Á + ˜ Ë 4 2¯ (v) (cos A + cos B)2 + (sin A + sin B)2 Ê A - Bˆ = 4 cos 2 Á Ë 2 ˜¯ 171. Prove that sin 2 (24°) - sin 2 (6°) =

( 5 - 1) 8

( 5 + 1) 8 1 173. Prove that sin (12°) ◊ sin (48°) ◊ sin (54°) = 8 172. Prove that sin 2 (48°) - cos 2 (12°) =

1 174. Prove that sin (6°) ◊ sin (42°) ◊ sin (66°) ◊ sin (78°) = 16 175. Prove that 4(sin (24°) + cos (6°)) = (1 + 5) 176. Prove that tan (6°) ◊ tan (42°) ◊ tan (66°) ◊ tan (78°) = 1 177. Prove that pˆÊ 3p ˆ Ê 5p ˆ Ê 5p ˆ 1 Ê ÁË1 + cos ˜¯ ÁË1 + cos ˜ Á1 + cos ˜ Á1 + cos ˜= 8 8 ¯Ë 8 ¯Ë 8 ¯ 8 178. Prove that Êpˆ Ê 3p ˆ Ê 5p ˆ Ê 7p ˆ 3 sin 4 Á ˜ + sin 4 Á ˜ + sin 4 Á ˜ + sin 4 Á ˜ = Ë 8¯ Ë 8 ¯ Ë 8 ¯ Ë 8 ¯ 2 179. Prove that Êpˆ Ê 3p ˆ Ê 5p ˆ Ê 7p ˆ 3 cos 4 Á ˜ + cos 4 Á ˜ + cos 4 Á ˜ + cos 4 Á ˜ = Ë 8¯ Ë 8 ¯ Ë 8 ¯ Ë 8 ¯ 2 180. Prove that tan 20∞ tan 80∞ = 3 tan 50∞ 1 181. Prove that tan (10°) tan (70°) = ¥ tan (40°) 3 182. Prove that sin 55° – sin 19° + sin 53° – sin 17° = cos 1° 2p 4p 6p 1 ◊ cos ◊ cos = 183. Prove that cos 7 7 7 8

2/10/2017 4:08:50 PM

1.18 2p 4p 6p 1 + cos + cos =7 7 7 2 Ê 1 ˆ Ê 1 ˆ 185. Find the value of tan Á 7 ∞˜ + cot Á 7 ∞˜ . Ë 2 ¯ Ë 2 ¯ Ê xˆ Ê xˆ 186. If cos Á ˜ - 3 sin Á ˜ takes its minimum value then Ë 2¯ Ë 2¯ find its x. 187 If a and b be two different roots a cos q + b sin q = c, 2ab . then prove that sin (a + b) = 2 a + b2 184. Prove that: cos

CONDITIONAL IDENTITIES

188. If A + B + C = p, then prove that, sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C 189. If A + B + C = p, then prove that, cos 2A + cos 2B + cos 2C = –1 – 4 cos A cos B cos C 190. If A + B + C = p, then prove that sin2 A + sin2 B – sin2 C = 2 sin A sin B cos C 191. If A + B + C = p, then prove that, (i) sin2 A + sin2 B + sin2 C = 2 + cos A cos B cos C (ii) cos2 A + cos2 A + cos2 A = 2 + sin A sin B sin C 192 If A + B + C = p, then prove that sin 2 A + sin 2 B + sin 2C cos A + cos B + cos C - 1 A B C = 8 cos cos cos 2 2 2 193. In a DABC prove that (i) tan A + tan B + tan C = tan A ◊ tan B ◊ tan C (ii) cot A ◊ cot B + cot B ◊ cot C + cot C ◊ cot A = 1. 194. If A, B, C and D be the angles of a quadrilateral, then prove that tan A + tan B + tan C + tan D cot A + cot B + cot C + cot D = tan A ◊ tan B ◊ tan C ◊ tan C 194. In a DABC prove that (cot A + cot B) (cot B + cot C) (cot C + cot A) =cosec A cosec B cosec C. 195. If xy + yz + zx = 1, then prove that x y z 4xyz + + = 1 - x 2 1 - y 2 1 - z 2 (1 - x 2 )(1 - y 2 )(1 - z 2 ) 196. If xy + yz + zx = 1, then prove that, x y z 2 + + = 2 1 + x2 1 + y 2 1 + z 2 (1 + x )(1 + y 2 )(1 + z 2 ) 197 Prove that tan (a – b) + tan (b – g) + tan (g – a) = tan (a – b) tan (b – g) tan (g – a) 198 In a DABC, if cot A + cot B + cot C = 3 , then prove that the triangle is an equilateral.

TR_01.indd 18

Trigonometry Booster

199 If x + y + z = xyz, then prove that 3x - x3 3y - y 3 3z - z 3 + + 1 - 3x 2 1 - 3y 2 1 - 3z 2 =

3x - x3 3y - y 3 3z - z 3 ◊ ◊ 1 - 3x 2 1 - 3y 2 1 - 3z 2

200. Prove that 1 + cos 56° + cos 58° – cos 66° = 4 cos 28° cos 29° sin 33°. TRIGONOMETRICAL SERIES

201. Prove that sin a + sin (a + b) + sin (a + 2b) + sin (a + 3b) + … + sin (a + (n – 1)b) Ê nb ˆ sin Á ˜ Ë 2¯ b˘ È ¥ sin Ía + (n - 1) ˙ = 2˚ Ê bˆ Î sin Á ˜ Ë 2¯ 202. Prove that sin a + sin 2a + sin 3a + … + sin na Ê na ˆ sin Á ˜ Ë 2 ¯ Êaˆ = ¥ sin (n + 1) Á ˜ Ë 2¯ Êaˆ sin Á ˜ Ë 2¯ 203. Prove that sin q + sin 3q + º + sin (2n - 1)q =

sin 2 nq sin q

204 Prove that cos a + cos (a + b) + cos (a + b) + cos (a + b) + … + cos (a + (n – 1)b) Ê nb ˆ sin Á ˜ Ë 2¯ bˆ Ê = ¥ cos Á a + (n - 1) ˜ Ë 2¯ Ê bˆ sin Á ˜ Ë 2¯ 205. Prove that cos a + cos 2a + cos 3a + … Ê na ˆ sin Á ˜ Ë 2 ¯ Ê (n + 1)a ˆ .... + cos na = ¥ cos Á ˜ Ë 2 ¯ Êaˆ sin Á ˜ Ë 2¯ 206. Find the sum of n-terms of the series sin x sin x + sin 2x ◊ sin 3x sin 3x ◊ sin 4x sin x + + .... to - n - terms sin 4x ◊ sin 5x 207. Prove that +

1 1 + cos q + cos 2q cos q + cos 5q

1 + .... to - n - terms cos q + cos 7q = cosec q [tan (n + 1) q – tan q]

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1.19

The Ratios and Identities

208. Prove that tan x ◊ tan 2x + tan 2x ◊ tan 3x + … + tan nx ◊ tan (n + 1)x = cot x [tan (n + 1)x – tan x ] – n 209. Prove that Ê x ˆ Ê x ˆ tan -1 Á + tan -1 Á Ë 1 + 2x 2 ˜¯ Ë 1 + 6x 2 ˜¯ x ˆ Ê + tan -1 Á Ë 1 + 12x 2 ˜¯

6. Which of the following is greatest? (a) sin 1 (b) cos 1 (c) tan 1 (d) cot 1 7. If A= cos (cos x) + sin (cos x), then the least and greatest value of A are (a) 0, 2 (b) –1, 1 (d) 0, 2 (c) - 2, 2 p 8. If A + B = , A, B > 0 then the maximum value of tan 3 A ◊ tan B is (a) 1/3 (b) 1 (c) 1/2 (d) 2/3 9. The maximum value of a sin 2x + b cos 2x for all real x is 2 2 (a) a + b (b) a + b (c) maximum {|a|, |b|} (d) maximum {a, b} 10. Which of the following is/are true? (a) sin 1 > sin 1°(b) tan 1 > tan 1° (c) sin 4 > sin 4°(d) tan 4 > tan 4° 11. If cos 5x = a cos5 x + b cos3 x + c cos x + d, then (a) a = 16 (b) b – 20 (c) c = 5 (d) d = 2 12. If sin3 x sin 3 x = c0 + c1 cos x + c2cos 2x + c3 cos 3x +.......... + cn cos nx, then (a) Highest value of n is 6 (b) c0 = 1/8 (c) c2 = –c4 (d) c1 = c3 = c5. 13. If f (x) = cos[p] x + sin [p] x, where [,] is the greatest integer function, then Êpˆ f Á ˜ is Ë 2¯

x Ê ˆ + tan -1 Á Ë 1 + n(n + 1)x 2 ˜¯

= tan–1 (n + 1) x – tan–1 x 210. Prove that Ê 1ˆ Ê 1ˆ Ê 1ˆ tan -1 Á ˜ + tan -1 Á ˜ + tan -1 Á ˜ Ë 7¯ Ë 13 ¯ Ë 19 ¯ 1 Ê ˆ + ..... + tan -1 Á 2 Ë n + 3n + 3 ˜¯ = tan–1 (n + 2) – tan–1 2. Ê 2 - 1ˆ Ê 1 ˆ + sin -1 Á 211. Prove that sin -1 Á ˜ ˜ Ë 2¯ 2 ¯ Ë Ê 3 - 2ˆ sin -1 Á ˜ 12 ¯ Ë

......to

p 2

Ê 1ˆ Ê 2ˆ 212. Prove that tan -1 Á ˜ + tan -1 Á ˜ Ë 3¯ Ë 9¯ Ê 4ˆ tan -1 Á ˜ Ë 23 ¯

LEVEL II

...... to

p 4

(a) 0 14. Let f (x)

(Mixed Problems)

1 1. If sec x = p + , then sec x + tan x is p 1 (a) p (b) 2p (c) 4p

= (d)

(a) 1/2 (b) –1/2 (c) 0 5. Which of the following is smallest? (a) sin 1 (b) sin 2 (c) sin 3

TR_01.indd 19

cos 2 x

sin 2 x

1 + cos 2 x

sin x

4 p

3

2 2

(d) None (d) sin 4

2

(c) cos 4

1 + sin 2 x 2

2. If cosec x – sin x = a , sec x – cos x = b , then a b (a + b2) is (a) 0 (b) 1 (c) –1 (d) ab 3 3. If sec x + cos x = 2, then the value of sec x(1+ sec3 x) + cos3 x(1 + cos3 x) is (a) 2 (b) 4 (c) 6 (d) 8 4. The value of Ê 2p ˆ Ê 4p ˆ Ê 6p ˆ cos Á ˜ + cos Á ˜ + cos Á ˜ is Ë 7 ¯ Ë 7 ¯ Ë 7 ¯ 3

(b) cos 3

2

cos x

(d) None

4 sin 2x 4 sin 2x , 1 + 4 sin 2x

then the maximum value of f (x) is (a) 0 (b) 2 (c) 6 (d) None 15. For any real x, the maximum value of cos2 (cos x) + sin2 (sin x) is (a) 1 (b) 1 + sin2 1 2 (c) 1 + cos 1 (d) None Êpˆ Ê 5p ˆ Ê 7p ˆ 16. If a = sin Á ˜ sin Á ˜ sin Á ˜ and x is the soluË 18 ¯ Ë 18 ¯ Ë 18 ¯ tion of the equation y = 2[x] + 2 and y = 3 [x – 2 ], where [,] = GIF, then a is (a) [x] (b) 1/[x] (c) 2[x] (d) [x]2 17. The minimum value of sin8 x + cos8 x is (a) 0 (b) 1 (c) 1/8 (d) 2

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1.20

Trigonometry Booster

sin 4 q cos 4 q 1 sin 8q cos8q + = + 3 is , then a b a+b a3 b 1 1 (b) (a) 3 3 a +b (a + b)3 1 (c) (d) None (a - b)3 Êpˆ Ê 2p ˆ Ê 3p ˆ 19. The value of tan Á ˜ tan Á ˜ tan Á ˜ is Ë 7¯ Ë 7 ¯ Ë 7 ¯ 1 (c) 7 (d) None (a) 1 (b) 7 20. If a and b are the solutions of sin2 x + a sin x + b = 0 as well as that of cos2 x + c cos x + d = 0, then sin (a + b) is a2 + c2 2bd (b) (a) 2 2ac b + d2 18. If

b2 + d 2 2ac (d) 2 2 2bd a +c If sec q + tan q = 1, then one of the roots of the equation a(b – c)x2 + b(c – a)x + c(a – b) = 0 is (a) tan q (b) sec q (c) cos q (d) sin q If a is the common positive root of the equation x2 – ax + 12 = 0, x2 – bx + 15 = 0 and x2 – (a –b)x + 36 = 0 and cos x + cos 2x + cos 3x = 0, then sin x + sin 2x + sin 3x is (a) 3 (b) –3 (c) 0 (d) 2 For any real q, the maximum value of cos2 (cos q) + sin2 (sin q) is (a) 1 (b) 1 + sin2 1 2 (d) 1 – cos2 1 (c) 1 + cos 1 If a + b = p/2 and b + g = a, then tan a is (a) 2(tan b + tan g) (b) (tan b + tan g) (c) (tan b + 2 tan g) (d) (2 tan b + tan g) The maximum value of cos a1 ◊ cos a2 ◊ cos a3 … cos an

28. The minimum value of the expression sin a + sin b + sin g, where a, b, g are real positive angles satisfying a + b + g = p, is (a) positive (b) negative (c) 0 (d) –3 29. The value of 4 cos 20∞ - 3 cot 20∞ is (a) 1 (b) –1 (c) –1/2 (d) 1/4 30. The maximum value of Ê xˆ Ê xˆ 4 sin 2 x + 3 cos 2 x + sin Á ˜ + cos Á ˜ is Ë 2¯ Ë 2¯ (b) 3 + 2 (a) 4 + 2 (c) 4 - 2 (d) 4 31. The value of the expression ( 3 sin 75° - cos 75°) is 1 1 (c) 2 (a) (b) 2 2 32. The value of (4 + sec 20°) sin 20° is

(c)

21.

22.

23.

24.

25.

under the restriction p and 2 cot a1 ◊ cot a2, cot a3 … cot an = 1, is

0 £ a1 , a 2 , a 3 , ................, a n £

1

1 (b) n 2

1 (a) n /2 (c) (d) 1 2 2n p 26. If A > 0 and B > 0 and A + B = , then the maximum 3 value of tan A. tan B is 1 1 (d) 3 2 27. If tan b = 2 sin a × sin g × cosec (a + g), then cot a, cot b, cot g are in (a) AP (b) GP (c) HP (d) AGP (a)

TR_01.indd 20

1 2

(b)

1 3

(c)

33.

34.

35.

36.

(d) 2

(a) 1 (b) 2 (c) 3 (d) 2 3 If (1 + tan 1°) (1 + tan 2°) … (1 + tan 45°) = 2° then the value of n is (a) 20 (b) 21 (c) 22 (d) 23 The number of all possible triplets (a1, a2, a3) such that a1 + a2 cos (2x) + a3 sin2 x = 0 is (a) 0 (b) 1 (c) 2 (d) infinite If sin (p cos q) = cos (p sin q), then the value of sin (2q) is (a) –1/2 (b) –1/3 (c) –2/3 (d) –3/4 A real root of the equation 8x3 – 6x – 1 = 0 is Êpˆ Êpˆ (a) cos Á ˜ (b) cos Á ˜ Ë 5¯ Ë 9¯ Êpˆ (c) cos Á ˜ Ë 18 ¯

Êpˆ (d) cos Á ˜ Ë 36 ¯

37. The value of ( 3 cot (20°) - 4 cos (20°)) is (a) 1

(b) –1

(c) -

3 2

(d)

Êp qˆ a 38. If tan 2 Á + ˜ = , then sin (q) is Ë 4 2¯ b Ê a – bˆ (a) Á Ë a + b ˜¯

Ê a – bˆ (b) - Á Ë a + b ˜¯

Ê a + bˆ (c) Á Ë a - b ˜¯

Ê a + bˆ (d) - Á Ë a - b ˜¯

3 2

39. The least value of cosec2 x + 25 sec2 x is (a) 26 (b) 36 (c) 16 (d) 12 sin 3 x cos3 x p ,0< x< cos x sin x 2 then the minimum value of y is (a) 0 (a) 1 (c) 3/2 (d) 2 41. The expression tan (55°) tan (65°) tan (75°) simplifies to cot (x°), 0 < x < 90, then x is (a) 5 (b) 8 (c) 9 (d) 10 40. Let y =

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The Ratios and Identities

42. If x1 and x2 are the roots of x2 + (1 – sin q) x –

1 cos2q 2

2 2 = 0, then the maximum value of x1 + x2 is (a) 2 (b) 3 (c) 9/4 (d) 4 43. The value of the expression Êpˆ Ê 3p ˆ Ê 5p ˆ Ê 7p ˆ cos 2 Á ˜ + cos 2 Á ˜ + cos 2 Á ˜ + cos 2 Á ˜ is Ë 8¯ Ë 8 ¯ Ë 8 ¯ Ë 8 ¯

(a) rational (c) prime

(b) integral (d) composite Êp ˆ 44. If a = tan x, then the value of cot Á - a˜ is Ë4 ¯ Ê a - 1ˆ (a) Á Ë a + 1˜¯

Ê a 2 - 1ˆ (b) Á 2 ˜ Ë a + 1¯

Ê a 2 + 1ˆ (c) Á 2 ˜ Ë a - 1¯

Ê a + 1ˆ (d) Á Ë a - 1˜¯

1 45. If sin q + cos q = , 0 £ q £ p , then tan q is 5 (a) 3/4 (b) 4/3 (c) –3/4 (d) –4/3

LEVEL III

x = cos q

where a, b, c Œ I+, find the value of (a + b + c – 2) 12. Find the value of 3 cot (20∞) - 4 cos (20∞) . 13. Prove that sin (2°) + sin (4°) + sin (6°) sin (8°) + … + sin (180°) = cot (1°) 14. Find the value of Ê p ˆ Ê 3p ˆ Ê 5p ˆ sin Á + sin Á + sin Á Ë 2013 ˜¯ Ë 2013 ˜¯ Ë 2013 ˜¯ Ê 7p ˆ + sin Á + ... upto (2013) trems Ë 2013 ˜¯ Ê1 + 1 + y ˆ 15. If tan y = Á ˜, Ë1 + 1 - y ¯ then prove that sin (4y) = y

(Problems for JEE Advanced)

1. Prove that tan a + 2 tan 2a + 4 tan 4a + 8 cot 8a = cot a 2. Prove that tan 9° – tan 27° – tan 63° + tan 81° = 4 3. Prove that sin x sin 3x sin 9x 1 + + = (tan 27x - tan x) cos 3x cos 9x cos 27x 2 sin x 1 cos x 3 = , = where x, y Œ R, then find the sin y 2 cos y 2 value of tan (x + y). 5. Prove that Êpˆ Ê 3p ˆ Ê 5p ˆ Ê 7p ˆ 3 sin 4 Á ˜ + sin 4 Á ˜ + sin 4 Á ˜ + sin 4 Á ˜ = Ë 16 ¯ Ë 16 ¯ Ë 16 ¯ Ë 16 ¯ 2 3 6 If cos (a - b ) + cos (b - g ) + cos (g - a ) = - , then 2 prove that cos a + cos b + cos g = 0 and sin a + sin b + sin g = 0.

4. If

7. If sin a sin b – cos a cos b + 1 = 0, then prove that 1 cot a tan b = 0 8. If a + b = 90° and b + g = a, then prove that tan a = tan b + 2 tan g Êpˆ 9. If tan Á ˜ = ( a - b )( c - d ) where a, b, c, d Ë 24 ¯ are positive integers, then find the value of (a + b + c + d + 2)

TR_01.indd 21

y z = , then find 2p ˆ 2p ˆ Ê Ê cos Á q + cos Á q ˜ ˜ Ë Ë 3 ¯ 3 ¯ the value of x + y + z. a+ b 11. If sin (25°) sin (35°) sin (85°) = c

10. If

Ê p yˆ Ê p xˆ 16. If tan Á + ˜ = tan 3 Á + ˜ , Ë 4 2¯ Ë 4 2¯ prove that 17. If tan b =

sin y 3 + sin 2 x = sin x 1 + 3 sin 2 x tan a + tan g , 1 + tan a tan g

prove that sin (2b ) =

sin (2a ) + sin (2g ) 1 + sin (2a ) ◊ sin (2g )

18. If 4 sin (27°) = (a + b )1/2 - (c - d )1/2 where a, b, c, d Œ N, find the value of (a + b + c + d + 2) 5 19. If (1 + sin q )(1 + cos q ) = , find the value of (1 – sin 4 q) (1 – cos q) Ê pˆ 20. If 3 sin x + 4 cos x = 5 where x ŒÁ 0, ˜ , then find the Ë 2¯ value of 2 sin x + cos x + 4 tan x 21. If cos A = tan B, cos B = tan C, cos C = tan A, prove that sin A = sin B = sin C = 2 sin (18°) 22. Find the value of Êpˆ Ê 3p ˆ Ê 5p ˆ Ê 7p ˆ tan 2 Á ˜ + tan 2 Á ˜ + tan 2 Á ˜ + tan 2 Á ˜ Ë 16 ¯ Ë 16 ¯ Ë 16 ¯ Ë 16 ¯ 23 If

sin (1°) ◊ sin (3°) ◊ sin (5°) .... sin (89°) =

1 2n

then

find the value of n 24. If (1 + tan (1°)) (1 + tan (2°)) (1 + tan (3°)) … (1 + tan (45°)), then find n

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Trigonometry Booster

25. Prove that Ê Ê p ˆˆ Ê Ê 3p ˆ ˆ ÁË1 + cos ÁË 10 ˜¯ ˜¯ ÁË1 + cos ÁË 10 ˜¯ ˜¯ Ê Ê 7p ˆ ˆ Ê Ê 9p ˆ ˆ 1 ÁË1 + cos ÁË 10 ˜¯ ˜¯ ÁË1 + cos ÁË 10 ˜¯ ˜¯ = 16 26 Prove that 1 cos (60°) cos (36°) cos (42°) cos (78°) = 16 k k 27. Let fk(q) = sin (q) + cos (q). 1 1 Then find the value of f 6 (q ) - f 4 (q ) 6 4 28. Find the maximum and minimum values of f (q) = sin2 (sin q) + cos2 (cos q) 29. Find the minimum value of f (q) = (3 sin (q) – 4 cos (q) – 10) (3 sin (q) + 4 cos (q) – 10) 30. Find the range of A = sin2010 q + cos2014 q sin A cos A 31. If = p, = q , prove that sin B cos B p Ê q2 - 1 ˆ tan A ◊ tan B = Á . q Ë 1 - p 2 ˜¯ tan (a - b ) sin 2 g + = 1 , then prove that tan2 g = tan tan a sin 2 a a ◊ tan b

32. If

1- e Êqˆ Êjˆ tan Á ˜ , then 33. If tan Á ˜ = Ë 2¯ Ë 2¯ 1+ e cos q - e prove that cos j = 1 - e cos q a cos j + b 34. If cos q = , then prove that a + b cos j Êqˆ tan Á ˜ = Ë 2¯

a-b Êjˆ tan Á ˜ . Ë 2¯ a+b

35. If sin x + sin y = a, cos x + cos y = b 4 - a 2 - b2 Ê x - yˆ then prove that tan Á =± ˜ Ë 2 ¯ a 2 + b2 Êpˆ 36. If tan Á ˜ = (a + b 2)1/2 - ( c + d ) , where Ë 16 ¯ a, b, c, d are +ve integers, then find the value of (a + b + c + d + 1). 37. If a and b are two values of q satisfying the equation cos q sin q 1 Êa + bˆ b = + = . Prove that cot Á Ë 2 ˜¯ a a b c Êpˆ 38. Prove that sin Á ˜ is a root of Ë 14 ¯ 8x3 – 4x2 – 4x + 1 = 0

TR_01.indd 22

39 If x + y + z = xyz, prove that 2x 2y 2z + + 1 - x2 1 - y 2 1 - z 2 2x 2y 2z = ◊ ◊ 2 2 1 - x 1 - y 1 - z2

[Roorkee, 1983]

40. If cos a + cos b + cos g = 0 and sin a + sin b + sin g = 0, then prove that cos (3a) + cos (3b) + cos (3g) = 3 cos (a + b + g) and sin (3a) + sin (3b) + sin (3g) = 3 sin (a + b + g) [Roorkee, 1985] 41. Show that (without using tables) tan 9° – tan 27° – tan 63° + tan 81° = 4 [Roorkee, 1989] 42. Find ‘a’ and ‘b’ such that the inequality pˆ Ê a £ cos x + 5 sin Á x - ˜ £ b holds good for all x. Ë 6¯ [Roorkee, 1989] 43. If A = cos2 q + sin4 q, then for all values of q, find range of A. [Roorkee, 1992] 44. Given the product p of sines of the angles of a triangle and the product q of their cosines, find the cubic equation, whose co-efficients are functions of p and q and whose roots are the tangents of the angles of the triangle. [Roorkee, 1992] 45 If x = cos (10°) cos (20°) cos (40°), then find the value of x. [Roorkee, 1995] 46. Find the real values of x for which 27cos 2x ◊ 81sin 2x is minimum and also find its minimum value. [Roorkee, 2000] 47. If eiq – log cos (x – iy) = 1, then find the values of x and y in terms of q. [Roorkee, 2001]

LEVEL IV

(Tougher Problems for JEE Advanced)

1. Prove that the sum of tan x tan 2x + tan 2x tan 3x + … + tan x tan (n + 1) x = cot x tan (n + 1) x – (n – 1) 2. Prove that cosec x + cosec 2x + cosec 4x + 5 to n terms Ê xˆ n -1 = cot Á ˜ - cot (2 x). Ë 2¯ 3. Prove that cot (16°) cot (44°) + cot (44°) cot (76°) –cot (76°) cot (16°) = 3 p , prove that 2n cos q 5. If q = n 2 -1 cos (2q) ◊ cos (4q) ◊ cos (8q) … cos (2n – 1 q) = –1 6. Prove that 7 Ê 2p ˆ Ê 4p ˆ Ê 8p ˆ sin Á ˜ + sin Á ˜ + sin Á ˜ = Ë 7 ¯ Ë 7 ¯ Ë 7¯ 2

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The Ratios and Identities

7. Prove that Êpˆ Ê 2p ˆ Ê 7p ˆ tan 2 Á ˜ + tan 2 Á ˜ + .... + tan 2 Á ˜ = 35 Ë 16 ¯ Ë 16 ¯ Ë 16 ¯ Ê Êpˆ Ê 2p ˆ Ê 3p ˆ ˆ 8. Prove that Á tan 2 Á ˜ + tan 2 Á ˜ + tan 2 Á ˜ ˜ Ë ¯ Ë ¯ Ë 7 ¯¯ Ë 7 7 Ê Êpˆ Ê 2p ˆ Ê 3p ˆ ˆ ¥ Á cot 2 Á ˜ + cot 2 Á ˜ + cot 2 Á ˜ ˜ = 105 Ë ¯ Ë ¯ Ë Ë 7 7 7 ¯¯ 3 + cot (76°) cot (16°) = cot (44°) 9. Prove that cot (76°) + cot (16°) 10. If cos x + cos y + cos z = 0, then prove that cos (3x) + cos (3y) + cos (3z) = 12 cos x cos y cos z Êpˆ Êpˆ Êpˆ 11. Prove that tan 6 Á ˜ - 33 tan 4 Á ˜ + 27 tan 2 Á ˜ = 3 Ë 9¯ Ë 9¯ Ë 9¯ 12. If cos A + cos B + cos C = 0 = sin A + sin B + sin C then prove that sin 2 A + sin 2 B + sin 2 C = cos2 A + cos2 B + 3 cos2 C = . 2 p 13. Let A, B, C be three angles such that A = and tan B. 4 tan C = p. Find all possible values of p such that A, B, C are three angles of a triangle. tan 3A sin 3A 2k = k , show that and k cannot = tan A sin A k - 1 lie between 1/3 and 3. 15. If A + B + C = p, then prove that cot A + cot B + cot C – cosec A cosec B cosec C = cot A ◊ cot B ◊ cot C 1 16. If tan a = x( x 2 + x + 1) 14. If

tan b =

and tan g =

x ( x + x + 1) 2

( x 2 + x + 1)

x x then prove that a + b = g 17. If a and b are acute angles and 3 cos 2b - 1 cos 2a = , prove that tan a : tan b = 2 :1 3 - cos 2b 3Êa

pˆ Êb pˆ 18. If tan Á + ˜ = tan Á + ˜ , prove that Ë 2 4¯ Ë 2 4¯ sin b =

(3 + sin 2 a ) sin a 1 + 3 sin 2 a

1 sin (2a + b ) , prove that 5 3 tan (a + b ) = tan a 2

19. If sin b =

TR_01.indd 23

20. If sin x + sin y = 3 (cos x – cos y), prove that sin (3x) + sin (3y) = 0 21. If sec (j – a), sec j, sec (j + a) are in AP then prove Êaˆ that cos (j ) = 2 cos Á ˜ Ë 2¯ Ê x + yˆ Ê x - yˆ 22. If tan Á , tan z , tan Á Ë 2 ˜¯ Ë 2 ˜¯

are in GP then

prove that cos (x) = cos (y) cos (2z) sec 2 q - tan q lies between 1/3 and 3 for all 23. Prove that sec 2 q + tan q real q p , find the value of 2n + 1 2n cos (q) cos (2q) cos (22 q) … cos (2n – 1 q) 25. Find the value of tan (6°) tan (42°) tan (66°) tan (78°) 24. If q =

tan (a + b - g ) tan g , prove that sin (b – g) = 0 or = tan (a - b + g ) tan b sin 2a + sin 2b + sin 2g = 0

26. If

27. If A + B + C = p, prove that sin A sin B = cot B + cot A + sin B sin C sin A sin C sin C = cot C + sin A sin B 28. If

sin (q + A) sin (2A) , then = sin (q + B ) sin (2B )

prove that tan2 q = tan A tan B 29. If cos (X – y) = –1, then prove that cos x + cos y = 0 and sin x + sin y = 0. 30. If

2 cos A = cos B + cos3 B

2 sin A = sin B - sin 3 B, 1 prove that sin ( A - B) = ± 3 1 31. Prove that sin (9°) = ( 3 + 5 - 5 - 5 ) 4 Ê p2 ˆ - x2 ˜ 32. Find the range of f (x) = sin Á Ë 36 ¯ and

6

33. Find the value of where i =

-1

Ê

Ê 2kp ˆ Ê 2kp ˆ ˆ ˜ - i cos ÁË ˜ 7 ¯ 7 ¯ ˜¯

 ÁË sin ÁË

k =1

34. If cos q + cos j = a and sin q + sin j = b, find the value Êqˆ Êjˆ of tan Á ˜ + tan Á ˜ Ë 2¯ Ë 2¯ tan q 1 = , find the value of 35. If tan q - tan 3q 3 cot q cot (q ) - cot (3q )

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1.24

Trigonometry Booster

Integer Type Questions y z = , find the 2p ˆ 2p ˆ Ê Ê cos Á q cos Á q + ˜ ˜ Ë Ë 3 ¯ 3 ¯ value of (x + y + z + 4)

1. If

x = cos q

Comprehensive Link Passages Passage-I Increasing product with angles are in GP cos a × cos 2a × cos 22a … cos 2n – 1 a Ï sin 2na : if a π np Ô n Ô 2 sin a Ô p Ô1 = Ì n : if a = n , 2 +1 Ô2 Ô Ô - 1 : if a = p ÔÓ 2n 2n - 1

2. Find the numerical value of Ê prˆ

9

 sin 2 ÁË 18 ˜¯

r =0

3. If

sin x 1 cos x 3 Ê pˆ = and = , where x, y ŒÁ 0, ˜ , Ë 2¯ sin y 2 cos y 2

find the value of

tan 2 ( x + y ) 5

4. If cos (x – y), cos x, cos (x + y) are in HP such that Ê yˆ sec x ◊ cos Á ˜ = m , find the value of (m2 + 2). Ë 2¯ Êp ˆ Ê 2p ˆ 5. If tan x + tan Á + x˜ + tan Á + x˜ Ë3 ¯ Ë 3 ¯ = k tan 3x, find k Ê 2p ˆ Ê 4p ˆ 6. Let f (q ) = sin 2 q + sin 2 Á + q ˜ + sin 2 Á + q˜ , Ë 3 ¯ Ë 3 ¯ Êpˆ find the value of 2f Á ˜ Ë 15 ¯ 7. If m = 3 cosec (20°) - sec (20°) and n = sin (12°) sin (48°) sin (5°) where m, n Œ N, find the value of (m + 8n + 2) 8. Let tan (15°) and tan (30°) are the roots of x2 + px + q = 0, find the value of (2 + q – p) 44

9. Let x =

 cos (n°)

n =1 44

(a) –1/2 (b) 1/2 p 2. If a = , the value of 13

, find the value [x + 3], where

 sin (n°)

[,] = GIF 10. If the value of the expression sin (25°) sin (35°) sin (85) a+ b where a, b, c Œ N and c Ê c ˆ + 2˜ are in their lowest form, find the value of Á Ëa+b ¯ 17 Ê kp ˆ 11. Let m =  cos Á ˜ , find the value of (m2 + m + 2) Ë 9 ¯ k =1 can be expressed as

12. If the expression tan (55°) tan (65°) tan (75°) simplifies to cot (x°) and m is the numerical value of the expression tan (27°) + tan (18°) + tan (27°) tan (18°), find the value of (m + x + 1)

(c) ¼

(d) 1/8

(c) 1/32

(d) –1/8

6

’ (cos (ra )) is r =1

(a) 1/64

(b) –1/64

Êpˆ Ê 3p ˆ Ê 5p ˆ Ê 7p ˆ 3. The value of sin Á ˜ sin Á ˜ sin Á ˜ sin Á ˜ Ë 14 ¯ Ë 14 ¯ Ë 14 ¯ Ë 14 ¯ Ê 9p ˆ Ê 11p ˆ Ê 13p ˆ is sin Á ˜ sin Á sin Á Ë 14 ¯ Ë 14 ˜¯ Ë 14 ˜¯ (a) 1

(b) 1/8

(c) 1/32

(d) 1/64

Êpˆ Ê 5p ˆ Ê 7p ˆ 4. The value of sin Á ˜ sin Á ˜ sin Á ˜ is Ë 18 ¯ Ë 18 ¯ Ë 18 ¯ (a) 1/16

n =1

TR_01.indd 24

where n is an integer. On the basis of above information, answer the following questions: Ê 2p ˆ Ê 4p ˆ Ê 6p ˆ 1. The value of cos Á ˜ cos Á ˜ cos Á ˜ is Ë 7 ¯ Ë 7 ¯ Ë 7 ¯

(b) 1/8

(c) – 1/8

(d) –1

5. The value of Êpˆ Êpˆ Êpˆ Êpˆ Êpˆ 64 3 sin Á ˜ cos Á ˜ cos Á ˜ cos Á ˜ cos Á ˜ Ë 48 ¯ Ë 48 ¯ Ë 24 ¯ Ë 12 ¯ Ë 6¯ is (a) 8 (b) 6 (c) 4 (d) –1 Passage-II Êpˆ Ê 3p ˆ Ê 5p ˆ If cos Á ˜ , cos Á ˜ , cos Á ˜ are the roots of the equaË 7¯ Ë 7 ¯ Ë 7 ¯ tion 8x3 – 4x2 – 4x + 1 = 0, on the basis of the above information, answer the following questions. Êpˆ Ê 3p ˆ Ê 5p ˆ 1. The value of sec Á ˜ + sec Á ˜ + sec Á ˜ is Ë 7¯ Ë 7 ¯ Ë 7 ¯ (a) 2

(b) 4

(c) 8

(d) None

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The Ratios and Identities

Êpˆ Ê 3p ˆ Ê 5p ˆ 2. The value of sin Á ˜ sin Á ˜ sin Á ˜ is Ë 14 ¯ Ë 14 ¯ Ë 14 ¯ (a) ¼

(b) 1/8

(c)

7 4

(d)

7 8

Êpˆ Ê 3p ˆ Ê 5p ˆ 3. The value of cos Á ˜ cos Á ˜ cos Á ˜ is Ë 14 ¯ Ë 14 ¯ Ë 14 ¯ (a) ¼

(b) 1/8

(c)

7 4

(d)

7 8

Êpˆ Ê 3p ˆ Ê 5p ˆ tan 2 Á ˜ , tan 2 Á ˜ , tan 2 Á ˜ is Ë 7¯ Ë 7 ¯ Ë 7 ¯ (a) x – 35x + 7x – 21 = 0 (b) x3 – 35x2 + 21x – 7 = 0 (c) x3 – 35x2 + 35x – 7 = 0 (d) x3 – 21x2 + 7x – 35 = 0 5. The value of 3 3 Ï Ï Ê 2r - 1ˆ ¸ Ê 2r - 1ˆ ¸  Ìtan 2 ÁË 7 ˜¯ p ˝ ¥  Ìcot 2 ÁË 7 ˜¯ p ˝ is ˛ r =1 Ó ˛ r =1 Ó (a) 15

2

(b) 105

(c) 21

(d) 147

Passage-III Let x2 + y2 = 1 for every x, y in R. Then, 1. The value of P = (3x – 4x3)2 + (3y – 4y3)2is (a) 2 (b) 1 (c) 0 (d) –1 2. The minimum value of Q = x6 + y6 is (a) 1 (b) 1/2 (c) 1/4 (d) –1 3. The maximum value of R = x2 + y4 IS (a) 0 (b) 1 (c) 1/2 (d) 3/4 Passage – IV Consider the polynomial P(x) = (x – cos 36°)(x – cos 84°)(x – cos 156°) Then, 1. The co-efficient of x2 is (a) 0 (b) 1 Ê 5 - 1ˆ (c) –1/2 (d) Á Ë 2 ˜¯ 2. The co-efficient of x is (a) 3/2 (b) –3/2 (c) –3/4 (d) 2 3. The constant term in P(x) is Ê 5 - 1ˆ Ê 5 - 1ˆ (a) Á (b) Á Ë 16 ˜¯ Ë 4 ˜¯ Ê 5 + 1ˆ (c) Á Ë 16 ˜¯

(d)

1 16

Passage-V If a sin x + b cos x = 1 such that a2 + b2 = 1 for all a, b Î Œ(0, 1) then,

TR_01.indd 25

(c) a/b

(d) b/a

(c) a/b

(d) b/a

(c) a/b

(d) b/a

Passage-VI

4. The equation whose roots are

3

1. The value of sin x is (a) a (b) b 2. The value of cos x is (a) a (b) b 3. The value of tan x is (a) a (b) b Let sec x + tan x =

p 22 , where 0 < x < 2 7

then Ê xˆ 1. The value of tan Á ˜ is Ë 2¯ (a) 15/29

(b) 13/25 (c) 14/29 Ê 1 – cos x ˆ 2. The value of Á1 – is 1 + cos x ˜¯ Ë

(d) –15/29

(a) 15/29 (b) 14/29 (c) 0 3. The value of (cosec x + cot x) is (a) 29/14 (b) 15/28 (c) 29/15

(d) 12/25 (d) 15/29

Matrix Match (For JEE-Advanced Examination Only) 1. Match the following columns: Column I (A)

p , where q 2 and f are positive, then Êpˆ (sin q + sin j ) sin Á ˜ is alË 4¯ ways less than If q + j =

Column II (P) 1

(B) If sin q – sin j = a and cos q + cos (Q) j = b, then a2 b2 can not exceed (C) If 3 sin q + 5 cos q = 5, (q π 0), (R) then the value of 5 sin q – 3 cos q is (S) (T)

2 3

4 5

2. Match the following columns: Column I (A) The value of cos (20°), cos (40°), cos (80°) is (B) The value of cos (20°), cos (40°), cos (60°), cos (80°) is (C) The value of sin (20°), sin (40°), sin (80°) is (D) The value of sin (20°), sin (40°), sin (60°), sin (80°) is

Column II (P) 3/8 (Q) (R) (S)

3/16 3/32 1/16

(T) 1/8

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Trigonometry Booster

6. Match the following columns :

3. Match the following columns : Column I Column II (A) If maximum and minimum val- (P) l + m = 2 7 + 6 tan q - tan 2 q ues of 1 + tan 2 q Ê p ˆ (Q) l – m = 6 For all real values of q Á π ˜ Ë 2¯ are l and m respectively, then (R) l + m = 6 (B) If maximum and minimum values of pˆ Ê 5 cos q + 3 cos Á q + ˜ + 3 Ë 3¯ For all real values of q are l and (S) m respectively, then (C) If maximum and (T) minimum values of Êp ˆ Êp ˆ 1 + sin Á + q ˜ + 2 cos Á - q ˜ Ë4 ¯ Ë4 ¯

l–m = 14

(B)

5. Match the following columns: Column II (P) 1/8

Ê 5p ˆ Ê 7p ˆ cos 4 Á ˜ + cos 4 Á ˜ is Ë 8 ¯ Ë 8 ¯

(C)

(D)

TR_01.indd 26

The value of sin (12°), sin (48°), sin (54°) is The value of sin (6°), sin (42°), sub(66°), sin (78°) is The value of tan (6°), tan (42°), tan (66°), tan (78°) is

4 8 –8

(D) The value of (1 + tan 235°), (1– (T) tan 190°) is

(A)

Column I Column II In a triangle ABC (A) sin 2A + sin 2B + (P) 4 sin A ◊ sin B ◊ sin sin 2C is C (B) cos 2A + cos 2B + (Q) –1 – 4 cos A ◊ cos B cos 2C is ◊ cos C 2 2 (C) sin A + sin B + (R) 2 + 2 cos A ◊ cos B sin2 C is ◊ cos C 2 2 (D) cos A + cos B + (S) 1 – 2 cos A ◊ cos B cos2 C is ◊ cos C

(B)

p , then the value of (1 4 + tan A) (1 + tan B) is 2 (A) The value of (1 + tan 21°) (1 + tan (Q) 22°) (1 + tan 23°) (1 + tan 24°) is (B) The value of ( 1 + tan 2058°), (1 (R) – tan 2013°) is (C) The value of (S) Ê Êp ˆˆ 1 + tan Á - x˜ ˜ ◊ Ë8 ¯¯ ËÁ If A + B =

–4

7. Match the following columns:

4. Match the following columns:

Column I The value of Êpˆ Ê 3p ˆ cos 4 Á ˜ + cos 4 Á ˜ + Ë 8¯ Ë 8 ¯

Column II (P) 2

Ê p ˆˆ Ê ÁË 1 + tan ËÁ x + 8 ¯˜ ¯˜ . is

For all real values of q are l and m respectively, then

(A)

Column I

(C) (D)

Column I The value of 2 tan (50°) + tan (20°) is The value of tan (40°) + 2 tan (10°) is The value of tan (20°) tan (40°), tan (60°), tan (80°) is If 3 sin x + 4 cos x = 5, then the value of 2 sin x + cos x + 4 tan x is

Column II (P) 3 (Q)

5

(R)

tan (70°) tan (50°)

(S)

8. Match the following columns: Column I

Column II (P) 1

(A)

The minimum value of 2 sin q + 3 cos2 q is

(B)

The maximum value of sin2 q + cos4 q is

(Q)

3/4

(C)

The least value of sin4 q + cos2 q is

(R)

2

(D)

The greatest value of sin2014 q + cos2010 q is

(S)

4

2

9. Match the following columns: (Q) (R)

–3/2 3/2

(S)

1/16

(T)

1

Column I

Column II

If a and b are the solutions of a cos q + b sin q = c, then (A) the value of sin a + sin b is (P) (B)

the value of sin a ◊ sin b is

(Q)

c2 - b2 a 2 + b2 2ac a + b2 2

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The Ratios and Identities

(C)

the value of cos a + cos b is

(R)

c2 - a2 a 2 + b2

(D)

the value of cos a ◊ cos b is

(S)

2bc a + b2 2

10. Match the following columns: Column I Column II (A) The value of cos (12°) + cos (84°) (P) 0 + cos (156°) + cos (132°) is (Q) 1 (B) The value of Êpˆ Êpˆ 2 tan Á ˜ + 3 sec Á ˜ Ë 10 ¯ Ë 10 ¯ Êpˆ - 4 cos Á ˜ is Ë 10 ¯ (C) The value of (R) 3 cot (20∞) - 4 cos (20∞) is (D) The value of tan (20°) + 2 tan (S) (50°) – tan (70°) is (T)

2

5. Assertion (A): pˆ 4p ˆ Ê Ê cos 2 a + cos 2 Á a + ˜ + cos 2 Á a + ˜ Ë ¯ Ë 3 3 ¯ 2p ˆ 4p ˆ Ê Ê = 3 cos a cos Á a + ˜ cos ÁË a + ˜ Ë 3 ¯ 3 ¯ Reason (R): If a + b + c = 0, then a3 + b3 + c3 = 3abc (a) A (b) B (c) C (d) D 6. Assertion (A): tan (5q) – tan (3q) – tan (q) = tan (5q), tan (3q), tan (q) Reason (R): If x = y + z, then tan x – tan y – tan z = tan x ◊ tan y ◊ tan z (a) A (b) B (c) C (d) D 7. Assertion (A): The maximum value of sin q + cos q is 2 Reason (R): The maximum value of sin q is 1 and that of cos q is also 1. (a) A (b) B (c) C (d) D n

8. Assertion (A): The maximum value of

i =1

–1/2 –1

under the restriction 0 £ a1 , a 2 , a 3 , ..., a n £ n

Reason (R):

Assertion and Reason Codes (A) Both A and R are individually true and R is the correct explanation of A. (B) Both A and R are individually true and R is not the correct explanation of A. (C) A is true but R is false. (D) A is false but R is true. 1 1. Assertion (A): sin q = x + is impossible if x Œ R – x {0}. Reason (R): AM ≥ GM (a) A (b) B (c) C (d) D 2. Assertion (A): A is an obtuse angle in DABC, then tan B. tan C > 1 tan B + tan C Reason (R): In DABC, tan A = tan B tan C - 1 (a) A (b) B (c) C (d) D 1 Ê 2p ˆ Ê 4p ˆ Ê 6p ˆ 3. Assertion (A): sin Á ˜ + sin Á ˜ + sin Á ˜ = Ë 7 ¯ Ë 7 ¯ Ë 7 ¯ 2 Ê 2p ˆ Ê 2p ˆ Reason (R): cos Á ˜ + i sin Á ˜ is complex 7th Ë 7 ¯ Ë 7 ¯ root of unity. (a) A (b) B (c) C (d) D 4. Assertion (A): tan a + 2 tan (2a) + 4 tan (4a) + 8 tan (8a) – 16 cot (16a) = cot a Reason (R): cot a – tan a = 2 cot 2a (a) A (b) B (c) C (d) D

TR_01.indd 27

’ cos (a i ) 1 p is n /2 2 2

’ cot (a i ) = 1 i =1

(a) A (b) B (c) C (d) D 9. Assertion (A): If A + B + C = p, then the maximum value of tan A ◊ tan B ◊ tan C is 3 3 Reason (R): AM ≥ GM (a) A (b) B (c) C (d) D 4xy 2 10. Assertion (A): sec q = is positive for all real ( x + y)2 values of x and y only when x = y Reason (R): sec2 q ≥ 1 (a) A (b) B (c) C (d) D

Questions Asked In Previous Years’ JEE-Advanced Examinations 1. Prove that sin x sin ◊ sin (x – y + sin y sin z sin (y – z) + sin z sin x sin (z – x) + sin (x– y) sin (y – z) sin (z – x) = 0 [IIT-JEE, 1978] 4 5 2. If cos (a + b ) = , sin (a - b ) = and a, b lie be5 13 p tween 0 and , find tan 2a [IIT-JEE, 1979] 4 3. Given A = sin2 q + cos4 q for all values of q, then [IIT-JEE, 1980] (a) 1 £ A £ 2 (b) 3/4 £ A £ 1 (c) 13/6 £ A £ 1 (d) 3/4 £ A £ 13/6 1 – cos B 4. If tan A = , then tan 2A = tan B. Is it true/ sin B false ? [IIT-JEE, 1980]

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1.28

Trigonometry Booster

5. Suppose sin x sin 3x = 3

n

 Cm cos (mx)

is an identity

m=0

in x, when C0, C1, … Cn are constants and Cn π 0 is the value of n = … [IIT-JEE, 1981] 6. Without using the tables, prove that 1 sin 12° sin 54° sin 48° = 8 [IIT-JEE, 1982] 7. If a + b + g = p, then prove that sin2 a + sin2 b – sin2 g = 2 sin a × sin b ◊ sin g [IIT-JEE, 1983] 8. Prove that Ê 2p ˆ Ê 4p ˆ Ê 8p ˆ Ê 16p ˆ 16 cos Á ˜ cos Á ˜ cos Á ˜ cos Á =1 Ë 15 ¯ Ë 15 ¯ Ë 15 ¯ Ë 15 ˜¯ [IIT-JEE, 1983] 9. The value of Ê Ê p ˆˆ Ê Ê 3p ˆ ˆ ÁË 1 + cos ÁË 8 ˜¯ ˜¯ ÁË 1 + cos ÁË 8 ˜¯ ˜¯ Ê Ê 5p ˆ ˆ Ê Ê 7p ˆ ˆ ÁË 1 + cos ÁË 8 ˜¯ ˜¯ ÁË1 + cos ÁË 8 ˜¯ ˜¯

(b) cos

p 8

(c) 1/8

1+ 2 2 2 [IIT-JEE, 1984] (d)

10. No questions asked in 1985. È 3p ˘ ˆ 11. The expression 3 Ísin 4 ÊÁ - a ˜ + sin 4 (3p + a ) ˙ – Ë ¯ 2 Î ˚ È 6Êp ˘ ˆ 2 Ísin Á + a ˜ + sin 6 (5p - a ) ˙ is equal to Ë2 ¯ Î ˚ (a) 0 (c) 3

12. 13.

14.

15. 16.

(b)* 1 (d) sin 4a + cos a [IIT-JEE, 1986] No questions asked in 1987. The value of the expression 3 cosec (20°) - sec (20°) is equal to 2 sin 20∞ (a) 2 (b) sin 40∞ 4 sin 20∞ (c) 4 (d) sin 40∞ [IIT-JEE, 1988] Prove that tan a + 2 tan 2a + 4 tan 4a + 8 cot 8a = cot a [IIT-JEE, 1988] No questions asked between 1989 – 1990. Find the value of Êpˆ Ê 3p ˆ Ê 5p ˆ Ê 7p ˆ sin Á ˜ sin Á ˜ sin Á ˜ sin Á ˜ Ë 14 ¯ Ë 14 ¯ Ë 14 ¯ Ë 14 ¯ Ê 9p ˆ Ê 11p ˆ Ê 13p ˆ sin Á ˜ sin Á sin Á ˜ Ë 14 ¯ Ë 14 ¯ Ë 14 ˜¯ [IIT-JEE, 1991]

TR_01.indd 28

Column I (i)

Positive

(ii) Negative

Column II (A)

Ê 13p 14p ˆ , ÁË ˜ 48 48 ¯

(B)

Ê 14p 18p ˆ , ÁË ˜ 48 48 ¯

(C)

Ê 18p 23p ˆ , ÁË ˜ 48 48 ¯

(D)

Ê pˆ ÁË 0, ˜¯ 2 [IIT-JEE, 1992]

is equal to (a) 1/2

17. If f (x) = cos [p2] x + cos [–p2], where [,] = G.I.F., then Êpˆ (a) f Á ˜ = 1 (b) f (p) = 1 Ë 2¯ Êpˆ (c) f (–p) = 0 (d) f Á ˜ = 1 Ë 4¯ [IIT-JEE-1991] 18. Match the following columns:

Êpˆ Ê 5p ˆ Ê 7p ˆ 19. If k = sin Á ˜ sin Á ˜ sin Á ˜ , the numerical valË 18 ¯ Ë 18 ¯ Ë 18 ¯ ue of k is ______ [IIT-JEE, 1993] p 20. If A > 0, B > 0 and A + B = , then the maximum val3 ue of tan A tan B is ______ [IIT-JEE, 1993] p 21. Let 0 < x < , then (sec 2x – tan 2x) equals 4 pˆ Ê Êp ˆ (a) tan Á x - ˜ (b) tan Á - x˜ Ë Ë4 ¯ 4¯ p p Ê ˆ Ê ˆ (c) tan Á + x˜ (d) tan 2 Á + x˜ Ë4 ¯ Ë4 ¯ [IIT-JEE, 1994] 22. The value of the expression 3(sin x – cos x)4 + 6(sin x + cos x)2 + 4(sin6 x + cos6 x) is (a) 11 (b) 12 (c) 13 (d) 14 [IIT-JEE, 1995] 23. The minimum value of the expression sin a + sin b + sin g, where a, b, g are real numbers satisfying a + b + g = p is (a) positive (b) 0 (c) negative (d) –3 [IIT-JEE, 1995] Ê 4xy ˆ 2 24. sec q = Á is true if and only if Ë ( x + y ) 2 ˜¯ (a) x + y = 0 (b) x = y, x π 0 (c) x = y (d) x π 0, y π 0 [IIT-JEE, 1996] 25. The graph of the function cos x cos (x + 2) – cos2 (x + 1) is (a) a straight line passing through (0, –sin2 q) with slope 2.

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The Ratios and Identities

(b) a straight line passing through (0, 0) (c) a parabola with vertex (1, – sin2 1) (d) a straight line passing through the point Êp 2 ˆ ÁË , - sin 1˜¯ and parallel to x-axis. 2 [IIT-JEE,1997] 26. Which of the following numbers is/are rational? (a) sin 15° (b) cos 15° (c) sin 15° cos 15° (d) sin 15° cos 75° [IIT-JEE, 1998] 27. For a positive integer n, let

34.

35.

Êqˆ f n (q ) = tan Á ˜ (1 + sec q )(1 + sec 2q )(1 + sec 22q ) Ë 2¯ … (1 + sec (2n q)), then p (a) f 2 ÊÁ ˆ˜ = 1 Ë 16 ¯

p (b) f3 ÊÁ ˆ˜ = 1 Ë 32 ¯

Ê p ˆ (d) f5 Á =1 Ë 128 ˜¯ [IIT-JEE, 1998] 28. Let f (q) = sin q (sin q + sin 3q), then f (q) (a) ≥ 0 when q ≥ 0 (b) £ 0 for all real q (c) ≥ 0 for all real q (d) £ 0 only when q £ 0 [IIT-JEE, 2000] 29. The maximum value of cos a1 × cos an under the rep striction 0 £ a1 , a 2 , ......., a n £ and 2 cot a1, ◊ cot a2 …, cot an = 1 is

36.

p (c) f 4 ÊÁ ˆ˜ = 1 Ë 64 ¯

(a)

1 2

n /2

(b)

1 2n

(c)

1 2n

(d) 1 [IIT-JEE-2001]

30. No questions asked in 2002. p 31. If a + b = and a = b + g, then tan a is 2 (a) 2 (tan b + tan g) (b) tan b + tan g (c) (tan b + 2 tan g) (d) 2 tan b + tan g [IIT-JEE, 2003] Ê pˆ 32. If a ŒÁ 0, ˜ , then the expression Ë 2¯ y = x2 + x +

tan 2 a x2 + x

to (a) 2 tan a (b) 2

is always greater than or equal

(d) sec3 a [IIT-JEE, 2003] 33 Given that both q and j are acute angles and 1 1 sin q = , cos j = , then the value of (q + j) belongs 2 3 to the interval Êp pˆ Ê p 2p ˆ (a) Á , ˜ (b) Á , Ë 3 2¯ Ë 2 3 ˜¯

TR_01.indd 29

(c) 1

Ê 5p ˆ (d) Á , p ˜ Ë 6 ¯ [IIT-JEE, 2004] Find the range of values of t for which 1 - 2x + 5x 2 Ê p pˆ 2 sin t = 2 , t ŒÁ - , ˜ Ë 2 2¯ 3x - 2x - 1 [IIT-JEE, 2005] 1 cos (a - b ) = 1 and cos (a + b ) = , where a, b Œ e [–p, p]. Values of a, b which satisfy the equations is/ are (a) 0 (b) 1 (c) 2 (d) 4. [IIT-JEE, 2005] Ê pˆ Let q ŒÁ 0, ˜ and Ë 4¯ t1 = (tan q) tan q, t2 = (tan q) cot q t3 = (cot q) tan q, t4 = (cot q) cot q, then (a) t1 > t2 > t3 > t4 (b) t4 > t3 > t1 > t2 (c) t3 > t1 > t2 > t4 (d) t2 > t3 > t1 > t4 [IIT-JEE, 2006] Note: No questions asked in 2007, 2008. sin 4 x cos 4 x 1 If + = , then 2 3 5 2 sin 8 x cos8 x 1 2 (b) (a) tan x = + = 3 8 27 125 sin 8 x cos8 x 2 1 (c) + = (d) tan 2 x = 8 27 125 3 [IIT-JEE, 2009] The maximum value of the expression 1 is ______ sin 2 q + 3 sin q cos q + 5 cos 2 q [IIT-JEE, 2010] 2p 5p ˆ (c) ÊÁ , Ë 3 6 ˜¯

37.

38.

39. Let P = {q : sin q - cos q = 2 cos q } and Q = {q : sin q + cos q = 2 sin q } be two sets. Then, (a) P Ã Q and Q – P π j (b) Q À P (c) P À Q (d) P = Q [IIT-JEE, 2011] 40. The positive integral value of n > 3 satisfying the equa1 1 1 tion is… = + Êpˆ Ê 2p ˆ Ê 3p ˆ sin Á ˜ sin Á ˜ sin Á ˜ Ë n¯ Ë n ¯ Ë n ¯ [IIT-JEE, 2011] 2 2 - sec 2 q Ê 1ˆ . Then the values of f Á ˜ is Ë 3¯

41. Let f : (–1, 1) Æ R be such that f (cos 4q ) = Ê pˆ Êp pˆ for q ŒÁ 0, ˜ » Á , ˜ Ë 4¯ Ë 4 2¯ /are 3 (b) 1 + 2

3 (c) 1 2

2 2 (d) 1 + 3 3 [IIT-JEE, 2012] No questions asked in between 2013 to 2016. (a) 1 -

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1.30

Trigonometry Booster

A NSWERS

LEVEL II 1. 6. 10. 13. 18. 23. 28. 33. 38. 43.

6. 3 11. 2

(b) 2. (c) 7. (a,b, d) (c) 14. (b) 19. (b) 24. (a) 29. (d) 34. (a) 39. (a, b, c)44.

(b) (c)

3. 8. 11. 15. 20. 25. 30. 35. 40. 45.

(c) (c) (c) (b) (d) (b) (d)

(b) (a, b) (a,c) (b) (d) (a) (a) (d) (b) (d)

4. 9. 12. 16. 21. 26. 31. 36. 41.

(b) 5. (d) (a,c, d) (b) 17. (b, c) 22. (b) 27. (c) 32. (b) 37. (a) 42.

(d)

(a) (c) (c) (c) (a) (d)

, 3 2 2]

[3 2 2, )

2

È 1˘ 32. R f = Í0, ˙ Î 2˚ 33. i 4b Ê ˆ 34. Á 2 2˜ Ë a + 2a + b ¯ 35. 2/3

2. 5

3. 3

4. 4

Passage-I: Passage-II: Passage-III: Passage-IV: Passage-V: Passage-VI:

1. 1. 1. 1. 1. 1.

(d) (b) (b) (a) (a) (a)

2. 2. 2. 2. 2. 2.

(a) (b) (c) (c) (b) (b)

3. 3. 3. 2. 3. 3.

(d) (d) (b) (b) (c) (c)

4. (b) 4. (c)

5. (b) 5. (b)

(A) Æ (P, Q, R, S, T): (B) Æ (S, T); (C) Æ (R) (A) Æ T; (B) Æ S; (C) Æ P; (D) Æ Q (A) Æ (R, S); (B) Æ (R, T); (C) Æ (P, Q) (A) Æ P; (B) Æ Q; (C) Æ R, (D) Æ S (A) Æ (R); (B) Æ (P);(C) Æ (S), (D) Æ (T) (A) Æ (Q); (B) Æ (P); (C) Æ (P), (D) Æ (P) (A) Æ (R); (B) Æ (S); (C) Æ (P), (D) Æ (Q) (A) Æ (R); (B) Æ (P); (C) Æ (Q); (D) Æ (P) (A) Æ (S); (B) Æ (R); (C) Æ (Q), (D) Æ (P) (A) Æ (S); (B) Æ (P); (C) Æ (Q), (D) Æ (P)

1. (a) 6. (a)

5. 3

H INTS

AND

22 ¥ 4900 7 = 44 ¥ 700 = 30800 km 2. Let the three angles be 3x, 4x and 5x, respectively Thus, 3x + 4x + 5x = 180° fi 12x = 180° fi x = 15° Therefore, the smallest angle = 3x = 3 ¥ 15° = 45° and the greatest angle = 5x = 5 ¥ 15° = 75°

3. (d) 8. (a)

4. (a) 9. (a)

5. (a) 10. (a)

p ˆ Ê = Á 75 ¥ ˜ radians Ë 180 ¯

1. Given r = 4900 km Circumference = 2p r = 2¥

2. (d) 7. (d)

S OLUTIONS

LEVEL I

TR_01.indd 30

10. 4

ASSERTION AND REASON

INTEGER TYPE QUESTIONS

1. 4

9. 5

COMPREHENSIVE LINK PASSAGES

1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

16. qx – px + (1 + q) x – p = 0 3

8. 3

MATRIX MATCH

LEVEL IV 13. p (

7. 7 12. 7

Ê 5p ˆ = Á ˜ radians Ë 12 ¯ 3. Let the three angles be a + d, a, a – d Thus, a + d + a + a – d = 180° fi 3a = 180° 180° fi a= = 60° 3 It is given that, p 60 = 180 p (a - d ) 180 60 ¥ = p p (a + d )

(a - d )° : (a + d ) ¥ fi

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The Ratios and Identities

fi fi fi fi

(a - d ) 1 = (a + d ) 3 a + d = 3a – 3d 4d = 2a a d = = 30° 2

Hence, the three angles are 90°, 60°, 30°. 4. Let the number of sides of the given polygons be 5x and 4x respectively. It is given that, Ê 2 ¥ 5x - 4 2 ¥ 4x - 4 ˆ ÁË ˜¯ ¥ 90 = 9 5x 4x fi

Ê 10x - 4 2x - 1ˆ 1 ÁË ˜= 5x x ¯ 10

fi

Ê 10x - 4 – 10x + 5 ˆ 1 ÁË ˜¯ = 5x 10

fi

Ê 1ˆ 1 ÁË ˜¯ = x 2

fi x=2 Hence, the number of sides of the polygons would be 10 and 8 respectively. 5. Let the angles of the quadrilateral be a – 3d, a – d, a + d, a + 3d It is given that, a + 3d = 2(a – 3d) fi a + 3d = 2a – 6d fi a = 9d Also, a + 3d + a – d + a + d + a + 3d = 360 fi 4a = 360 fi a = 90 and d = 10 Hence, the smallest angle = 90° – 30° = 60° Êpˆ = Á ˜ Ë 3¯

c

1 6. Clearly, at half past 4, the hour hand will be at 4 and 2 minute hand will be at 6. In 1 hour angle made by the hour hand 30°. 1 In 4 hours angle made by the hour hand 2 9 ¥ 30° = 135° 2 In 1 minute angle made by the minute hand = 6° In 30° minutes, angle made by the minute hand = 6 ¥ 30° = 180° Thus, the angle between the hour hand and the minute hand = 180° – 135° = 45° =

TR_01.indd 31

7. Angle subtended at the centre p ˆ p Ê = 30° = Á 30 ¥ ˜= Ë 180 ¯ 6 Hence, l = 10 ¥

p 5p = . 6 3

8. The angle traced by a minute hand in 60 minutes = 360° = 2p radians Thus the angle traced by minute hand in 18 minutes 18 3p = 2p ¥ = radians 60 5 Hence, the distance moved by the tip in 18 minutes 3p 22 = l = 35 ¥ = 21 ¥ = 66 cm 5 7 9. Let AB be the height of the man and the required distance be x, where BC = x A

B

x

10°

C

2 180 10 ¥ = x p 60 2 180 fi x= ¥ ¥ 60 10 p 12 ¥ 180 fi x= p 12 ¥ 180 12 ¥ 180 ¥ 7 fi x= = 22 22 7 42 ¥ 180 fi x= = 687.3 11 10. Let the required distance be x cm. According to the question, 11 180 6° = ¥ 2¥ x p 6 11 180 = ¥ fi 60 2 ¥ x p Therefore,

fi

x=

11 180 60 ¥ ¥ 2 p 6

fi

x=

11 180 ¥ 7 ¥ ¥ 10 2 22

fi x = 45 ¥ 7 ¥ 10 = 3150 Hence, the required distance be 3150 cms. 11. Let the radius of the moon be x km 16 2x 180 = ¥ It is given that, 60 60 ¥ 6400 p

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Trigonometry Booster

fi fi fi

16 ¥ 6400 ¥ p x= 180 ¥ 2 4 ¥ 640 ¥ p x= 9 4 ¥ 640 ¥ 22 x= 9¥7

fi x = 894 Hence, the radius of the moon be 894 km. 12. The difference between the the acute angles of a right 2p angled triangle is radians. Express the angles in 3 degrees. Ans. 81o, 9o 13. The angles of a quadrilateral are in AP and the greatest angle is 120o. Find the angles in radians. p 4p 5p 2p Ans. , , , . 3 9 9 3 1 14. At what distance does a man 5 ft in height, subtend 2 an angle of 15°? Ans. 14.32 miles 15. Find the angle between the hour hand and minute-hand in circular measure at 4 O’clock. 4p Ans. 3 16. Given sec q + tan q = 3 …(i) 1 1 fi …(ii) (sec q - tan q ) = = (sec q + tan q ) 3 Adding (i) and (ii) we get, 2 sec q = 3 + fi fi

1 10 = 3 3

5 3 5 cos q = 3

sec q =

1 17. Given, cosec q - cot q = 5 1 =5 cosec q + cot q Adding (i) and (ii) we get,

fi

cosec q - cot q =

2 cosec q = 5 + fi

cosec q =

TR_01.indd 32

We have …(i) …(ii)

1 26 = 5 5

13 5

5 13 18. Given, a = c cos q + d sin q and b = c sin q – d cos q Squaring and adding (i) and (ii) we get, a2 + b2 = (c cos q + d sin q)2 + (c sin q – d cos q)2 fi a2 + b2 = (c2 cos2 q + d2 sin2 q) + (c2 sin2 q + d2 cos2 q) fi

fi a2 + b2 = c2 + d2 fi m = 2, n = 2, p = 2, q = 2 Hence, the value of m + n + p + q + 42 = 50 19. Let x = 3 cos q – 4 sin q …(i) and 5 = 3 sin q – 4 cos q …(ii) Squaring and adding (i) and (ii) we get, x2 + 52 = (3 cos q + 4 sin q)2 + (3 sin q – 4 cos q)2 fi x2+52 = (9 cos2 q + 16 sin2 q + 24 sin q cos q) + (9 sin2 q + 16 cos2 q – 24 sin q cos q) = (9 cos2 q + 16 sin2 q) + (9 sin2 q + 16 cos2 q) = 9(cos2 q + sin2 q) + 16(cos2 q + sin2 q) fi x2 + 25 = 25 fi x2 = 0 fi x=0 fi 3 cos q – 4 sin q = 0 20. We have x2 + y2 + z2 = (r cos q cos j)2 + (r cos q sin j)2 + (r sin q)2 fi x2 + y2 + z2 = (r2 cos2 q cos2 j) + (r2 cos2 q sin2 j) + (r2 sin2 q) fi x2 + y2 + z2 = r2 cos2 q (cos2 j + sin2 j) + (r2 sin2 q) = r2 cos2 q + r2 sin2 q = r2(cos2 q + sin2 q) = r2 2 fi x + y2 + z2 = r2 fi m = 2, n = 2, p = 2 Thus, the value of (m + n + p – 4)(m + n + p + 4) = 210 = 1024 2 sin a 21. Given, x = 1 + cos a + 3 sin a

sin a + 3(1 – cos a ) 2 (1 – cos a ) sin a 3 = + 2 (1 – cos a ) 2 sin a (1 + cos a ) 3 = + 2 2 (1 – cos 2 a ) =

=

sin q =

…(i) …(ii)

sin a - 3 cos a + 3 2 - 2 cos a

sin a (1 + cos a )

+

3 2

2 sin a (1 + cos a ) 3 = + 2 sin a 2 (1 + cos a + 3 sin a ) = 2 sin a 1 = x 2

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22. We have P = sec6 q – tan6 q – 3 sec2 q tan2 q = (sec2 q – tan2 q)3 = 1 Q = cosec6 q – cot6 q – 3 cosec2 q cot2 q = (cosec2 q – cot2 q)3 = 1 and R = sin6 q + cos6 q + 3 sin2 q cos2 q = (sin2 q + cos2 q)3 = 1 Hence, the value of (P + Q + R)(P + Q + R) = 33 = 27 23. We have 3 sin x + 4 cos x = 5 Let y = 3 cos x – 4 sin x Now, y2 + 52 = (3 cos x – 4 sin x)2 + (3 sin x + 4 cos x)2 2 fi y + 25 = 9 cos2 x + 16 sin2 x – 24 sin x cos x + 9 sin2 x + 16 cos2 x + 24 sin x cos x 2 fi y + 25 = 25(cos2 x + sin2 x) = 25 fi y2=0 fi y=0 fi 3 cos x – 4 sin x = 0 fi 3 cos x = 4 sin x fi tan x = 3/4 Hence, the value of 2 sin x + cos x + 4 tan x Ê 3ˆ Ê 4 ˆ Ê 3ˆ = 2Á ˜ + Á ˜ + 4Á ˜ = 2 + 3 = 5 Ë 5¯ Ë 5 ¯ Ë 4¯ 24. Given, sin A + sin B + sin C = – 3 fi sin A = –1, sin B = –1, sin C = –1 p p p A= - , B = - ,C = fi 2 2 2 Hence, the value of cos A + cos B + cos C + 10 = 0 + 0 + 0 + 10 = 10. 5 25. We have (1 + sin q )(1 + cos q ) = 4 5 fi 1 + sin q + cos q + sin q cos q = 4 Ê t 2 - 1ˆ 5 fi 1+ t + Á = (sin q + cos q = t, say) Ë 2 ˜¯ 4 fi fi fi fi

fi fi

TR_01.indd 33

Ê t 2 - 1ˆ 1 t +Á = Ë 2 ˜¯ 4 1 t 2 + 2t - 1 = 2 2t2 + 4t – 3 = 0 - 4 ± 16 + 24 4 - 4 ± 2 10 1 = = -1 ± 10 4 2 1 t = -1 + 10 2 1 sin q + cos q = -1 + 10 2 t=

Now, (1 – sin q)(1 – cos q) = 1 – sin q – cos q + sin q cos q = 1 – (sin q + cos q) + sin q cos q Ê 10 ˆ 1 Ê 10 ˆ = 1 - Á –1 + ˜¯ + Á - 10 ˜ Ë Ë ¯ 2 2 4 Ê = Á2 + Ë

5ˆ ˜ - 10 4¯

Ê 13 ˆ = Á - 10 ˜ Ë4 ¯ 9x 2 sin 2 x + 4 x sin x 4 = 9x sin x + x sin x Applying AM ≥ GM we get, 4 ˆ Ê 9x sin x + Á 4 x sin x ˜ ÁË ˜¯ ≥ 9x sin x ¥ x sin x 2

26. Given, f (x) =

fi

4 Ê 9x sin x + Á x sin ÁË 2

ˆ x˜ ˜¯ ≥ 6

fi

Ê 4 ˆ ÁË 9x sin x + x sin x ˜¯ ≥ 12

Hence, the minimum value of f (x) is 12. 27. We have, cos q + sin q = 2 cos q fi fi fi

sin q = ( 2 - 1) cos q sin q cos q = ( 2 - 1) cos q = ( 2 + 1) sin q

fi cos q – sin q = 2 sin q 28. We have, tan2 q = 1 – e2 fi 1 + tan2 q = 1 + 1 – e2 = 2 – e2 fi sec2 q = 2 – e2 fi sec q = 2 - e 2 Now, sec q + tan3 q ◊ cosec q = sec q +

sin 3q 1 ◊ 3 cos q sin q

sin 2q cos3q sin 2q 1 = sec q + ◊ cos 2q cos q = sec q +

= sec q + tan2 q ◊ sec q = sec q (1 + tan2 q) = sec3 q = (2 – e2)3/2

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Trigonometry Booster

29. Given, sin q + sin2 q + sin3 q = 1 fi (sin q + sin3 q) = 1 – sin2 q = cos2 q fi (sin q + sin3 q)2 = (cos2 q)2 fi (sin q + sin3 q)2 = (cos2 q)2 fi (1 – cos2 q)(2 – cos2 q)2 = cos4 q fi (1 – cos2 q)(4 – 4 cos2 q + cos4 q) = cos4 q fi 4 – 4 cos2 q + cos4 q – 4 cos2 q + 4 cos4 q – cos6 q = cos4 q fi cos6 q – 4 cos4 q + 8 cos2 q = 4 2 sin q 30. Given, x = 1 + cos q + sin q

2

2 sin q ((1 + sin q ) - cos q )

fi

81sin q = 3

(1 + sin 2 q + 2 sin q - cos 2 q ) 2 sin q ((1 + sin q ) - cos q )

fi fi

34 sin q = 3 4 sin2 q = 1

(sin 2 q + 2 sin q + (1 - cos 2 q ))

fi

Ê 1ˆ sin 2q = Á ˜ Ë 2¯

2 sin q (1 + sin q ) + cos q

=

2 sin q ((1 + sin q ) - cos q ) ((1 + sin q ) + cos q )((1 + sin q ) - cos q )

= = =

2 sin q ((1 + sin q ) - cos q ) ((1 + sin q ) 2 - cos 2 q )

2 sin q ((1 + sin q ) - cos q ) (2 sin q + 2 sin 2 q )

=

2 sin q ((1 + sin q ) - cos q ) 2 sin q (1 + sin q )

=

((1 + sin q ) - cos q ) (1 + sin q )

=

(1 - cos q + sin q ) (1 + sin q )

31. We have 3(sin x – cos x)4 + 6(sin x + cos x)2 + 4(sin6 x + cos6 x) = 3(sin4 x – 4 sin3 x cos x + 6 sin2 x cos2 x – 4 sin x cos3 x + cos4 x) + 6(sin2 x + cos2 x + 2 sin x cos x) + 4{(sin2 x)3 + (cos2 x)3} = 3(sin4 x + cos4 x – 4 sin x cos x (sin2 x + cos2 x) + 6 sin2 x cos2 x) + 6(1 + 2 sin x cos x) + 4(sin2 x + cos2 x) 2 – 12 sin2 x cos2 x = 3 – 6 sin2 x cos2 x – 12 sin x cos x + 18 sin2 x cos2 x + 6 + 12 sin x cos x + 4 – 12 sin2 x cos2 x =3+6+4 =13

TR_01.indd 34

2

2

33. We have 81sin q + 81cos q = 30 fi

=

=

32. We have sin x + sin2 x = 1 fi sin x = 1 – sin2 x = cos2 x Now, cos8 x + 2 cos6 x + cos4 x = (cos4 x)2 + 2 ◊ cos4 x ◊ cos2 x + (cos2 x)2 = (cos4 x + cos2 x)2 = (sin2 x + sin x)2 = (1)2 = 1

2

81sin q + 811– sin q = 30 2 81 fi = 30 81sin q + 2 81sin q 2 81 fi a + = 30, a = 81sin q a fi a2 – 30a + 81 = 0 fi (a – 27)(a – 3) = 0 fi a = 3, 27 When a = 3 2

2

2

Êpˆ sin 2q = sin 2 Á ˜ Ë 6¯ pˆ Ê fi q = Á np ± ˜ Ë 6¯ p 5p fi q= , 6 6 When a = 27 fi

2

fi

81sin q = 27

fi fi

34 sin q = 33 4 sin2 q = 3

2

2

fi fi

Ê 3ˆ sin q = Á ˜ Ë 2 ¯ Êpˆ sin 2q = sin 2 Á ˜ Ë 3¯ 2

pˆ Ê q = Á np ± ˜ Ë 3¯ p 2p fi q= , 3 3 p p 2p 5p Hence, the values of q are , , , . 6 3 3 6 34. We have f6(q) = sin6 q + cos6 q = 1 – 3 sin2 q cos2 q Also f4(q) = sin4 q + cos4 q = 1 – 2 sin2 q cos2 q fi

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The Ratios and Identities

Now,

1 1 f 6 (q ) - f 4 (q ) 6 4

1 1 = (1 - 3 sin 2q cos 2q ) - (1 - 2 sin 2q cos 2q ) 6 4 1 1 1 1 = - sin 2q cos 2q - + sin 2q cos 2q 6 2 4 2 1 1 = 6 4 1 =12 35. Given, x sin a = y cos a x y = =k fi cos a sin a Also, x sin a + y cos a = sin a cos a fi k sin3 a cos a + k sin a cos3 a = sin a cos a fi k sin a cos a (sin2 a + cos2 a) = sin a cos a fi k sin a cos a = sin a cos a fi k=1 Thus, x = k cos a = cos a, y = k sin a = sin a Now, x2 + y2 = cos2 a + sin2 a = 1 36. Given, tan q + sin q = m, tan q – sin q = n, fi mn = tan2 q – sin2 q 3

3

Ê 1 - cos 2q ˆ sin 4q = sin 2q Á = Ë cos 2q ˜¯ cos 2q Now, m2 – n2 = (tan q + sin q)2 – (tan q – sin q)2 = 4 tan q sin q =4

sin 2q cos q

= 4 mn 37. Given,

Ê cos 2 x sin 2 x ˆ (cos 2 x – cos 2 y ) Á ˜ =0 Ë cos 2 y sin 2 y ¯

fi

Ê cos 2 x sin 2 x ˆ (cos 2 x – cos 2 y ) = 0, Á ˜ =0 Ë cos 2 y sin 2 y ¯

fi

cos2 x = cos2 y, sin2 x = sin2 y

Now,

cos 4 y sin 4 y + cos 2 x sin 2 x =

cos 4 y sin 4 y + cos 2 y sin 2 y

= cos2 y + sin2 y =1 38. We have 2f6(q) – 3f4(q) + 1 = 2(sin6 q + cos6 q) – 3(sin4 q + cos4 q) + 1 = 2(1 – 3 sin2 q cos2 q) –(1–2 sin2 q cos2 q) + 1 =2–3+1 =0 39. We have fi fi

sin A cos A = p, =q sin B cos B

sin A sin B p / = cos A cos B q tan A p = tan B q

fi

tan A tan B = =l p q

Also,

sin A cos A = p, =q sin B cos B

fi

sin A cos A = pq sin B cos B

fi

2 sin A cos A = pq 2 sin B cos B

fi

cos 4 x sin 4 x + = cos 2 x + sin 2 x 2 2 cos y sin y

fi

fi

cos 4 x sin 4 x 2 2 cos sin – x = x cos 2 y sin 2 y

sin 2A = pq sin 2B

fi

fi

Ê cos 2 x ˆ Ê sin 2 x ˆ cos 2 x Á - 1˜ = sin 2 x Á1 – ˜ 2 sin 2 y ¯ Ë cos y ¯ Ë

2 tan A 2 tan B / = pq 1 + tan 2 A 1 + tan 2 B

fi

fi

cos 2 x (cos 2 x - cos 2 y ) 2 cos y

2pl 2ql / = pq 2 2 1 + p l 1 + q2l 2

fi

p (1 + q 2 l 2 ) ¥ = pq 2 2 q (1 + p l )

fi

(1 + q 2 l 2 ) = q2 (1 + p 2 l 2 )

fi

(1 + q2l2) = q2(1 + p2l2)

= =

TR_01.indd 35

cos 4 x sin 4 x + =1 cos 2 y sin 2 y

fi

sin 2 x (sin 2 y – sin 2 x) 2 sin y sin 2 x (cos 2 x – cos 2 y ) 2 sin y

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Trigonometry Booster

fi

l2(1 – p2) q2 = q2 – 1

fi

l2 =

fi

l=±

=

(q 2 - 1) (1 - p 2 )q 2

=

1 (q 2 - 1) q (1 - p 2 )

=

p (q 2 - 1) Therefore, tan A = ± q (1 - p 2 ) and tan B = ±

40. We have

(q 2 - 1) (1 - p 2 )

= =

sin 4 a cos 4 a 1 + = a b a+b

fi

Ê a + bˆ Ê a + bˆ 4 cos 4a = 1 ÁË ˜¯ sin a + Á Ë b ˜¯ a

fi

Ê ÁË1 +

fi

a Êb 4 4 ˆ 4 4 ÁË sin a + cos a ˜¯ + (sin a + cos a ) = 1 a b

fi

a Êb 4 4 ˆ 2 2 ÁË sin a + cos a ˜¯ + (1 – 2 sin a ◊ cos a ) = 1 a b

fi

a Êb 4 4 2 2 ˆ ÁË sin a + cos a – 2 sin a ◊ cos a ˜¯ = 0 a b

fi

Ê b Ê a 2 ˆ 2 ˆ Á a sin a˜ + Á b cos a ˜ Ë ¯ Ë ¯

bˆ Ê 4 ˜ sin a + ÁË1 + a¯

aˆ 4 ˜ cos a = 1 b¯

2

2

b a sin 2a ◊ cos 2a = 0 a b

-2

2

fi

Ê b sin 2a – ÁË a

ˆ a cos 2a ˜ = 0 ¯ b

fi

Ê b 2 Á a sin a – Ë

ˆ a cos 2a ˜ = 0 b ¯

fi

b a sin 2a = cos 2a a b

fi

sin 2a cos 2a 1 = = a b a+b a b , cos 2a = a+b a+b

fi

sin 2a =

Now,

sin a cos a + a3 b3 8

TR_01.indd 36

=

(sin 2a ) 4 (cos 2a ) 4 + a3 b3 Ê a ˆ ÁË a + b ˜¯

4

Ê b ˆ ÁË a + b ˜¯

4

+ a3 b3 a4 b4 + a 3 ( a + b) 4 b3 ( a + b) 4 a b + ( a + b) 4 ( a + b) 4 a+b ( a + b) 4 1 ( a + b )3

41. (i) sin (120°) = sin (90 ¥ 1 + 30°) = cos (30°) =

3 2

(ii) sin (150°) = sin (90 ¥ 2 – 30°) = sin (30°) =

1 2

(iii) sin (210°) = sin (90 ¥ 2 + 30°) = - sin (30°) = -

1 2

(iv) sin (225°) = sin (90 ¥ 2 + 45°) = - sin (45°) = -

1 2

(v) sin (300°) = sin (90 ¥ 3+ 30°) = - cos (30°) = -

3 2

(vi) sin (330°) = sin (90 ¥ 3 + 60°) = - cos (60°) = -

1 2

(vii) sin (405°) = sin (90 ¥ 4 + 45°) = sin (45°) =

1 2

(viii) sin (660°) = sin (90 ¥ 7 + 30°) = sin (30°) =

1 2

(ix) sin (1500°) = sin (90 ¥ 16 + 60°)

8

= sin (60°) =

3 2

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42.

43.

44.

45.

46.

TR_01.indd 37

(x) sin (2013°) = sin (90 ¥ 22 + 33°) = –sin (33°) We have cos (1°) cos (2°) cos (3°) … cos (189°). = cos (1°) cos (2°) cos (3°) … cos (89°) cos (90°) cos (91°) … cos (189°) = cos (1°) cos (°) cos (3°) … cos (89°) ¥ 0 ¥ cos (91°) … cos (189°) =0 We have tan (1°) tan (2°) tan (3°) … (89°) = tan (1°) tan (2°) tan (3°) … tan (44°) tan (45°) tan (46°) … tan (87°) tan (88°) tan (89°) = {tan (1°) ¥ tan (89°)} ◊ {tan (2°) ¥ tan (88°) … {tan (44°) ¥ tan (46°)}, tan (45°) =1 We have tan 35° ◊ tan 40° ◊ tan 45° ◊ tan 50° ◊ tan 55° = {tan 35° ¥ tan 55°} {tan 40° ¥ tan 50°} ◊ tan 45° = {tan 35° ¥ cot 35°} ◊ {tan 40° ¥ cot 40°} ¥ tan 45° =1 We have sin (10°) + sin (20°) + sin (30°) + sin (40°) + … + sin (360°) = sin (10°) + sin (20°) + sin (30°) + sin (40°) + … + sin (150°) + sin (340°) + sin (350°) + sin (360°) = sin (10°) + sin (20°) + sin (30°) + sin (40°) + … + sin (80°) + sin (90°) + sin (100°) + sin (360° – 40°) + sin (360° – 30°) + sin (360° – 20) + sin (360° – 10°) + sin (360°) = sin (10°) + sin (20°) + sin (30°) + sin (40°) + … + sin (80°) + sin (90°) + sin (100°) –sin (40°) – sin (30°) –sin (20°) – sin (10°) + sin (180°) =0 We have cos (10°) + cos (20°) + cos (30°) + cos (40°) + … + cos (360°) = cos 20° + cos 30° + cos 40°+ … + cos 140° + cos 150° + cos 160° + cos 170° + cos 180° + (cos 190° + cos 200° + cos 210° + cos 220° + … + cos 360°) = cos 10° + cos 20° + cos 30° + cos 40° +… – cos 40° – cos 50° – cos 60° – cos70° + cos 180° + (cos 190°

+ cos 200° + cos 210° + cos 220° + … + cos 360°) =cos 180° + cos 360° = –1 + 1 =0 47. We have sin2 5° + sin2 10° + sin2 15° + … + sin2 90° = sin2 5° + sin2 10° + sin2 15° + … + sin2 40 + sin2 45 + sin2 50 + sin2 80 + sin2 85 + sin2 90° = (sin2 5° + sin2 85°) + (sin2 10° + sin2 80°) + (sin2 15° + sin2 75°) + … + (sin2 40° + sin2 50°) + (sin2 45° + sin2 90°) = (sin2 5° + cos2 5°) + (sin2 10° + cos2 10°) + (sin2 15° + cos2 15°) + … … + (sin2 40° + cos2 40°) + (sin2 45° + sin2 90°) Ê1 ˆ = (1 + 1 + … 8 times) + Á + 1˜ Ë2 ¯ 1ˆ Ê = Á8 + 1 + ˜ Ë 2¯ =9

1 2

Êpˆ Êpˆ 48. We have sin 2 Á ˜ + sin 2 Á ˜ Ë 18 ¯ Ë 9¯ Ê 4p ˆ Ê 7p ˆ + sin 2 Á ˜ + sin 2 Á ˜ Ë 9 ¯ Ë 18 ¯ Êpˆ Ê 2p ˆ = sin 2 Á ˜ + sin 2 Á ˜ Ë 18 ¯ Ë 18 ¯ Ê 8p ˆ Ê 7p ˆ + sin 2 Á ˜ + sin 2 Á ˜ Ë 18 ¯ Ë 18 ¯ Êpˆ Ê p 7p ˆ = sin 2 Á ˜ + sin 2 Á Ë 18 ¯ Ë 2 18 ˜¯ Êp p ˆ Ê 7p ˆ + sin 2 Á - ˜ + sin 2 Á ˜ Ë 2 18 ¯ Ë 18 ¯ Êpˆ Ê 7p ˆ = sin 2 Á ˜ + cos 2 Á ˜ Ë 18 ¯ Ë 18 ¯ Ê 7p ˆ Êpˆ + cos 2 Á ˜ + sin 2 Á ˜ Ë 18 ¯ Ë 18 ¯ Ï Êpˆ Ê p ˆ¸ = Ìsin 2 Á ˜ + cos 2 Á ˜ ˝ Ë ¯ Ë 18 ¯ ˛ 18 Ó Ï Ê 7p ˆ Ê 7p ˆ ¸ + Ìcos 2 Á ˜ + sin 2 Á ˜ ˝ Ë ¯ Ë 18 ¯ ˛ 18 Ó =1+1 =2

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Trigonometry Booster

49. We have tan (20°) tan (25°) tan (45°) tan (65°) tan (70°) = tan (20°) tan (25°) tan (45°) tan (90° – 25°) tan (90° – 20°) = tan (20°) tan (25°) tan (45°) cot (25°) cot (20°) = tan (45°) =1 50. Given, sin (q1) + sin (q2) + sin (q3) = 3 It is possible only when each term of the above equation will provide the maximum value Thus, sin (q1) = 1, sin (q2) = 1, sin (q3) = 1 p p p , q 2 = , q3 = 2 2 2 Hence, the value of cos (q1) + cos (q2) + cos (q3) Êpˆ Êpˆ Êpˆ = cos Á ˜ + cos Á ˜ + cos Á ˜ Ë 2¯ Ë 2¯ Ë 2¯ =0 We have sin2 6° + sin2 12° + … + sin2 90° = (sin2 6° + sin2 84°) + (sin2 12° + sin2 78°) + … + (sin2 42° + sin2 48°) + sin2 90° =7¥1+1 =8 We have sin2 10° + sin2 20° + … + sin2 90° = (sin2 10° + sin2 80°) + (sin2 20° + sin2 70°) + … + (sin2 40° + sin2 50°) + sin2 90° =4¥1+1 =5 We have sin2 9° + sin2 18° + … + sin2 90° = (sin2 9° + sin2 81°) + (sin2 18° + sin2 72°) + … + (sin2 36° + sin2 54°) + sin2 45° + sin2 90° 1 = 4 ¥1+ +1 2 1 =5 2 We have tan 1° ◊ tan 2° ◊ tan 3° … tan 89° = (tan 1° ◊ tan 89°) (tan 2° ◊ tan 88°) (tan 3° ◊ tan 87°) … (tan 44° ◊ tan 46°) tan 45° = 1.1.1........1 =1 We have cos 1° ◊ cos 2° ◊ cos 3° … cos 189° = cos 1° ◊ cos 2° ◊ cos 3° … cos 89° cos 90° … cos 180° =0 Given equation is 2 sin2 q + 2 cos q = 0 fi 2 – 2 cos2 q + 3 cos q = 0 fi 2 cos2 q – 3 cos q – 2 = 0 fi 2 cos2 q – 4 cos q + cos q – 2 = 0 fi

51.

52.

53.

54.

55.

56.

TR_01.indd 38

q1 =

fi fi fi

2 cos q (cos q – 2) + 1(cos q – 2) = 0 (2 cos q + 1) (cos q – 2) = 0 1 cos q = - , 2 2

fi q = 120°, 240° 57. Given equation is cos q + 3 sin q = 2 fi fi

1 3 cos q + sin q = 1 2 2 pˆ Ê cos Á q – ˜ = 1 Ë 3¯

pˆ Ê ÁË q – ˜¯ = 0 3 p fi q= 3 58. Given, 4na = p p 2na = 2 Now, tan a tan a ◊ tan ◊ tan 3a tan 3 tan (2n – 1)a = tan a tan a (2n – 1) a) (tan 2a tan 2 (2n – 2) a) ◊ (tan 3a ◊ tan (2n – 3) a) … = (tan a ◊ tan (2na – a)) (tan 2a ◊ tan (2na – 2a)) (tan 3a ◊ tan (2na – 3a)) … fi

Ê Êp ˆˆ Ê Êp ˆˆ = Á tan a ◊ tan Á - a ˜ ˜ ◊ Á tan 2a ◊ tan Á - 2a ˜ ˜ Ë2 ¯¯ Ë Ë2 ¯¯ Ë Ê Êp ˆˆ ◊ Á tan 3a ◊ tan Á - 3a ˜ ˜ ... Ë2 ¯¯ Ë = (tan a × cot a) ◊ (tan 2a ◊ cot 2a) ◊ (tan 3a ◊ cot 3a) =1 59. We have cos (18°) + cos (234°) + cos (162°) + cos (306°) = = cos (18°) – cos (54°) – cos (18°) + cos (54°) =0 60. We have cos (20°) + cos (40°) + cos (60°) + … + cos (180°) = cos (20°) + cos (160°) + cos (40°) + cos (140°) + cos (60°) + cos (120°) + cos (80°) + cos (100°) + cos (180°) = cos (180°) = –1 61. We have sin (20°) + sin (40°) + sin (60°) + … + sin (360°) = sin (20°) + sin (340°) + sin (40°) + sin (320°) + … + sin (180°) + sin (360°) =0 63. 1 (i) Given equation is sin x = 2

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As we know the period of sin x is 2p So, there is two solutions in [0, 2p] 3 (ii) Given equation is cos x = 2 As we know the period of cos x is 2p For each 2p. there is 2 solutions So, it has 3 solutions (iii) Given equation is 4 sin2 x – 1 = 0 1 fi sin 2 x = 4 The period of sin2 x is p. For each p, there is two solutions. So, it has 6 solutions. (iv) Given equation is sin2 x – 3 sin x + 2 = 0 fi (sin x – 1)(sin x – 2) = 0 fi sin x = 1, 2 fi sin x = 1 For each 2p, there is 2 solutions. So, it has 3 solutions. (v) Given equation is cos2 x – cos x – 2 = 0 fi cos2 x – 2 cos x + cos x – 2 = 0 fi cos x(cos x – 2) + 1(cos x – 2) = 0 fi (cos x + 1)(cos x – 2) = 0 fi cos x = –1, 2 fi cos x = –1 For each 2p, there are 2 solutions. So, it has 3 solutions. 64. We have, (i) sin (15°) = sin (45° – 30°) = sin (45°) cos (30°) – cos (45°) sin (30°) = =

1 3 1 1 ◊ - ◊ 2 2 2 2 3 -1 2 2

(ii) cos (15°) = cos (45° – 30°) = cos (45°) cos (30°) + sin (45°) sin (30°) =

1 3 1 1 ◊ + ◊ 2 2 2 2

3 +1 2 2 (iii) tan (15°) = tan (45° – 30°) tan 45∞ - tan 30∞ = 1 + tan 45∞ ◊ tan 30∞ 1 13 = 1 1+ 3 =

TR_01.indd 39

= =

3 -1 3 +1 ( 3 - 1) 2 3 -1

3+1- 2 3 2 4-2 3 = 2 =2- 3 =

Note: (i) cot (15°) =

1 1 = =2+ 3 tan (15°) 2 - 3 (ii) tan (105°) = - cot (15∞) = - (2 + 3) (iii) cot (105°) = - tan (15∞) = - (2 - 3) = 3-2

65. We have tan (75°) + cot (75°) = cot (15°) + tan (15°) = (2 + 3) + (2 - 3) =4 66. We have cos (9°) + sin (9°) 1 Ê 1 ˆ cos (9°) + sin (9°)˜ = 2Á Ë 2 ¯ 2 = 2 (sin (45∞) cos (9°) + cos (45∞) sin (9°)) = 2 (sin (45∞ + 9∞)) = 2 sin (54∞) 67. We have tan (70°) = tan (50° + 20°) tan 50∞ + tan 20∞ tan (70∞) = fi 1 - tan 50∞ tan 20∞ fi

tan (70°) – tan (70°) ◊ tan (50°) ◊ tan (20°) = tan 50° + tan 20° fi tan (70°) – tan (70°) tan (50°) ◊ cot (70°) = tan 50° + tan 20° fi tan (70°) – tan (50°) = tan 50° + tan 20° fi tan (70°) = 2 tan 50° + tan 20° cos 20° - sin 20° 68. We have cos 20° + sin 20° 1 - tan 20° = 1 + tan 20° tan 45° - tan 20° = 1 + tan 45° tan 20° = tan (45° – 20°) = tan (25°) cos 7∞ + sin 7∞ 1 + tan (7°) 69. We have = cos 7∞ - sin 7∞ 1 – tan (7°) = tan (45° – 7°) = tan (52°)

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Trigonometry Booster

70. We have tan (45°) fi tan (25° + 20°) = 1 tan 25° + tan 20° fi =1 1 - tan 25∞ tan 20∞ fi tan 25° + tan 20° = 1 – tan 25° tan 20 fi tan 25° + tan 20° + tan 25° tan 20° = 1 71. We have A + B = 45° fi tan (A + B) = tan (45°) fi tan (A + B) = 1 tan A + tan B fi =1 1 - tan A ◊ tan B fi tan A + tan B = 1 = tan A ◊ tan B fi tan A + tan B + tan A ◊ tan B = 1 fi 1 + tan A + tan B + tan A ◊ tan B = 1 + 1 = 2 fi (1 + tan A) + tan B (1 + tan A = 2) fi (1 + tan A) (1 + tan B = 2 72. We have (1 + tan 245°) (1 + tan 250°) (1 + tan 260°) (1 – tan 200°) (1 – tan 205°) (1 – tan 215°) = {(1 + tan 245°) (1 + tan (–200°))} {(1 + tan 250°) (1 + tan (–205°))} {(1 + tan 260°) (1 + tan (–215°))} =2¥2¥2 =8 73. Now, tan 13 A = tan (9A + 4A) tan 9A + tan 4A = 1 - tan 9A tan 4A tan 13A – tan 4A tan 9A tran 13A = tan 9A + tan 4A tan 13A – tan 9A – tan 4 A = tan 4A ◊ tan 9A ◊ tan 13A 74. Do yourself. 75. We have tan (a + b) tan a + tan b = 1 - tan a ◊ tan b m 1 + m + 1 2m + 1 = m 1 ◊ 1m + 1 2m + 1 2m 2 + m + m + 1 (m + 1)(2m + 1) = 2m 2 + 3m + 1 - m (m + 1)(2m + 1) =

2m 2 + 2m + 1 =1 2m 2 + 2m + 1

tan (a + b) = 1 p (a + b ) = 4

TR_01.indd 40

76. We have sin2 A + sin2 (A – B) – 2sin A cos B sin (A – B) = sin2 A + sin (A – B) (sin (A – B) – 2 sin A cos B) = sin2 A – sin (A – B) sin (A + B) = sin2 A – (sin2 A – sin2 B) = sin2 B 77. We have cos 2x cos 2y + cos2 (x + y) – cos2 (x – y) 1 = [2 cos 2x cos 2y + 2 cos 2 ( x + y ) - 2 cos 2 ( x - y )] 2 1 [cos (2x + 2y ) + cos (2x - 2y ) + 1 2 + cos (2x + 2y ) - 1 - cos (2x - 2y )] 1 = [cos (2x + 2y ) + cos (2x + 2y )] 2 1 = [2 cos (2x + 2y )] 2 = cos (2x + 2y) x- y 78. Given, sin q = x+ y Applying componendo and dividendo, we get, sin q + 1 x - y + x + y = sin q - 1 x - y - x - y sin q + 1 x fi =sin q - 1 y fi

1 + sin q x = 1 – sin q y

fi

Ê Êqˆ Êqˆ ˆ Á cos ËÁ 2 ¯˜ + sin ËÁ 2 ¯˜ ˜ x Á ˜ = q q y Ê ˆ Ê ˆ Á cos - sin Á ˜ ˜˜ ËÁ 2 ¯˜ Ë 2¯ ¯ ËÁ

fi

Ê Êqˆ ˆ Á 1 + tan ÁË 2 ˜¯ ˜ x Á ˜ = q y Ê ˆ Á 1 - tan ˜ ÁË ˜¯ ˜¯ ÁË 2

fi

Ê Êqˆ ˆ Á 1 + tan ÁË 2 ˜¯ ˜ x Á ˜ =± q y Ê ˆ Á 1 - tan ˜ ÁË ˜¯ ˜¯ ÁË 2

fi

x Êp qˆ tan Á + ˜ = ± Ë 4 2¯ y

2

2

79. If tan a =

Q sin b , P + Q cos b

prove that tan (b - a ) =

P sin a Q + P cos a

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The Ratios and Identities

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The Ratios and Identities

80. We have 3 cos (a – b ) + cos (b - g ) + cos (g - a ) = 2 fi

2(cos (a – b) + cos (b – g) + cos (g – a)) + 3 = 0

fi

2(cos a cos b + cos b cos g + cos g cos a) + 2(sin a sin b + sin b sin g + sin g sin a) + 3 = 0

fi

(cos2 a + sin2 a) + (cos2 b + sin2 b) + (cos2 g + sin2 g) + 2(cos a cos b + cos b cos g + cos g cos a) + 2(sin a sin b + sin b sin g + sin g sin a) = 0

fi

cos a + cos b + cos g 2

2

2

+ 2 cos a cos b + 2 cos b cos g + 2 cos g cos a + sin2 a + sin2 b + sin2 g + 2 sin g sin a = 0

Êa + bˆ 2 tan Á Ë 2 ˜¯ = Êa + bˆ 1 + tan 2 Á Ë 2 ˜¯ 2(b /a ) 2ab = 1 + (b /a ) 2 a 2 + b 2

Ê 1 - tan 2 (q /2) ˆ Ê 2 tan (q /2) ˆ aÁ =c ˜ + bÁ 2 Ë 1 + tan 2 (q /2) ˜¯ Ë 1 + tan (q /2) ¯

(cos a + cos b + cos g) = 0

a(1 – tan2 (q/2)) + 2b tan (q/2) = c(1 + tan2 (q/2))

and (sin a + sin b + sin g) = 0

(a + c) tan2 (q/2) – 2b tan (q/2) + (c – a) = 0

81. We have tan (a + q) = n tan (a – q) fi

tan (a + q ) n = tan (a – q ) 1

Êaˆ Ê bˆ Let its roots be tan Á ˜ , tan Á ˜ Ë 2¯ Ë 2¯

fi

tan (a + q ) + tan (a – q ) n + 1 = tan (a + q ) - tan (a – q ) n – 1

2b Êaˆ Ê bˆ Thus, tan Á ˜ + tan Á ˜ = Ë 2¯ Ë 2¯ a + c

fi

sin (a + q + a – q ) n + 1 = sin (a + q - a + q ) n – 1

Êaˆ Ê bˆ c – a tan Á ˜ tan Á ˜ = Ë 2¯ Ë 2¯ a + c

fi

sin (2a ) n + 1 = sin (2q ) n – 1

fi

sin (2q ) n - 1 = sin (2a ) n + 1

tan (a /2) + tan (b /2) Êa + bˆ = tan Á ˜ Ë 2 ¯ 1 - tan (a /2) tan ( b /2) 2b 2b b a+c = = = c - a 2a a 1c+a

Hence, the result. 82. Given sin a + sin b = a and cos a + cos b = b (i) divides by (ii), we get, sin a + sin b a = cos a + cos b b

TR_01a.indd 41

a2 2 2 b2 = b - a = 2 2 a a + b2 1+ 2 b (ii) sin (a + b)

83. Given equation is a cos q + b sin q = c

(cos a + cos b + cos g)2 + (sin a + sin b + sin g)2 = 0

fi

+ bˆ ˜ 2 ¯ + bˆ ˜ 2 ¯

1-

=

+ 2 sin a sin b + 2 sin b sin g fi

(i) Now, cos (a + b) Êa 1 - tan 2 Á Ë = Êa 1 + tan 2 Á Ë

fi

Êa + bˆ Êa - bˆ 2 sin Á cos Á ˜ Ë 2 ¯ Ë 2 ˜¯ a = Êa + bˆ Êa – bˆ b 2 cos Á cos Á Ë 2 ˜¯ Ë 2 ˜¯

fi

Êa + bˆ a tan Á = Ë 2 ˜¯ b

…(i) …(ii)

(i) Now, cos (a + b) Êa 1 - tan 2 Á Ë = Êa 1 + tan 2 Á Ë

+ bˆ ˜ 2 ¯ + bˆ ˜ 2 ¯

b2 2 2 a2 = a - b = b2 a 2 + b2 1+ 2 a 1-

Êa - bˆ (ii) Now, tan Á Ë 2 ˜¯

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Trigonometry Booster

85. Given, tan (p cos q) = cot (p sin q)

Êaˆ Ê bˆ tan Á ˜ - tan Á ˜ Ë 2¯ Ë 2¯ = Êaˆ Ê bˆ 1 + tan Á ˜ ◊ tan Á ˜ Ë 2¯ Ë 2¯

fi fi 2

Ê Êaˆ Ê b ˆˆ Êaˆ Ê bˆ ÁË tan ÁË 2 ˜¯ + tan ÁË 2 ˜¯ ˜¯ - 4 tan ÁË 2 ˜¯ tan ÁË 2 ˜¯ = Êaˆ Êbˆ 1 + tan Á ˜ ◊ tan Á ˜ Ë 2¯ Ë 2¯

fi fi

2

Ê 2b ˆ Ê c - aˆ ÁË a + c ˜¯ - 4 ÁË c + a ˜¯ = c-a 1+ c+a =

fi

fi

a 2 + b2 - c2 c Now, =

=

=

- bˆ ˜ 2 ¯ + bˆ ˜ 2 ¯

Ê a 2 + b2 - c2 ˆ 1- Á ˜¯ Ë c2

1 Êp ˆ cos Á - q ˜ = ± Ë4 ¯ 2 2 x sin j 1 - x cos j sin q x sin j = cos q 1 - x cos j

fi fi

sin q – x sin q cos j = x cos q sin j x sin (q + j) = sin q

fi

x=

sin q sin (q + j )

Similarly, y =

sin j sin (q + j )

Dividing the above relations, we get, x sin q = y sin j

Êa +b -c 1+ Á ˜¯ Ë c2

fi

2c - (a + b )

Hence, the result.

2

2

2ˆ

2

2

2

a +b 84. Given, a tan q + b sec q = c fi (a tan q – c)2 = (–b sec q)2 fi a2 tan2 q – 2ac tan q + c2 = b2 sec2 q fi a2 tan2 q – 2ac tan q + c2 = b2 + b2 tan2 q fi (a2 – b2) tan2 q – 2ac tan q + (c2 + b2) = 0 Let its roots be tan a, tan b 2ac So, tan a + tan b = 2 a - b2 c2 - b2 and tan a ◊ tan b = 2 a - b2 Now, tan (a + b) tan a + tan b = 1 - tan a tan b 2ac 2 a - b2 = c2 - b2 1- 2 a - b2 2ac 2ac = 2 = a - b2 - c2 + b2 a 2 - c2

TR_01a.indd 42

1 1 1 cos q + sin q = ± 2 2 2 2

86. We have tan q =

4b 2 - 4(c 2 - a 2 ) 2c

Êa 1 - tan 2 Á Ë cos (a – b) = Êa 1 + tan 2 Á Ë

Ê p ˆ tan (p cos q ) = tan Á ± - p sin q ˜ Ë 2 ¯ p Ê ˆ (p cos q ) = Á ± - p sin q ˜ Ë 2 ¯ 1 cos q + sin q = ± 2

2

2

x sin j = y sin q

87. We have tan (a + b) = c fi fi fi

tan a + tan b =c 1 - tan a tan b a =c 1 - tan a tan b tan a tan b =

a a-c -1= c c

…(i)

Now, cot a + cot b = b fi

1 1 + =b tan a tan b

fi

tan a + tan b =b tan a ◊ tan b

fi

tan a ◊ tan b =

a b

From (i) and (ii), we get, fi

ac + bc = ab

…(ii) a-c a = c b

Which is the required relation.

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88. We have, tan a - tan b tan (a - b ) = 1 + tan a ◊ tan b sin a Ê n sin a cos a ˆ cos a ÁË 1 - n sin 2a ˜¯ = sin a Ê n sin a cos a ˆ ¥ 1+ cos a ÁË 1 – n sin 2a ˜¯ sin a (1 - n sin 2a ) - n sin a cos 2a = cos a (1 – n sin 2a ) + n sin 2a ◊ cos a sin a - n sin a (sin 2a + cos 2a ) cos a - n sin 2a ◊ cos a + n sin 2a ◊ cos a sin a - n sin a = cos a (1 – n) sin a = cos a = (1 – n) tan a 89. Let A= x + y – z, B = y + z – x, C = x + y – z Then, A + B + C = (x + y – z) + (y + z – x) + (x + y – z) = (x + y + z) = 0 fi A +B = – C fi cot (A + B) = cot (–C) cot A cot B - 1 fi = cot (- C ) cot A + cot B cot A cot B - 1 fi = - cot C cot A + cot B fi cot A cot B – 1 = – cot A cot C – cot B cot C fi cot A cot B + cot A cot C + cot B cot C = 1 fi cot (x + y – z) cot (y + z – x) + cot (y + z – x) cot (zx + x – y) + cot (z + x – y) cot (x + y – z) = 1 tan a - tan b 90. We have tan (a - b ) = 1 + tan a ◊ tan b 3 tan b - tan b = 2 3 1 + tan b tan b 2 tan b = 2 + 3 tan 2 b =

sin b cos b = sin 2 b 2+3 cos 2 b

TR_01a.indd 43

=

sin b cos b 2 cos 2 b + 3 sin 2 b

=

sin b cos b 2 + sin 2 b

=

2 sin b cos b 4 + 2 sin 2 b

=

sin 2b 4 + 1 - cos 2b

=

sin 2b 5 - cos 2b

91. We have sin 5 A - sin 3 A 2 cos 4A sin A = cos 5 A + cos 3 A 2 cos 4A cos A = tan A 92. We have sin A + sin 3 A 2 sin 2A cos A = cos A + cos 3 A 2 cos 2A cos A = tan 2A 93. We have Ê A + Bˆ Ê A – Bˆ 2 sin Á cos Á Ë 2 ˜¯ Ë 2 ˜¯ sin A + sin B = cos A + cos B Ê A + Bˆ Ê A – Bˆ 2 cos Á cos Á Ë 2 ˜¯ Ë 2 ˜¯ Ê A + Bˆ = tan Á Ë 2 ˜¯ 94. We have sin 38° + sin 22° = 2 sin (30°) cos (8°) 1 = 2 ¥ ¥ cos (90° - 82°) 2 = sin (82°) 95. We have sin 105° + cos 105° = sin (105°) – sin (15°) = 2 cos (60°) sin (45°) 1 = 2 ¥ ¥ sin (45°) 2 = sin (45°) = cos (45°) 96. We have cos 55° + cos 65° + cos 175° = cos (65°) + cos (55°) – cos (5°) = 2 cos (60°) cos (5°) – cos (5°) 1 = 2 ¥ ¥ cos (5°) - cos (5°) 2 = cos (5°) – cos (5°) =0 97. We have cos 20° + cos 100° + cos 140° = cos (100°) + cos (20°) – co(40°) = 2 cos (60°) cos (40°) – cos (40°) 1 = 2 ¥ ¥ cos (40°) - cos (40°) 2 = cos (40°) – cos (40°) = 0.

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Trigonometry Booster

98. We have sin 50° – sin 70 + sin 10° = sin (50°) + sin (10°) – sin (70°) = 2 sin (30°) cos (20°) – sin (70°) 1 = 2 ¥ ¥ cos (20°) - sin (90° - 20°) 2 = cos (20°) – cos (20°) = 0. 99. We have sin (47°) + cos (77°) = cos (17°) = sin (47°) + sin (13°) Ê 47∞ + 13∞ ˆ Ê 47∞ - 13∞ ˆ = 2 sin Á ˜¯ cos ÁË ˜¯ Ë 2 2 = 2 sin (30°) cos (17°) = 2¥

1 ¥ cos (17∞) 2

= cos (17°) 100. We have cos (80°) + cos (40°) – cos (20°) Ê 80° + 40° ˆ Ê 80° - 40° ˆ = 2 cos Á ˜¯ cos ÁË ˜¯ – cos (20°) Ë 2 2

= 2 + (1 + cos (a – b)) Êa - bˆ = 2 ¥ 2 cos 2 Á Ë 2 ˜¯ Êa - bˆ = 4 cos 2 Á Ë 2 ˜¯ 103. We have (cos a – cos b)2 + (sin a – sin b)2 = 2 – 2 cos (a – b) = 2(1 – cos (a – b)) Êa - bˆ = 2 ¥ 2 sin 2 Á Ë 2 ˜¯ Êa - bˆ = 4 sin 2 Á Ë 2 ˜¯ 104. Do yourself. 105. We have cos (20°) cos (40°) cos (80°) = cos (40°) cos (20°) cos (80°) 1 = [4 cos (60° - 20°) cos (20°) cos (60° + 20°)] 4 =

1 ¥ cos (20° ¥ 3) 4

=

1 ¥ cos (60°) 4

=

1 1 1 ¥ = 4 2 8

= 2 cos (60°) cos (20°) – cos (20°) 1 = 2 ¥ ¥ cos (20∞) - cos (20∞) 2 = cos (20°) – cos (20°) = 0. 101. We have sin (10°) + sin (20°) + sin (40°) + sin (50°) – sin (70°) – sin (80°) = {sin (50°) + sin (10°)} + {sin (40°) + sin (20°)} –sin (70°) – sin (80°) Ê 50∞ + 10∞ ˆ Ê 50∞ - 10∞ ˆ = 2 sin Á ˜¯ cos Á ˜¯ Ë Ë 2 2 Ê 40∞ + 20∞ ˆ Ê 40∞ - 20∞ ˆ + 2 sin Á ˜¯ cos Á ˜¯ Ë Ë 2 2 –sin (70°) – sin (80°) = 2 sin (30°) cos (20°) + 2 sin (30°) cos (10°) –sin (70°) – sin (80°) = cos (20°) + cos (10°) – sin (70°) – sin (80°) = cos (20°) + cos (10°) – cos (20°) – cos (10°) =0 102. We have (cos a + cos b)2 + (sin a + sin b)2 = 2 + 2 cos (a – b)

TR_01a.indd 44

106. We have cos 25° cos 35° cos 65° = cos 25° cos 35° cos 65° 1 = [4 cos (60° - 25) cos 25° cos (60° + 25)] 4 1 = ¥ cos (25° ¥ 3) 4 1 = ¥ cos (75°) 4 1 = ¥ sin (15°) 4 =

3 -1 8 2

107. We have sin (20°) sin (40°) sin (80°) = sin (20°) sin (40°) sin (80°) 1 = [4 sin (60° - 20°) sin (20°) sin (60° + 20°)] 4 1 = ¥ sin (60°) 4 =

3 8

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108. We have sin (10°) sin (50°) sin (60°) sin (70°) = = = = =

3 [sin (50°) sin (10°) sin (70°) 2 3 [sin (60° - 10°) sin (10°) sin (60° + 10°)] 2 3 [4 sin (60° - 10°) sin (10°) sin (60° + 10°)] 8 3 ¥ sin (30°) 8 3 1 3 ¥ = 8 2 16

109. We have cos (10°) cos (30°) cos (50°) cos (70°)

112. We have

sin A + sin 3A + sin 5A + sin 7A cos A + cos 3A + cos 5A + cos 7A =

(sin 7A + sin A) + (sin 5A + sin 3 A) (cos 7A + cos A) + (cos 5A + cos 3A)

=

2 sin 4A cos 3A + 2 sin 4A cos A 2 cos 4A cos 3A + 2 cos 4A cos A

=

2 sin 4A (cos 3A + cos A) 2 cos 4A (cos 3A + cos A)

= tan 4 A 113. We have

sin A + sin 2A + sin 4A + sin 5A cos A + cos 2A + cos 4A + cos 5A

=

3 [cos (50°) cos (10°) cos (70°)] 2

=

(sin 5 A + sin A) + (sin 4A + sin 2A) (cos 5 A + cos A) + (cos 4A + cos 2A)

=

3 [cos (60° - 10°) cos (10°) cos (60° + 10°)] 2

=

=

3 [4 cos (60° - 10°) cos (10°) cos (60° + 10°)] 8

2 sin 3A cos 2A + 2 sin 3A cos A 2 cos 3A cos 2A + 2 cos 3A cos A

=

=

3 ¥ cos (30°) 8

sin 3A (cos 2A + cos A) cos 3A (cos 2A + cos A)

=

sin 3A cos 3A

3 3 ¥ 8 2 3 = 16 sin A + sin 3 A + sin 5 A 110. We have cos A + cos 3 A + cos 5 A =

= tan 3 A Hence, the result. 114. Given sin A - sin B =

1 2

…(i)

cos A - cos B =

1 3

…(ii)

=

sin 5A + sin 3A + sin A cos 5A + cos 3A + cos A

and

(sin 5A + sin A) + sin 3A (cos 5A + cos A) + cos 3A

Dividing (i) by (ii) we get,

= =

2 sin 3A cos 2A + sin 3A 2 cos 3A cos 2A + cos 3A

=

sin 3A (2 cos 2A + 1) cos 3A (2 cos 2A + 1)

sin A - sin B 1/2 3 = = cos A - cos B 1/3 2

fi

Ê A + Bˆ Ê A - Bˆ 2 cos Á sin Á ˜ Ë 2 ¯ Ë 2 ˜¯ 3 = Ê A + Bˆ Ê A - Bˆ 2 - 2 sin Á sin Á Ë 2 ˜¯ Ë 2 ˜¯

fi

Ê A + Bˆ cos Á Ë 2 ˜¯ 3 =2 Ê A + Bˆ sin Á ˜ Ë 2 ¯

= tan 3 A 111. We have

cos 4x + cos 3x + cos 2x sin 4x + sin 3x + sin 2x (cos 4x + cos 2x) + cos 3x = (sin 4x + sin 2x) + sin 3x =

2 cos 3x cos x + cos 3x 2 sin 3x cos x + sin 3x

fi

3 Ê A + Bˆ cot Á =Ë 2 ˜¯ 2

=

cos 3x (2 cos x + 1) sin 3x (2 cos x + 1)

fi

2 Ê A + Bˆ tan Á =- . ˜ Ë 2 ¯ 3

= cot 3 x

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1 , 4

115. Given, sin A + sin B = and

cos A + cos B =

=

Ê 3 ˆ 2Á cos (20°) - 2 sin (20°) cos (20°)˜ Ë 2 ¯ = sin (20°) 2(sin (60∞) cos (20°) - sin (40°)) = sin (20°) (2 sin (60∞) cos (20°) - 2 sin (40°)) = sin (20°) (sin (80∞) + sin (40°) - 2 sin (40°)) = sin (20°) (sin (80∞) - sin (40°)) = sin (20°) 2 cos (60∞) sin (20∞) = sin (20°) =1

1 2

Dividing, the above relations, we get, sin A + sin B 1 = cos A + cos B 2 fi

Ê A + Bˆ Ê A – Bˆ 2 sin Á cos Á ˜ Ë 2 ¯ Ë 2 ˜¯ 1 = Ê A + Bˆ Ê A – Bˆ 2 2 cos Á cos Á Ë 2 ˜¯ Ë 2 ˜¯

fi

Ê A + Bˆ 1 tan Á = Ë 2 ˜¯ 2

Hence, the result. 116. Given, cosec A + sec A = cosec B + sec B fi cosec A + cosec B = sec B + sec A fi fi fi

fi

fi

1 1 1 1 = sin A sin B cos B cos A sin B – sin A cos A – cos B = sin A sin B cos A cos B sin A sin B sin B – sin A = cos A cos B cos A – cos B Ê A + Bˆ Ê A – Bˆ 2 cos Á sin Á ˜ Ë 2 ¯ Ë 2 ˜¯ tan A tan B = Ê A + Bˆ Ê A – Bˆ 2 sin Á sin Á Ë 2 ˜¯ Ë 2 ˜¯ Ê A + Bˆ tan A tan B = cot Á Ë 2 ˜¯

Hence, the result. 117. Given, sin 2A = l sin 2B fi fi fi fi fi

sin 2A l = sin 2B 1 sin 2A + sin 2B l + 1 = sin 2A – sin 2B l – 1 2 sin ( A + B) cos ( A – B) l + 1 = 2 cos ( A + B) sin ( A – B) l – 1 l +1 tan ( A + B) cot ( A – B) = l –1 tan ( A + B ) l + 1 = tan ( A – B) l – 1

118. We have 3 cot (20∞) - 4 cos (20∞)

TR_01a.indd 46

3 cos (20°) - 4 sin (20°) cos (20°) sin (20°)

119. Given, sin a + sin b = a and cos a + cos b = b (i) divides by (ii), we get, sin a + sin b a = cos a + cos b b fi

fi

…(i) …(ii)

Êa + bˆ Êa - bˆ 2 sin Á cos Á Ë 2 ˜¯ Ë 2 ˜¯ a = Êa + bˆ Êa - bˆ b 2 cos Á cos Á Ë 2 ˜¯ Ë 2 ˜¯ Êa + bˆ a tan Á = Ë 2 ˜¯ b

Now, Êa 1 - tan 2 Á Ë cos (a + b ) = Êa 1 + tan 2 Á Ë

+ bˆ ˜ 2 ¯ + bˆ ˜ 2 ¯

a2 2 2 b2 = b - a = a 2 a 2 + b2 1+ 2 b 1-

120. We have

1Ê 1ˆ sin A = 1 - Á x + ˜ x¯ 4Ë =i

1Ê Áx + 4Ë

=i

1Ê 1ˆ Á x - ˜¯ x 4Ë

2

2

1ˆ ˜ -1 x¯ 2

iÊ 1ˆ = Áx - ˜ x¯ 2Ë

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The Ratios and Identities

Ê a - b a + b + 2g + Á 2 Êa + bˆ 2 = 2 cos Á ¥ cos Á ˜ Ë Ë 2 ¯ 2

iÊ 1ˆ Similarly, sin B = Á y - ˜ 2Ë y¯ Now, cos (A – B) = cos A cos B + sin A sin B 1Ê 1ˆ Ê 1ˆ 1 Ê 1ˆ Ê 1ˆ = Áx + ˜Áy + ˜ + Áx - ˜Áy - ˜ 4Ë x¯ Ë y¯ 4 Ë x¯ Ë y¯ 1Ê 2ˆ = Á 2xy + ˜ 4Ë xy ¯ 1Ê 1ˆ = Á xy + ˜ 2Ë xy ¯ 121. We have sin (47°) + sin (61°) – sin (11°) – sin (25°) = (sin (61°) + sin (47°)) – (sin (25°) + sin (11°)) = 2 sin (54°) cos (7°) – 2 sin (18°) cos (7°) = 2 cos (7°) [sin (54°) – sin (18°)] = 2 cos (7°) [cos (36°) – sin (18°)] È 5 +1 5 - 1˘ = 2 cos (7°) Í ˙ 4 ˚ Î 4 1 = 2 cos (7°) ¥ 2 = cos (7°) 122. Do yourself. 123. We have

126. We have 1 + cos 2q 2 cos 2q = sin 2q 2 sin q cos q cos q = = cot q sin q Ê cos q sin q ˆ =Á Ë sin q cos q ˜¯ cos 2q – sin 2q sin q cos q 2 cos 2q = 2 sin q cos q 2 cos 2q = sin 2q = 2 cot 2q 128. We have tan q + 2 tan (2q) + 4 tan (4q) + 8 cot 8q = cot q (cot q – tan q) + 2 tan 2q + 4 tan 4q + 8 cot 8q = cot q – 2 cot 2q + 2 tan 2q + 4 tan 4q + 8 cot 8q = cot q – 2(cot 2q – tan 2q) + 4 tan 4q + 8 cot 8q = cot q – 4 cot 4q + 4 tan 4q + 8 cot 8q = cot q – 4(cot 4q tan 4q) + 8 cot 8q = cot q – 8 cot 8q + 8 cot 8q = cot q 129. We have a cos (2q) + b sin (2q) =

- 74 £ 2k + 1 £ 74 - 74 - 1 74 - 1 £k£ 2 2 k = –4, –3, –2, –1, 0, 1, 2, 3 Hence, the number of integral values of 124. We have cos a + cos b + cos g + cos (a + b + g) = (cos a + cos b) + (cos (a + b + g) + cos g) Êa + bˆ Êa - bˆ cos Á = 2 cos Á Ë 2 ˜¯ Ë 2 ˜¯ Êa + b + g + g ˆ Êa + b + g - g ˆ + 2 cos Á ˜¯ cos Á ˜¯ Ë Ë 2 2 Êa + bˆ Êa - bˆ cos Á = 2 cos Á Ë 2 ˜¯ Ë 2 ˜¯

Ê 1 – tan 2q ˆ Ê 2 tan q ˆ = aÁ + bÁ 2 ˜ Ë 1 + tan 2q ˜¯ Ë 1 + tan q ¯

Ê a + b + 2g ˆ Êa + bˆ + 2 cos Á ˜¯ cos ÁË ˜ Ë 2 2 ¯ Êa + bˆ = 2 cos Á ¥ Ë 2 ˜¯

TR_01a.indd 47

Ê a - b a + b + 2g ˆ Á ˜ 2 cos Á 2 ˜¯ Ë 2 Êa + bˆ Ï Êa + g ˆ Ê b + g ˆ¸ = 2 cos Á ¥ Ìcos Á cos Á ˝ ˜ ˜ Ë 2 ¯ Ó Ë 2 ¯ Ë 2 ˜¯ ˛ 125. We have 1 - cos 2q 2 sin 2q = sin 2q 2 sin q cos q sin q = = tan q cos q

127. We have (cot q – tan q)

- 7 2 + 52 £ 2k + 1 £ 7 2 + 52

Ï Êa - bˆ Ê a + b + 2g Ìcos Á ˜¯ + 2 cos ÁË Ë 2 2 Ó

ˆ ˜ ˜¯

ˆ¸ ˜¯ ˝ ˛

Ê Á1 – = aÁ Á1 + ÁË

b2 ˆ Ê b ˆ 2 2˜ a ˜ + bÁ a ˜ Á 2 b ˜ b 2 ˜˜ Á 1 + ˜ Ë a2 ¯ a2 ¯

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Ê sec 8A - 1ˆ 132. We have Á Ë sec 4A - 1˜¯

Ê a 2 - b2 ˆ Ê 2ab ˆ = aÁ 2 + bÁ 2 ˜ 2˜ Ë a + b2 ¯ Ëa +b ¯ = =

1 -1 cos 8A = 1 -1 cos 4A

a ( a 2 - b 2 + 2b 2 ) a 2 + b2 a(a 2 + b 2 ) a 2 + b2

=

=a 130. We have 3 cosec (20°) - sec (20°) = =

=

3 1 sin (20°) cos (20°) 3 cos (20°) - sin (20°) sin (20°) cos (20°)

Ê 3 ˆ 1 4Á cos (20°) - sin (20°)˜ 2 Ë 2 ¯ = 2 sin (20°) cos (20°) 4(sin (20°) cos (20°) - cos (60°) sin (20°)) = 2 sin (20°) cos (20∞) 4(sin (60∞ - 20∞)) = sin (40°) =4 131. We have tan (9°) + tan (27°) – tan (63°) + tan (81°) = {tan (9°) + tan (81°)} – {tan (27°) + tan (63°)} = {tan (9°) + cot (9°)} – {tan (27°) + cot (27°)} Ï sin (9∞) cos (9∞) ¸ Ï sin (27∞) cos (27∞) ¸ =Ì + + ˝-Ì ˝ Ó cos (9∞) sin (9∞) ˛ Ó cos (27∞) sin (27∞) ˛ ÏÔ sin (9∞) + cos (9∞) ¸Ô ÏÔ sin (27∞) + cos (27∞) ¸Ô =Ì ˝-Ì ˝ ÔÓ sin (9∞) cos (9∞) Ô˛ ÔÓ sin (27∞) cos (27∞) Ô˛ 2

2

2

2 Ï ¸ Ï 2 ¸ =Ô Ô Ô-Ô Ê ˆ Ì 5 -1 ˝ Ì 5 +1˝ Ô ÁË 4 ˜¯ Ô ÔÓ 4 Ô˛ Ó ˛ Ï 8 ¸ Ï 8 ¸ =Ì ˝-Ì ˝ Ó 5 - 1˛ Ó 5 + 1˛ ÔÏ 8( 5 + 1 - 5 + 1) Ô¸ =Ì ˝ 5 -1 ÓÔ ˛Ô =4

TR_01a.indd 48

2 sin 2 4A 2

2 sin 2A

¥

cos 4A cos 8A

=

2 sin 4A cos 4A sin 4A ¥ cos 8A 2 sin 2 2A

=

sin 8A 2 sin 2A cos 2A ¥ cos 8A 2 sin 2 2A

=

sin 8A cos 2A ¥ cos 8A sin 2A

= tan 8 A ¥ cot 2 A =

tan 8A tan 2A

133. We have Ê 2p ˆ Ê 2p ˆ cos 2 (q ) + cos 2 Á - q ˜ + cos 2 Á + q˜ Ë 3 ¯ Ë 3 ¯ =

1Ê Ê 2p ˆ Ê 2p ˆˆ 2 cos 2 (q ) + 2 cos 2 Á - q ˜ + 2 cos 2 Á + q˜˜ Ë 3 ¯ Ë 3 ¯¯ 2 ÁË

=

1 1Ê Ê 4p ˆˆ (1 + cos (2q )) + Á 1 + cos Á - 2q ˜ ˜ Ë 3 ¯¯ 2 2Ë +

2

2 2 Ï ¸ Ï ¸ =Ì ˝-Ì ˝ ∞ ∞ ∞ ∞ 2 sin (9 ) cos (9 ) 2 sin (27 ) co s (27 ) Ó ˛ Ó ˛ 2 2 Ï ¸ Ï ¸ =Ì ˝-Ì ˝ ∞ ∞ sin (18 ) sin (54 ) Ó ˛ Ó ˛

1 - cos 8A cos 4A ¥ 1 - cos 4A cos 8A

1Ê Ê 4p ˆˆ 1 + cos Á + 2q ˜ ˜ Á Ë ¯¯ 2Ë 3

=

Ê 1Ê Ê 4p ˆ Ê 4p ˆˆˆ - 2q ˜ + cos Á + 2q ˜ ˜ ˜ 3 + Á cos 2q + cos Á Á Ë ¯ Ë ¯¯¯ Ë 2Ë 3 3

=

Ê ˆˆ 1Ê Ê 4p ˆ 3 + Á cos 2q + 2 cos Á ˜ cos (2q )˜ ˜ Á Ë 3 ¯ Ë ¯¯ 2Ë

=

1Ê Ê ˆˆ Ê 1ˆ 3 + Á cos 2q + 2 Á - ˜ cos (2q )˜ ˜ Á Ë 2¯ Ë ¯¯ 2Ë

=

3 2

134. We have sin2 q + sin2(120° + q) + sin2(240° + q) 1Ê Ê 2p ˆˆ = Á 2 sin 2q + 2 sin 2 Á + q˜˜ Ë 3 ¯¯ 2Ë +

1Ê Ê 4p ˆˆ + q˜˜ 2 sin 2 Á Á Ë ¯¯ 2Ë 3

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The Ratios and Identities

=

1Ê Ê Ê 4p ˆˆˆ (1 – cos 2q ) + Á1 - cos Á + 2q ˜ ˜ ˜ Á Ë 3 ¯¯¯ Ë 2Ë +

=

1 ÊÊ Ê 8p ˆˆˆ 1 - cos Á + 2q ˜ ˜ ˜ Á Á Ë ¯¯¯ 2 ËË 3

3 1 1È Ê 4p ˆ Ê 8p ˆ˘ - cos 2q - Ícos Á + 2q ˜ + cos Á + 2q ˜ ˙ Ë 3 ¯ Ë 3 ¯˚ 2 2 2Î

3 1 1 - cos 2q - (2 cos (120°) cos 2q ) 2 2 2 3 1 1 = - cos 2q + cos (2q ) 2 2 2 3 = 2 =

135. We have Êp ˆ Êp ˆ 4 sin (q ) sin Á - q ˜ sin Á + q ˜ Ë3 ¯ Ë3 ¯

138. We have cos (10°) ◊ cos (50°) ◊ cos (70°) = cos (10°) ◊ cos (60° – 10°) ◊ cos (60° + 10°) 1 = (4 cos (10∞) ◊ cos (60° - 10∞) ◊ cos (60∞ + 10∞)) 4 1 = (cos (3 ◊ 10∞)) 4 3 = 8 139. We have tan (q) – tan (60° – q) + tan (60° + q) tan (60°) - tan (q ) = tan (q ) 1 + tan (60°) ◊ tan (q ) tan (60°) + tan (q ) + 1 - tan (60°) ◊ tan (q ) = tan (q ) -

Ê Êp ˆ Êp ˆˆ = 4 sin (q ) ¥ Á sin Á - q ˜ sin Á + q ˜ ˜ Ë3 ¯ Ë3 ¯¯ Ë

- 3 + 3 tan (q ) + tan (q ) - 3 tan 2 (q )

Ê ˆ Êpˆ = 4 sin (q ) ¥ Á sin 2 Á ˜ - sin 2q ˜ Ë 3¯ Ë ¯

+ 3 + 3 tan (q ) + tan (q ) + 3 tan 2 (q ) (1 - 3 tan 2q ) 8 tan (q ) = tan (q ) + 1 - 3 tan 2 (q ) =

Ê3 ˆ = 4 sin (q ) ¥ Á - sin 2q ˜ Ë4 ¯ Ê3 ˆ = 4 ¥ Á sin q - sin 3q ˜ Ë4 ¯ = (3 sin q – 4 sin3 q) = sin (3q) 136. We have sin (20°) sin (40°) sin (80°) 1 = (sin (3.20∞)) 4 1 = (sin (60∞)) 4 3 = 8 137. We have, 4 cos (q) ◊ cos (60° – q) ◊ cos (60° + q) = 4 cos (q) ◊ (cos (60° – q) ◊ cos (60° + q)) = 4 cos (q) ◊ (cos2(60°) – sin2 q) Ê1 ˆ = 4 cos (q ) ◊ Á - 1 + cos 2 q ˜ Ë4 ¯ Ê 3 ˆ = 4 cos (q ) ◊ Á - + cos 2 q ˜ Ë 4 ¯ = cos (q) ◊ (–3 + 4 cos2 q) = (4 cos3 q – 3 cos (q)) = cos (3q)

TR_01a.indd 49

3 - tan (q ) 3 + tan (q ) + 1 + 3 ◊ tan (q ) 1 - 3 ◊ tan (q )

=

tan (q ) – 3 tan 3 (q ) + 8 tan (q ) 1 - 3 tan 2 (q )

Ê 3 tan (q ) - tan 3 (q ) ˆ =3¥Á ˜ Ë 1 - 3 tan 2 (q ) ¯ = 3 tan (3q ) 140. We have cos (q) cos (2q) ◊ cos (22q) ◊ cos (23q) … cos (2n–1q) 1 = (2 sin q cos q )(cos 2q ◊ cos (22q )...cos (22n -1q )) 2 sin q 1 (2 sin 2q cos 2q )(cos (22q )...cos (22n -1q )) 2 sin q 1 = 3 (2 sin 4q cos 4q )(cos (23q )...cos (22n -1q )) 2 sin q =

=

2

1 (2 sin 23q cos 23q )(cos (24q )...cos (22n -1q )) 2 sin q 4

o 1 (2 sin 2n –1q cos 2n –1q ) 2 sin q 1 = n (sin 2nq ) 2 sin q =

=

n

sin (2nq ) 2n sin q

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Ê 2p ˆ Ê 4p ˆ Ê 8p ˆ 141. We have cos Á ˜ cos Á ˜ cos Á ˜ Ë 7 ¯ Ë 7 ¯ Ë 7¯ pˆ Ê 2p ˆ Ê 4p ˆ Ê = cos Á ˜ cos Á ˜ cos Á p + ˜ Ë 7 ¯ Ë 7 ¯ Ë 7¯ Êpˆ Ê 2p ˆ Ê 4p ˆ = – cos Á ˜ cos Á ˜ cos Á ˜ Ë 7¯ Ë 7 ¯ Ë 7 ¯ =–

=–

=–

1 Ê Êpˆ Ê p ˆˆ Ê 2p ˆ Ê 4p ˆ 2 sin Á ˜ cos Á ˜ ˜ cos Á ˜ cos Á ˜ Á Ë ¯ Ë ¯ Ë ¯ Ë 7 ¯ p Ë ¯ 7 7 7 Ê ˆ 2 sin Á ˜ Ë 7¯ Ê Ê 2p ˆ Ê 2p ˆ ˆ Ê 4p ˆ ÁË 2 sin ÁË 7 ˜¯ cos ÁË 7 ˜¯ ˜¯ cos ÁË 7 ˜¯ p Ê ˆ 22 sin Á ˜ Ë 7¯ 1

Ê Ê 4p ˆ Ê 4p ˆ ˆ ÁË 2 sin ÁË 7 ˜¯ cos ÁË 7 ˜¯ ˜¯ p Ê ˆ 23 sin Á ˜ Ë 7¯ 1

Ê 8p ˆ sin Á ˜ Ë 7¯ =– Êpˆ 3 2 sin Á ˜ Ë 7¯ pˆ Ê sin Á p + ˜ Ë 7¯ =– Êpˆ 3 2 sin Á ˜ Ë 7¯ Êpˆ sin Á ˜ Ë 7¯ = Êpˆ 8 sin Á ˜ Ë 7¯ =

1 8

Ê 2p ˆ Ê 4p ˆ Ê 6p ˆ 142. Let z = cos Á ˜ + cos Á ˜ + cos Á ˜ Ë 7 ¯ Ë 7 ¯ Ë 7 ¯ Êpˆ fi 2z sin Á ˜ Ë 7¯ Êpˆ Ê 2p ˆ Êpˆ Ê 4p ˆ = 2 sin Á ˜ cos Á ˜ + 2 sin Á ˜ cos Á ˜ Ë 7¯ Ë 7 ¯ Ë 7¯ Ë 7 ¯ Êpˆ Ê 6p ˆ + 2 sin Á ˜ cos Á ˜ Ë 7¯ Ë 7 ¯ Ê 3p ˆ Êpˆ Ê 5p ˆ Ê 3p ˆ = sin Á ˜ - sin Á ˜ + sin Á ˜ - sin Á ˜ Ë 7 ¯ Ë 7¯ Ë 7 ¯ Ë 7 ¯ Ê 7p ˆ Ê 5p ˆ + sin Á ˜ - sin Á ˜ Ë 7 ¯ Ë 7 ¯ Êpˆ = - sin Á ˜ Ë 7¯ Êpˆ Êpˆ Thus, 2z sin Á ˜ = - sin Á ˜ Ë 7¯ Ë 7¯

TR_01a.indd 50

1 2

fi

z=–

fi

1 Ê 2p ˆ Ê 4p ˆ Ê 6p ˆ cos Á ˜ + cos Á ˜ + cos Á ˜ = Ë 7 ¯ Ë 7 ¯ Ë 7 ¯ 2

143. Given M = a 2 cos 2q + b 2 sin 2q + a 2 sin 2q + b 2 cos 2q fi

M2 = a2 cos2 q + b2 sin2 q + a2 sin2 q + b2 cos2 q + 2 (a 2 cos 2q + b 2 sin 2q )(a 2 sin 2q + b 2 cos 2q )

fi

M 2 = a 2 + b 2 + 2 (a 2 cos 2q + b 2 sin 2q ) (a 2 sin 2q + b 2 cos 2q )

fi

M2 = a2 + b2 + 2[(a4 + b4) sin2 q cos2 q + a2b2(sin4 q + cos4 q)]1/2

fi

M2 = a2 + b2 + 2[(a4 + b4) sin2 q cos2 q + a2b2(1 – 2 sin2 q + cos2 q)]1/2

fi

M2 = a2 + b2 + 2((a4 + b4 – 2a2b2) sin2 q cos2 q + a2b2)1/2

fi

M 2 = a 2 + b2 +

(4(a 4 + b 4 - 2a 2b 2 ) sin 2q cos 2q + 4a 2b 2 )

fi

M 2 = a 2 + b2 +

((a 2 - b 2 ) 2 + (sin (2q )) 2 + 4a 2 b 2 )

Thus, maximum (M2) = a2 + b2 + (a2 + b2) = 2(a2 + b2) 2 and minimum (M ) = a2 + b2 + 2ab = (a + b)2 Hence, the value of m1 – m 2 = maximum (M 2) – minmum (M 2) = 2(a2 + b2) – (a + b)2 144. We have tan (4q) = tan (3q + q) =

tan 3q + tan q 1 - tan 3q tan q

3 tan q – tan 3q + tan q 1 - 3 tan 2q = 3 tan q – tan 3q 1tan q 1 - 3 tan 2q 3 tan q – tan 3q + tan q – 3 tan 3q 1 - 3 tan 2q = 2 1 – 3 tan q – 3 tan 2q + tan 4q 1 - 3 tan 2q =

4 tan q - 4 tan 3q 1 - 6 tan 2q + tan 4q

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145. We have Ê sin x sin 3x sin 9x ˆ ÁË cos 3x + cos 9x + cos 27x ˜¯

(i) Here, a = 3, b = 4 and c = 10 Thus, the minimum values of f (x)

=

1 Ê 2 sin x cos x 2 sin 3x cos 3x 2 sin 9x cos 9x ˆ + + 2 ÁË cos 3x cos x cos 3x cos 9x cos 27x cos 9x ˜¯

= - a 2 + b 2 + c = –5 + 10 = 5 and the maximum values of

=

ˆ 1 Ê sin 2x sin 6x sin 18x + + 2 ÁË cos 3x cos x cos 3x cos 9x cos 9x cos 27x ˜¯

f (x) =

=

1 Ê sin (3x – x) sin (9x – 3x) sin (27x – 9x) ˆ + + 2 ÁË cos 3x cos x cos 3x cos 9x cos 9x cos 27x ˜¯

=

1 Ê sin 3x cos x – cos 3x sin x 2 ÁË cos 3x cos x +

sin 9x cos 3x – cos 9x sin 3x cos 3x cos 9x

sin 27x cos 9x – cos 27x sin 9x ˆ + ˜¯ cos 9x cos 27x 1 (tan 3x - tan x + tan 9x - tan 3x + tan 27x - tan 9 x) 2 1 = (tan 27x - tan x) 2 =

146. We have Êqˆ tan Á ˜ (1 + sec q )(1 + sec 2q )(1 + sec 22q ) ... sec (2nq ) Ë 2¯ Êqˆ Now, tan Á ˜ (1 + sec q ) Ë 2¯ sin (q /2) 2 cos 2 (q /2) ¥ cos (q /2) cos q 2 sin (q /2) cos (q /2) = cos q sin q = = tan q cos q =

Also, tan q ¥ (1 + sec 2q) sin q Ê 1 + cos 2q ˆ = ¥ cos q ÁË cos 2q ˜¯ =

sin q Ê 2 cos 2q ˆ ¥ cos q ÁË cos 2q ˜¯

=

2 sin q cos q cos 2q

=

sin 2q = tan 2q cos 2q

Êqˆ Thus, tan Á ˜ (1 + sec q )(1 + sec 2q ) Ë 2¯ (1 + sec 22q ) ... sec (2nq ) = tan (2nq)

TR_01a.indd 51

147.

a 2 + b 2 + c = 5 + 10 = 15.

(ii) Maximum value =

33 + 42 + 10 = 15

2 2 Minimum value = - 3 + 4 + 10 = 5 (iii) Maximum value = 3 + 4 = 7 Minimum value = –3 + 4 = 1 (iv) Maximum value = 2 + 5 = 7 Minimum value = –2 + 5 = 3

(v) Maximum value =

2

Minimum value = - 2 (vi) Maximum value =

2

Minimum value = - 2 (vii) Maximum value = sin 1 Minimum value = –sin 1 (viii) Maximum value = cos 1 Minimum value = 0 (ix) Maximum value =

2

Minimum value = - 2 (x) Given f (x) = cos (sin x) + sin (cos x). = cos (sin x) + sin (sin x) + sin (cos x) – sin (sin x) Maximum value =

2 + sin 1

Minimum value = 2 - sin 1 148. Rf = [minimum f (x), maximum f (x)] = [– 2 + 3, 2 + 3] 149. Given 2 sin2 q + 3 cos2 q = 2(sin2 q + cos2 q) + cos2 q = 2 + cos2 q Maximum value = 2 + 1 = 3 Minimum value = 2 + 0 = 2 pˆ Ê 150. Let f (q ) = 5 cos q + 3 cos Á q + ˜ + 3 Ë 3¯ = 5 cos q + =

3 3 3 cos q – sin q + 3 2 2

13 3 3 cos q – sin q + 3 2 2

Maximum value =

169 27 196 + +3= + 3 = 10 4 4 4

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Minimum value = -

169 27 + + 3 = -7 + 3 = -4 4 4

151. Let f (q) = cos2 q + 3 sin2 q – 3 sin 2q + 2 = 1 + 2 sin2 q – 3 sin 2q + 2 = 3 + 1 – cos 2q – 3 sin 2q = 4 – (cos 2q + 3 sin 2q) Maximum value = 4 + 10 Minimum value = 4 - 10 152. Let f (x) = cosec2 x + 25 sec2 x = 1 + cot2 x + 25 + 25 tan2 x = 26 + cot2 x + 25 tan2 x ≥ 26 + 10 = 36 Hence, the minimum value is 26. 153. Given expression is 2 – cos x + sin2 x = 2 – cos x + 1 – cos2 x = 3 – cos x – cos2 x = –(cos2 x + cos x – 3) 2 ÊÊ 1ˆ = – Á Á cos x + ˜ - 3 ËË 2¯

1ˆ ˜ 4¯

2

13 Ê 1ˆ = - Á cos x + ˜ Ë 4 2¯ 13 Maximum value = 4 13 1 Minimum value = - =3 4 4 154. Given y = 4 sin2 q – cos 2q = 2(2 sin2 q) – cos 2q = 2(1 – cos 2q) – cos 2q = 2 – 3 cos 2q = 2 + 3(–cos 2q) Maximum value = 2 + 3 = 5 Minimum value = 2 – 3 = –1 Hence y lies in [–1, 5] 155. Here, m = –3 + 5 = 2 and n = 3 + 2 = 5 Hence, the value of (m + n + 2) = 9 156. Given f (x) = sin2 x + cos4 x 1 1 = (2 sin 2 x) + (2 cos 2 x) 2 2 4 1 1 = (1 - cos 2 x) + (1 + cos 2 x) 2 2 4 1 1 = (1 - cos 2x) + (1 + 2 cos 2x + cos 2 2x) 2 4 1 1 1 = + + cos 2 2x 2 4 4 3 1 = + cos 2 2x 4 4

TR_01a.indd 52

3 1 + ◊1 = 4 4 3 1 Minimum value = + ◊ 0 = 4 4

Maximum value =

3 1 4 + = =1 4 4 4 3 4

157. Given f (x) = cos2 x + sin4 x 1 1 = (2 cos 2 x) + (2 sin 2 x) 2 2 4 1 1 = (1 + cos 2x) + (1 – cos 2x) 2 2 4 1 1 = (1 + cos 2x) + (1 – 2 cos 2x + cos 2 2x) 2 4 1 1 1 = + + cos 2 x 2 4 4 3 1 = + cos 2 2x 4 4 3 1 3 1 4 Maximum value = + ◊ 1 = + = = 1 4 4 4 4 4 3 1 3 Minimum value = + ◊ 0 = 4 4 4 158. Given f (x) = sin4 x + cos4 x = (sin2 x + cos2 x)2 – 2 sin2 x cos2 x 1 = 1 - sin 2 2x 2 Maximum value = 1 + 0 = 1 1 1 Minimum value = 1 - = 2 2 159. We have, f (q) = sin6 q + cos6 q = (sin2 q)3 + (cos2 q)3 = (sin2 q + cos2 q)2 – 3 sin2 q cos2 q (sin2 q + cos2 q) = 1 – 3 sin2 q cos2 q 3 = 1 - (4 sin 2q cos 2q ) 4 =1-

3 (sin 2 2q ) 4

3 (- sin 2 2q ) 4 As we know, –1 £ (–sin2 2q) £ 0 =1+

fi

3 3(- sin 2 2q ) - £ £0 4 4

fi

1-

fi

1 £ f (q ) £ 1 4

3 3(- sin 2 2q ) £1+ £1 4 4

Hence, the maximum value = 1 and the minimum value 1 = 4

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160. Now, A = cos2 q + sin4 q 1 1 (2 sin 2q ) + (4 sin 4q ) 2 4 1 1 = (1 - cos (2q )) + (1 + cos (2q )) 2 2 4 1 1 = (1 + cos (2q )) + (1 - 2 cos (2q ) + cos 2 (2q )) 2 4 1 1 1 1 1 = + cos (2q ) + - cos (2q ) + (cos 2 (2q )) 2 2 4 2 4 3 1 = + (cos 2 (2q )) 4 4 3 1 Maximum value of A = m1 = + ◊ 1 = 1 4 4 Also, B = sin2 q + cos4 q =

1 1 = (2 sin 2q ) + (4 sin 4q ) 2 4 1 1 = (2 sin 2q ) + (2 cos 2q ) 2 2 4 1 1 = (1 - cos (2q )) + (1 + cos (2q )) 2 2 4 1 1 = (1 - cos (2q )) + (1 + 2 cos (2q ) + cos 2 (2q )) 2 4 1 1 1 = + + cos 2 (2q ) 2 4 4 3 1 = + cos 2 (2q ) 4 4 3 1 3 Thus, the minimum value of B = m = + ◊ 0 = 4 4 4 Now, the value of m12 + m22 + m1m2 =1+ =

9 3 + 16 4

37 16

161. Given f (x) = (sin x + cos x + cosec 2x) As we know that, AM ≥ GM

3

Ê sin x + cos x + cosec 2x ˆ 3 ÁË ˜¯ ≥ (sin x ◊ cos x ◊ cosec 2x) 3 Ê sin x + cos x + cosec 2x ˆ 3 1 ÁË ˜¯ ≥ 3 2 3

1 Ê sin x + cos x + cosec 2x ˆ ÁË ˜¯ ≥ 3 2 27 (sin x + cos x + cosec 2x)3 ≥ 2

TR_01a.indd 53

27 2 162. Find the maximum and minimum values of 5 f (x) = 2 sin q - 6 sin q cos q + 3 cos 2q Hence, the minimum value of f (x) is

163. Given f (x) =

a2 b2 + cos 2 x sin 2 x

= a2 sec2 x + b2 cosec2 x = a2 + a2 tan2 x + b2 + b2 cot2 x = a2 + b2 + (a2 tan2 x + b2 cot2 x) ≥ a2 + b2 + 2ab = (a + b)2 Hence, the minimum value of f (x) is (a + b)2. 164. We have, f (x) =

x 2 sin 2 x + 4 x sin x

= x sin x +

4 ≥4 x sin x

Hence, the minimum values of f (x) is 4 165. Given f (x) = logx y + logy y As we know, AM ≥ GM log x y + log y x 2 log x y + log y x

≥ log x y ◊ log y x = 1

≥1 2 logx y + logy x ≥ 2 Hence, the minimum value of f (x) is 2. 166. Given f (x) = 2 log10 x – logx (0.01), x > 1 = 2 log10 x – logx (10)–2 = 2 log10 x + 2 logx (10) = 2(log10 x + logx (10)) ≥ 2. 2 = 4 Hence, the minimum value of f (x) is 4 167. Given f ( x, y, z ) =

( x 2 + 1)( y 2 + 1)( z 2 + 1) xyz

Ê x 2 + 1ˆ Ê y 2 + 1ˆ Ê z 2 + 1ˆ =Á ˜Á ˜Á ˜ Ë x ¯Ë y ¯Ë z ¯ Ê =Áx + Ë

1ˆ Ê 1ˆ Ê ˜¯ Á y + ˜ ÁË z + x Ë y¯

1ˆ ˜ z¯

≥ 2.2.2 = 8 Hence, the minimum value is 2. 168. Given f ( x, y, z ) =

( x3 + 2)( y 3 + 2)( z 3 + 2) xyz

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Ê x3 + 2 ˆ Ê y 3 + 2 ˆ Ê z 3 + 2 ˆ =Á Ë x ˜¯ ÁË y ˜¯ ÁË z ˜¯ 2ˆ Ê Ê = Á x2 + ˜ Á y 2 + Ë x¯ Ë

2 ˆ Ê 2 2ˆ Á z + ˜¯ y ˜¯ Ë z

≥ 3.3.3 = 27 Hence, the minimum value is 27. 169. We have f (a, b, c, d) (a 2 + 1)(b 2 + 1)(c 2 + 1)(d 2 + 1) abcd 2 (a + 1) (b 2 + 1) (c 2 + 1) (d 2 + 1) = ¥ ¥ ¥ a b c d 1ˆ Ê 1ˆ Ê 1ˆ Ê 1ˆ Ê = Áa + ˜ Áb + ˜ Ác + ˜ Ád + ˜ Ë a¯ Ë b¯ Ë c¯ Ë d¯ =

≥ 2.2.2.2 = 16 Hence, the minimum value is 16. 170. Do yourself. 171. We have sin2(24°) – sin2(6°) = sin (24° + 6°) sin (24° – 6°) = sin (30°) ¥ sin (18°) 1 = ¥ sin (18∞) 2 1 5 -1 = ¥ 2 4 ( 5 - 1) = 8 172. We have sin2(48°) – cos2(12°) = cos (48° + 12°) ¥ cos (48° – 12°) = cos (60°) ¥ cos (36°) 1 5 +1 ¥ 2 4 5 +1 = 8 173. We have sin (12°) ◊ sin (48°) ◊ sin (54°) =

1 (sin (12∞) ◊ sin (48∞) ◊ sin (72∞))(sin (54∞)) sin (72∞) 1 = 4 sin (72∞) =

(4 sin (60° – 12°) ◊ sin (12°) ◊ sin (60° + 12°)) ¥ (cos (36°)) 1 = (sin (36∞) ◊ cos (36∞)) 4 sin (72∞) 1 = (2 sin (36∞) ◊ cos (36∞)) 8 sin (72∞) 1 = (sin (72∞)) 8 sin (72∞) 1 = 8

TR_01a.indd 54

174. We have sin (6°) ◊ sin (42°) ◊ sin (66°) sin (78°) 1 = 4 sin (54∞) (4 sin (6°) ◊ sin (60° – 6°) ◊ sin (60° + 6°)) ¥ (sin (78°) ◊ sin (42°)) 1 (sin (18∞) sin (72∞) ◊ sin (42∞)) 4 cos (36∞) 1 = 16 cos (36∞) =

(4 sin (18°) ◊ sin (60° + 18°) ◊ sin (60° – 18°)) 1 (sin (54∞)) 16 cos (36∞) 1 = (cos (36∞)) 16 cos (36∞) 1 = 16 175. We have 4(sin (24°) + cos (6°)) = 4(sin (24°) + sin (84°)) =

Ê Ê 24∞ + 84∞ ˆ Ê 24∞ - 84∞ ˆ ˆ = 4 Á 2 sin ÁË ˜¯ cos Á ˜¯ ˜¯ Ë Ë 2 2 = 8(sin (54°) cos (30°)) = 8(sin (36°) cos (30°)) Ê 5 +1 = 8Á ¥ Ë 4 =

(

)

1ˆ 2 ˜¯

5 +1

176. We have tan (6°) ◊ tan (42°) ◊ tan (66°) tan (78°) = (tan (6°) ◊ tan (66°)) ¥ (tan (42°) ◊ tan (78°)) 1 = ¥ (tan (6°) ◊ tan (54°) ◊ tan (66°)) tan (54∞) ¥ (tan (42°) ◊ tan (78°)) 1 ¥ (tan (6°) ◊ tan (60° – 6°) ◊ tan (60° = tan (54∞) + 6°)) ¥ (tan (42°) ◊ tan (78°)) 1 = (tan (18∞)) ¥ (tan (42∞) ◊ tan (78∞)) tan (54∞) 1 = (tan (18∞) ◊ tan (42∞) ◊ tan (78∞)) tan (54∞) 1 = tan (54∞) (tan (60° – 18°) ◊ tan (18°) ◊ tan (60° + 18°)) =

1 ¥ ( ◊ tan (54∞)) tan (54∞)

=1

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Ê Ê p ˆˆ Ê Ê 3p ˆ ˆ 177. We have Á1 + cos Á ˜ ˜ ◊ Á1 + cos Á ˜ ˜ Ë ¯ Ë 8 ¯¯ Ë 8 ¯ Ë

Ê Ê Ê p ˆˆ = Á 2 - Á1 - cos Á ˜ ˜ Ë 4¯¯ Ë Ë

Ê Ê 5p ˆ ˆ Ê Ê 7p ˆ ˆ ◊ Á1 + cos Á ˜ ˜ ◊ Á1 + cos Á ˜ ˜ Ë 8 ¯¯ Ë Ë 8 ¯¯ Ë

Ê 1 ˆ Ê 1 ˆˆ Ê = Á 2 - Á1 ◊ Á1 + ˜ ˜ Ë ¯ Ë Ë 2 2 ¯ ¯˜

Ê Ê p ˆˆ Ê Ê 7p ˆ ˆ = Á1 + cos Á ˜ ˜ ◊ Á1 + cos Á ˜ ˜ Ë 8 ¯¯ Ë Ë 8 ¯¯ Ë Ê Ê 3p ˆ ˆ Ê Ê 5p ˆ ˆ ◊ Á 1 + cos Á ˜ ˜ ◊ Á1 + cos Á ˜ ˜ Ë ¯ Ë 8 ¯¯ Ë 8 ¯ Ë Ê Ê p ˆˆ Ê Ê p ˆˆ = Á1 + cos Á ˜ ˜ ◊ Á1 - cos Á ˜ ˜ Ë 8 ¯¯ Ë Ë 8 ¯¯ Ë Ê Ê 3p ˆ ˆ Ê Ê 3p ˆ ˆ ◊ Á1 + cos Á ˜ ˜ ◊ Á1 - cos Á ˜ ˜ Ë 8 ¯¯ Ë Ë 8 ¯¯ Ë Ê Ê p ˆˆ Ê Ê 3p ˆ ˆ = Á1 - cos 2 Á ˜ ˜ ◊ Á1 - cos 2 Á ˜ ˜ Ë 8 ¯¯ Ë Ë 8 ¯¯ Ë Ê Ê p ˆˆ Ê Ê 3p ˆ ˆ = Á sin 2 Á ˜ ˜ ◊ Á sin 2 Á ˜ ˜ Ë ¯ Ë 8 ¯¯ Ë 8 ¯ Ë 1Ê Ê p ˆˆ Ê Ê 3p ˆ ˆ = Á 2 sin 2 Á ˜ ˜ ◊ Á 2 sin 2 Á ˜ ˜ Ë 8 ¯¯ Ë Ë 8 ¯¯ 4Ë 1Ê Ê p ˆˆ Ê Ê 3p ˆ ˆ = Á 1 - cos Á ˜ ˜ ◊ Á 1 - cos Á ˜ ˜ Ë 4¯¯ Ë Ë 4 ¯¯ 4Ë 1Ê 1 ˆ Ê 1 ˆ = Á1 ˜¯ ◊ ÁË 1 + ˜ 4Ë 2 2¯ 1Ê 1ˆ = Á1 - ˜ . 4Ë 2¯ 1 = . 8 178. We have Êpˆ Ê 3p ˆ Ê 5p ˆ Ê 7p ˆ sin 4 Á ˜ + sin 4 Á ˜ + sin 4 Á ˜ + sin 4 Á ˜ Ë 8¯ Ë 8 ¯ Ë 8 ¯ Ë 8 ¯ Êpˆ Ê 3p ˆ Ê 3p ˆ Êpˆ = sin 4 Á ˜ + sin 4 Á ˜ + sin 4 Á ˜ + sin 4 Á ˜ Ë 8¯ Ë 8 ¯ Ë 8 ¯ Ë 8¯ Ê Êpˆ Ê 3p ˆ ˆ = 2 Á sin 4 Á ˜ + sin 4 Á ˜ ˜ Ë 8¯ Ë 8 ¯¯ Ë

TR_01a.indd 55

Ê Ê p ˆˆˆ ◊ Á1 + cos Á ˜ ˜ ˜ Ë 4¯¯¯ Ë

Ê Ê = Á 2 - Á1 Ë Ë

1ˆ ˆ ˜ 2 ¯ ˜¯

3 2 179. We have =

Êpˆ Ê 3p ˆ Ê 5p ˆ Ê 7p ˆ cos 4 Á ˜ + cos 4 Á ˜ + cos 4 Á ˜ + cos 4 Á ˜ Ë 8¯ Ë 8 ¯ Ë 8 ¯ Ë 8 ¯ Êpˆ Ê 3p ˆ Ê 3p ˆ Êpˆ = cos 4 Á ˜ + cos 4 Á ˜ + cos 4 Á ˜ + cos 4 Á ˜ Ë 8¯ Ë 8 ¯ Ë 8 ¯ Ë 8¯ Ê 3p ˆ Êpˆ + cos 4 Á ˜ + cos 4 Á ˜ Ë 8 ¯ Ë 8¯ Ê Êpˆ Ê 3p ˆ ˆ = 2 Á cos 4 Á ˜ + cos 4 Á ˜ ˜ Ë 8¯ Ë 8 ¯¯ Ë Ê Êpˆ Ê p ˆˆ = 2 Á cos 4 Á ˜ + sin 4 Á ˜ ˜ Ë ¯ Ë 8 ¯¯ Ë 8 Ê Êpˆ Ê p ˆˆ = 2 Á 1 - 2 cos 2 Á ˜ ◊ sin 2 Á ˜ ˜ Ë 8¯ Ë 8 ¯¯ Ë Ê Ê Ê p ˆˆ = Á 2 - Á 2 cos 2 Á ˜ ˜ Ë 8 ¯¯ Ë Ë

Ê Ê p ˆˆˆ ◊ Á 2 sin 2 Á ˜ ˜ ˜ Ë 8 ¯¯¯ Ë

Ê Ê Ê p ˆˆ = Á 2 - Á 1 + cos Á ˜ ˜ Ë 4¯¯ Ë Ë

Ê Ê p ˆˆˆ ◊ Á 1 + cos Á ˜ ˜ ˜ Ë 4¯¯¯ Ë

Ê 1 ˆ Ê 1 ˆˆ Ê = Á 2 - Á1 + ◊ Á1 ˜ ˜ Ë Ë 2¯ Ë 2 ¯ ¯˜ Ê 1ˆ ˆ Ê = Á 2 - Á1 - ˜ ˜ Ë Ë 2¯ ¯ =

3 2

180. We have tan (20°) tan (80°)

Ê Êpˆ Ê p ˆˆ = 2 Á sin 4 Á ˜ + cos 4 Á ˜ ˜ Ë 8¯ Ë 8 ¯¯ Ë

=

1 [tan (40∞) tan (20∞) tan (80∞)] tan (40∞)

Ê Êpˆ Ê p ˆˆ = 2 Á1 - 2 sin 2 Á ˜ ◊ cos 2 Á ˜ ˜ Ë 8¯ Ë 8 ¯¯ Ë

=

1 ¥ tan (60∞) tan (40∞)

Ê Êpˆ Ê p ˆˆ = Á 2 - 2 sin 2 Á ˜ ◊ 2 cos 2 Á ˜ ˜ Ë 8¯ Ë 8 ¯¯ Ë

=

3 tan (40∞)

Ê Ê Ê p ˆˆ Ê Ê p ˆˆˆ = Á 2 - Á 2 sin 2 Á ˜ ˜ ◊ Á 2 cos 2 Á ˜ ˜ ˜ Ë 8 ¯¯ Ë Ë 8 ¯¯¯ Ë Ë

= 3 cot (40∞) = 3 tan (60∞)

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181. We have tan (10°) tan (70°) 1 = [tan (50°) tan (10°) tan (70°)] tan (50°) 1 = ¥ tan (30°) tan (50°) =

1 ¥ cot (50°) 3

1 = ¥ tan (40°) 3 182. We have sin 55° – sin 19° + sin 53° – sin 17° = (sin (55°) + sin (53°)) – (sin (19°) + sin (17°)) = 2 sin (54°) cos (1°) – 2 sin (18°) cos (1°) = 2 cos (1°)[sin (54°) – sin (18°)] È 5 +1 5 - 1˘ = 2 cos (1°) Í ˙ 4 ˚ Î 4 1 = 2 cos (1°) ¥ = cos (1°) 2 183. We have Ê 2p ˆ Ê 4p ˆ Ê 6p ˆ cos Á ˜ cos Á ˜ cos Á ˜ Ë 7 ¯ Ë 7 ¯ Ë 7 ¯ Êpˆ Ê 2p ˆ Ê 4p ˆ = – cos Á ˜ cos Á ˜ cos Á ˜ Ë 7¯ Ë 7¯ Ë 7¯ =–

=-

=–

=–

=–

=–

1 È Êpˆ Ê p ˆ˘ Ê 2p ˆ Ê 4p ˆ ¥ 2 sin Á ˜ cos Á ˜ ˙ cos Á ˜ cos Á ˜ Ë 7¯ Ë 7¯˚ Ë 7¯ Ë 7¯ Ê p ˆ ÍÎ 2 sin Á ˜ Ë 7¯ È Ê 2p ˆ Ê 2p ˆ ˘ Ê 4p ˆ ¥ Í2 sin Á ˜ cos Á ˜ ˙ cos Á ˜ Ë 7¯ Ë 7 ¯˚ Ë 7¯ Êpˆ 22 sin Á ˜ Î Ë 7¯

Êpˆ Êpˆ Ê 2p ˆ fi 2 sin Á ˜ S = 2 sin Á ˜ cos Á ˜ Ë 7¯ Ë 7¯ Ë 7 ¯ Êpˆ Ê 4p ˆ Êpˆ Ê 6p ˆ + 2 sin Á ˜ cos Á ˜ + 2 sin Á ˜ cos Á ˜ Ë 7¯ Ë 7 ¯ Ë 7¯ Ë 7 ¯ p Ê 3p p ˆ fi 2 sin ÊÁ ˆ˜ S = Á sin ÊÁ ˆ˜ - sin ÊÁ ˆ˜ ˜ Ë 7¯ Ë 7 ¯ Ë 7 ¯¯ Ë Ê Ê 5p ˆ Ê 3p ˆ ˆ Ê Ê 7p ˆ Ê 5p ˆ ˆ + Á sin Á ˜ - sin Á ˜ ˜ + Á sin Á ˜ - sin Á ˜ ˜ Ë 7 ¯ Ë 7 ¯¯ Ë Ë 7 ¯ Ë 7 ¯¯ Ë Êpˆ Êpˆ fi 2 sin Á ˜ S = - sin Á ˜ Ë 7¯ Ë 7¯ 1 fi S=2 Hence, the result. Ê 1 ˆ Ê 1 ˆ 185. We have tan Á 7 ∞˜ + cot Á 7 ∞˜ Ë 2 ¯ Ë 2 ¯ Ê 1 ˆ Ê 1 ˆ sin Á 7 °˜ cos Á 7 °˜ Ë 2 ¯ Ë 2 ¯ = + Ê 1 ˆ Ê 1 ˆ cos Á 7 °˜ sin Á 7 °˜ Ë 2 ¯ Ë 2 ¯ Ê 1 ˆ Ê 1 ˆ sin 2 Á 7 °˜ + cos 2 Á 7 °˜ Ë 2 ¯ Ë 2 ¯ = 1 Ê ˆ Ê 1 ˆ sin Á 7 °˜ cos Á 7 °˜ Ë 2 ¯ Ë 2 ¯ =

2 Ê 1 ˆ Ê 1 ˆ 2 sin Á 7 °˜ cos Á 7 °˜ Ë 2 ¯ Ë 2 ¯

=

2 sin (15°) 2

1

1 Êpˆ 23 sin Á ˜ Ë 7¯

È Ê 4p ˆ ˘ Ê 4p ˆ ¥ Í2 sin Á ˜ cos Á ˜ ˙ Ë 7 ¯˚ Ë ¯ 7 Î

Ê 8p ˆ ¥ sin Á ˜ Ë 7¯ Êpˆ 23 sin Á ˜ Ë 7¯ 1

Ê ¥ sin Áp Ë Êpˆ 3 2 sin Á ˜ Ë 7¯ 1

1 Êpˆ 23 sin Á ˜ Ë 7¯

1 1 = 3= 8 2

TR_01a.indd 56

Ê 2p ˆ Ê 4p ˆ Ê 6p ˆ 184. Let S = cos Á ˜ + cos Á ˜ + cos Á ˜ Ë 7 ¯ Ë 7 ¯ Ë 7 ¯

pˆ + ˜ 7¯

Êpˆ ¥ - sin Á ˜ Ë 7¯

=

= =

3 -1 2 2 4 2 3 -1 4 2( 3 + 1) 2

= 2 2( 3 + 1)

Ê xˆ Ê xˆ 186. Let y = cos Á ˜ - 3 sin Á ˜ Ë 2¯ Ë 2¯ Ê1 3 Ê xˆ Ê xˆˆ sin Á ˜ ˜ = 2 Á cos Á ˜ Ë 2¯ Ë 2¯ ¯ 2 Ë2 Ê x pˆ = 2 cos Á + ˜ Ë 2 3¯ dy Ê x pˆ = – sin Á + ˜ Now, Ë 2 3¯ dx

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For maximum or minimum,

dy =0 dx

Ê x pˆ sin Á + ˜ = 0 Ë 2 3¯ Ê x pˆ fi ÁË + ˜¯ = p 2 3 x p 2p fi =p – = 2 3 3 4p x= fi 3 4p Hence, the value of x is 3 187. If a and b be two different roots a cos q + b sin q = c, then prove that 2ab . sin (a + b) = 2 a + b2 fi

188. We have, sin 2A + sin 2B + sin 2C = (sin 2A + sin 2B) + sin 2C = 2(sin (A + B) cos (A – B)) + sin 2C = 2(sin (p – C) ◊ cos (A – B)) + 2 sin C cos C = 2(sin C ◊ cos (A – B)) + 2 sin C cos C = 2(sin C(cos (A – B)) + cos C) = 2 sin C(cos (A – B) + cos (p – (A + B))) = 2 sin C(cos (A – B) – cos (A + B)) = 2 sin C(2 sin A sin B) = 4 sin A ◊ sin B ◊ sin C 189. We have cos 2A + cos 2B + cos 2C = (cos 2A + cos 2B) + cos 2C = 2 cos (A + B) cos (A – B) + cos 2C = 2 cos {p – C} cos (A – B) + cos 2C = –2 cos C cos (A + B) + 2 cos2 C – 1 = –1 – 2 cos C(cos (A – B) – cos C) = –1 – 2 cos C(cos (A – B) + cos (A + B)) = –1 – 2 cos C(2 cos A ◊ cos B) = –1 – 4 cos A ◊ cos B ◊ cos C 190. We have sin2 A + sin2 B – sin2 C = sin2 A + (sin2 B – sin2 C) = sin2 A + sin (B + C) sin (B – C) = sin2 A + sin (p – A) sin (B – C) = sin2 A + sin A sin (B – C) = sin A(sin A + sin (B – C)) = sin A(sin (B + C) + sin (B – C)) = sin A(2 sin B cos C) = 2 sin A ◊ sin B ◊ cos C 191.(i) We have sin2 A + sin2 B + sin2 C = 1 – cos2 A + sin2 B + sin2 C = 1 – (cos2 A – sin2 B) + (1 – cos2 C)

TR_01a.indd 57

= 2 – (cos2 A – sin2 B) – cos2 C = 2 – (cos (A + B) ◊ cos (A – B)) – cos2 C = 2 – (cos (p – C) ◊ cos (A – B)) – cos2 C = 2 + cos C(cos (A – B) – cos C) = 2 + cos C(cos (A – B) + cos (A + B)) = 2 + cos C(2 cos A ◊ cos B) = 2 + 2 cos A ◊ cos B ◊ cos C 192. We have sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C Also, cos A + cos B + cos C – 1 Ê A + Bˆ Ê A - Bˆ Ê Cˆ cos Á = 2 cos Á - 2 sin 2 Á ˜ Ë 2 ˜¯ Ë 2 ˜¯ Ë 2¯ Êp Cˆ Ê A - Bˆ Ê Cˆ = 2 cos Á - ˜ cos Á - 2 sin 2 Á ˜ Ë 2 2¯ Ë 2 ˜¯ Ë 2¯ Ê Cˆ Ê A - Bˆ Ê Cˆ = 2 sin Á ˜ cos Á - 2 sin 2 Á ˜ Ë 2¯ Ë 2 ˜¯ Ë 2¯ Ê Cˆ Ê Ê A - Bˆ Ê Cˆˆ = 2 sin Á ˜ Á cos Á - sin Á ˜ ˜ Ë 2¯Ë Ë 2 ˜¯ Ë 2 ¯¯ Ê A + Bˆˆ Ê Cˆ Ê Ê A - Bˆ = 2 sin Á ˜ Á cos Á - cos Á ˜ Ë 2 ˜¯ ˜¯ Ë 2¯Ë Ë 2 ¯ Ê Cˆ Ê Aˆ Ê Bˆ = 2 sin Á ˜ 2 sin Á ˜ sin Á ˜ Ë 2¯ Ë 2¯ Ë 2¯ Ê Aˆ Ê Bˆ ÊCˆ = 4 sin Á ˜ sin Á ˜ sin Á ˜ Ë 2¯ Ë 2¯ Ë 2¯ Thus,

sin 2A + sin 2B + sin 2C cos A + cos B + cos C - 1 =

4 sin A sin B sin C 4 sin ( A/2) sin ( B /2) sin ( B /2)

Ê Aˆ Ê Bˆ Ê Cˆ = 8 cos Á ˜ cos Á ˜ cos Á ˜ Ë 2¯ Ë 2¯ Ë 2¯ 193. (i) We have A + B + C = p fi A+B=p–C fi tan (A + B) = tan (p – C) tan A + tan B fi = - tan C 1 - tan A ◊ tan B fi tan A + tan B = –tan C(1 – tan A ◊ tan B) fi tan A + tan B = –tan C + tan A ◊ tan B ◊ tan C fi tan A + tan B + tan C = tan A ◊ tan B ◊ tan C (ii) As we know that, tan A + tan B + tan C = tan A ◊ tan B ◊ tan C Dividing both the sides by ‘tan A ◊ tan B ◊ tan C’, we get, tan A tan B = + tan A ◊ tan B ◊ tan C tan A ◊ tan B ◊ tan C tan C + =1 tan A ◊ tan B ◊ tan C

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fi

1 1 1 + + =1 tan B ◊ tan C tan A ◊ tan C tan A ◊ tan B

fi cot B ◊ cot C + cot A ◊ cot C + cot A ◊ cot B = 1 194. We have A + B + C = 2p fi A + B + C = 2p fi A + B = 2p – (C + D) fi tan (A + B) = tan{2p – (C + D)} fi tan (A + B) = –tan (C + D) tan A + tan B tan C + tan D fi =1 - tan A ◊ tan B 1 - tan C ◊ tan D fi fi

fi

fi

fi

fi

fi

(tan A + tan B)(1 – tan C ◊ tan D) = –(tan C + tan D)(1 – tan A ◊ tan B) tan A + tan B – tan A ◊ tan C ◊ tan D – tan B ◊ tan C ◊ tan D = –tan C – tan D + tan A ◊ tan B ◊ tan C + tan A ◊ tan B ◊ tan D tan A + tan B + tan C + tan D = tan A ◊ tan C ◊ tan D + tan B ◊ tan C ◊ tan D + tan A ◊ tan B ◊ tan C + tan A ◊ tan B ◊ tan D

Thus, (cot A + cot B)(cot B + cot C) (cot C + cot A) sin C sin B sin A = ¥ ¥ sin A sin B sin A sin C sin B sin C 1 = sin A sin B sin C = cosec A cosec B cosec C 195. Put x = tan A, y = tan B and z = tan C Given, xy + yz + zx = 1 fi tan A ◊ tan B + tan B ◊ tan C + tan C ◊ tan A = 1 fi tan B ◊ tan C + tan C ◊ tan A = 1 – tan A ◊ tan B fi tan C (tan B + tan A) = 1 – tan A ◊ tan B tan A + tan B 1 = fi 1 – tan A ◊ tan B tan C

tan A + tan B + tan C + tan D tan A ◊ tan B ◊ tan C ◊ tan D tan A ◊ tan B ◊ tan C = tan A ◊ tan B ◊ tan C ◊ tan D

fi

Êp ˆ tan ( A + B) = cot C = tan Á - C ˜ Ë2 ¯

fi

Êp ˆ ( A + B) = Á - C ˜ Ë2 ¯

tan A ◊ tan C ◊ tan D + tan A ◊ tan B ◊ tan C ◊ tan D tan A ◊ tan B ◊ tan D + tan A ◊ tan B ◊ tan C ◊ tan D tan B ◊ tan C ◊ tan D + tan A ◊ tan B ◊ tan C ◊ tan D

fi

( A + B + C) =

tan A + tan B + tan C + tan D tan A ◊ tan B ◊ tan C ◊ tan D 1 1 1 1 = + + + tan A tan B tan C tan D tan A + tan B + tan C + tan D tan A ◊ tan B ◊ tan C ◊ tan D = cot A + cot B + cot C + cot D tan A + tan B + tan C + tan D cot A + cot B + cot C + cot D

= tan A ◊ tan B ◊ tan C ◊ tan D cos A cos B + 194. We have (cot A + cot B) = sin A sin B cos A sin B + sin A ◊ cos B sin A sin B sin ( A + B ) sin C = = sin A sin B sin A sin B =

TR_01a.indd 58

sin A sin B sin C sin B and (cot C + cot A) = sin A sin C Similarly, (cot B + cot C ) =

p 2

Now, x y z + + 1 - x2 1 - y 2 1 - z 2 tan A tan B tan C = + + 2 2 1 - tan A 1 - tan B 1 - tan 2C

LHS =

=

1 Ê 2 tan A 2 tan B 2 tan C ˆ + + 2 ÁË 1 - tan 2 A 1 - tan 2 B 1 - tan 2C ˜¯

1 (tan 2A + tan 2B + tan 2C ) 2 1 = (tan 2A ◊ tan 2B ◊ tan 2C ) 2 1 Ê 2 tan A 2 tan B 2 tan C ˆ = Á ◊ ◊ 2 2 2 Ë 1 – tan A 1 – tan B 1 – tan 2C ˜¯ =

4 tan A ◊ tan B ◊ tan C Ê ˆ =Á 2 2 2 Ë (1 – tan A)(1 – tan B)(1 – tan C ) ˜¯ =

4xyz (1 – x 2 )(1 – y 2 )(1 – z 2 )

Hence, the result. 196. Put x = tan A, y = tan B and z = tan C Given, xy + yz + zx = 1 fi tan A ◊ tan B + tan B ◊ tan C + tan C ◊ tan A = 1

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fi

A+ B+C =

p 2

Now, LHS =

x y z + + 2 2 1+ x 1+ y 1 + z2

=

1 Ê 2x 2y 2z ˆ + + 2 2 Á 2 Ë1 + x 1+ y 1 + z 2 ˜¯

=

1 Ê 2 tan A 2 tan B 2 tan C ˆ + + 2 ÁË 1 + tan 2 A 1 + tan 2 B 1 + tan 2C ˜¯

1 (sin 2A + sin 2B + sin 2C ) 2 1 = (4 cos A ◊ cos B ◊ cos C ) 2 = 2 cos A ◊ cos B ◊ cos C 2 = sec A ◊ sec B ◊ sec C 2 = 2 (1 + tan A)(1 + tan 2 B)(1 + tan 2C ) =

=

2 (1 + x 2 )(1 + y 2 )(1 + z 2 )

Hence, the result. 197. Let A = p – b, B = b – g, C = g – a Now, A + B + C = 0 fi A + B = –C fi tan (A + B) = tan (–C) = – tan C tan A + tan B fi = - tan C 1 - tan A tan B fi fi fi

tan A + tan B = –tan C + tan A tan B tan C tan A + tan B + tan C = tan A tan B tan C tan (a – b) + tan (b – g) + tan (g – a) = tan (a – b) tan (b – g) tan (g – a)

198. We have cot A + cot B + cot C = 3 fi (cot A + cot B + cot C)2 = 3 fi cot2 A + cot2 B + cot2 C + 2(cot A cot B + cot B cot C + cot C cot A) = 3 fi cot2 A + cot2 B + cot2 C + 2 = 3 fi cot2 A + cot2 B + cot2 C = 1 fi cot2 A + cot2 B + cot2 C = (cot A cot B + cot B cot C + cot C cot A) 1 fi [(cot A – cot B)2 + (cot A – cot B)2 + 2 (cot A – cot B)2] = 0 fi fi

TR_01a.indd 59

(cot A – cot B)2 = 0, (cot B – cot C)2 = 0, (cot C – cot A)2 = 0 cot A = cot B, cot B = cot C, cot C = cot A

fi cot A = cot B = cot C So, D is an equilateral. 199. Given expression is x + y + z = xyz Put x = tan A, y = tan B and Z = tan C So, tan A + tan B + tan C = tan A tan B tan C tan A + tan B = –tan C(1 – tan A tan B) tan A + tan B = tan (p – C ) (1 - tan A tan B) tan (A + B) = tan (p – C) (A + B) = (p – C) (A + B + C) = p (3A + 3B + 3C) = 3p (3A + 3B) = 3p – 3C tan (3A + 3B) = tan (3p – 3C) tan 3A + tan 3B = - tan 3C 1 - tan 3A tan 3B tan 3A + tan 3B + tan 3C = tan 3A ◊ tan 3B ◊ tan 3C 3 tan A - tan 3 A 3 tan B - tan 3 B 3 tan C - tan 3C + + 1 - 3 tan 2 A 1 - 3 tan 2 B 1 - 3 tan 2C =

3 tan A - tan 3 A 3 tan B - tan 3 B 3 tan C - tan 3C ◊ ◊ 1 - 3 tan 2 A 1 - 3 tan 2 B 1 - 3 tan 2C 3x - x3 3y - y 3 3z - z 3 + + 1 - 3x 2 1 - 3y 2 1 - 3z 2

3x - x3 3y - y 3 3z - z 3 ◊ ◊ 1 - 3x 2 1 - 3y 2 1 - 3z 2 Hence, the result. 200. We have 1 + cos 56° + cos 58° – cos 66° = (1 – cos (66°)) + (cos (58°)) + cos (56°)) = 2 sin2(33°) + 2 cos (57°) cos (1°) = 2 sin2(33°) + 2 sin (33°) cos (1°) = 2 sin (33°)(sin (33°) + cos (1°)) = 2 sin (33°)(cos (57°) + cos (1°)) = 2 sin (33°)(2 cos (29°) cos (28°)) = 4 cos (29°) sin (28°) sin (33°) Hence, the result. 201. Let S = sin a + sin (a + b) + sin (a + 2b) + sin (a + 3b) + … + sin (a + (n – 1)a) Now, bˆ bˆ Ê bˆ Ê Ê 2 sin a sin Á ˜ = cos Á a - ˜ - cos Á a + ˜ Ë 2¯ Ë ¯ Ë 2 2¯ =

bˆ 3b ˆ Ê bˆ Ê Ê 2 sin (a + b ) ◊ sin Á ˜ = cos Á a + ˜ - cos Á a + ˜ Ë 2¯ Ë Ë 2¯ 2¯ 3b ˆ 5b ˆ Ê bˆ Ê Ê 2 sin (a + 2b ) ◊ sin Á ˜ = cos Á a + ˜¯ - cos ÁË a + ˜ Ë 2¯ Ë 2 2¯ o

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Adding,we get Ê bˆ 2 sin (a + (n - 1)b ) ◊ sin Á ˜ Ë 2¯ (2n - 3)b ˆ (2n - 1)b ˆ Ê Ê = cos Á a + ˜¯ - cos ÁË a + ˜¯ Ë 2 2 (n - 1)b ˆ Ê Ê bˆ = 2 sin Á a + ˜ ¥ sin Á ˜ Ë Ë 2¯ 2 ¯ Ê nb ˆ sin Á ˜ Ë 2¯ (n - 1)b ˆ Ê ¥ sin Á a + Thus S = ˜ Ë 2 ¯ Ê bˆ sin Á ˜ Ë 2¯ Ê nb ˆ sin Á ˜ Ë 2¯ n -1 ˆ Ê S= ¥ 2 cos Á a + b˜ Ë 2 ¯ Ê bˆ sin Á ˜ Ë 2¯

fi

202. Do yourself. 203. Do yourself. 204. Let S = cos a + cos (a + b) + cos (a + b) + cos (a + b) + … + cos (a + (n – 1)a) Ê bˆ Now, 2 cos a sin Á ˜ Ë 2¯ bˆ bˆ Ê Ê = sin Á a + ˜ - sin Á a - ˜ Ë Ë 2¯ 2¯ Ê bˆ 2 cos (a + b ) sin Á ˜ Ë 2¯ 3b ˆ bˆ Ê Ê = sin Á a + ˜¯ - sin ÁË a + ˜¯ Ë 2 2 Ê bˆ 2 cos (a + 3b ) sin Á ˜ Ë 2¯ 5b ˆ 3b ˆ Ê Ê = sin Á a + ˜ - sin ÁË a + ˜ Ë 2¯ 2¯ o

Ê bˆ 2 cos (a + (n - 1)b ) sin Á ˜ Ë 2¯ (2n - 1)b ˆ (2n - 3)b ˆ Ê Ê = sin Á a + ˜ - sin ËÁ a + Ë ¯ ¯˜ 2 2

Adding all we get, 2n - 1 ˆ Ê bˆ Ê 2 sin Á ˜ ¥ S = sin Á a + b ˜ - sin (a - b ) Ë 2¯ Ë ¯ 2 fi

Ê bˆ 2 sin Á ˜ ¥ S Ë 2¯ n -1 ˆ Ê Ê nb ˆ b ˜ ¥ sin Á ˜ = 2 cos Á a + Ë Ë 2¯ 2 ¯

205. Do yourself

TR_01a.indd 60

sin x sin (n + 1) x ◊ sin (n + 2) x sin [(n + 2)x - (n + 1)x] fi tn = sin (n + 1)x ◊ sin (n + 2)x sin (n + 2)x cos (n + 1)x = sin (n + 1)x ◊ sin (n + 2)x cos (n + 2)x sin (n + 1)x sin (n + 1)x ◊ sin (n + 2)x = cot (n + 1)x – cot (n + 2)x Thus, t1 = cot 2x – cot 3x t2 = cot 3x – cot 4x t3 = cot 4x – cot 5x o tn = cot (n + 1)x – cot (n + 2)x Adding all we get, S = cot 2x – cot (n + 2)x

206. Let

tn =

LEVEL III 1. We have tan a + 2 tan 2a + 4 tan 4a + 8 cot 8a = cot a – (cot a – tan a) + 2 tan 2a + 4 tan 4a + 8 cot 8a = cot a – 2(cot 2a – tan 2a) + 4 tan 4a + 8 cot 8a = cot a – 2.2cot 4a + 4 tan 4a + 8 cot 8a = cot a – 4(cot 4a – tan 4a) + 8 tan 8a = cot a – 4.2 cot 8a + 8 cot 8a = cot a – 8 cot 8a + 8 cot 8a = cot a 2. We have tan 9° – tan 27° – tan 63° + tan 81° = (tan 9° + tan 81°) – (tan 27° + tan 63°) = (tan 9° + cot 9°) – (tan 27° + cot 27°) 1 1 Ê ˆ Ê ˆ =Á Ë sin 9° cos 9° ¯˜ ËÁ sin 27° cos 27° ¯˜ 2 2 Ê ˆ Ê ˆ =Á -Á ˜ Ë 2 sin 9° cos 9° ¯ Ë 2 sin 27° cos 27° ˜¯ 2 ˆ Ê 2 =Á Ë sin 18° sin 54° ˜¯ 2 ˆ Ê 2 =Á Ë sin 18° cos 36° ˜¯ 8 ˆ Ê 8 =Á 5 + 1˜¯ Ë 5 -1 Ê 8( 5 + 1 - 5 + 1) ˆ =Á ˜ Ë ( 5 - 1) ( 5 + 1) ¯ Ê 8 ¥ 2ˆ =Á =4 Ë 4 ˜¯

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3. We have sin x sin 3x sin 9x + + cos 3x cos 9x cos 27x

fi

1 Ê 2 sin x cos x 2 sin 3x cos 3x 2 sin 9x cos 9xˆ = Á + + 2 Ë cos 3x cos x cos 3x cos 9x cos 27x cos 9x˜¯ 1 Ê sin 2x sin 6x sin 18x ˆ = Á + + 2 Ë cos 3x cos x cos 3x cos 9x cos 9x cos 27x˜¯ 1 Ê sin (3x – x) sin (9x – 3x) sin (27x – 9x) ˆ = Á + + 2 Ë cos 3x cos x cos 3x cos 9x cos 9x cos 27x˜¯ 1 Ê sin 3x cos x – cos 3x sin x = Á 2Ë cos 3x cos x +

sin 9x cos 3x – cos 9x sin 3x cos 3x cos 9x

+

sin 27x cos 9x – cos 27x sin 9xˆ ˜¯ cos 9x cos 27x

From (i) and (ii) we get, 4 5 4 5 ¥ 27 tan ( x + y ) = 3 3 = = 15 15 12 ¥ 3 3 127 5. We have Êpˆ Ê 3p ˆ Ê 5p ˆ Ê 7p ˆ sin 4 Á ˜ + sin 4 Á ˜ + sin 4 Á ˜ + sin 4 Á ˜ Ë 16 ¯ Ë 16 ¯ Ë 16 ¯ Ë 16 ¯

Êpˆ Ê 3p ˆ Ê 3p ˆ Êpˆ = sin 4 Á ˜ + sin 4 Á ˜ + cos 4 Á ˜ + cos 4 Á ˜ Ë16¯ Ë 16 ¯ Ë 16 ¯ Ë16¯ Ê Êpˆ Ê p ˆˆ Ê Ê 3p ˆ Ê 3p ˆ ˆ = Ásin 4 Á ˜ + cos 4 Á ˜˜ + Ásin 4 Á ˜ + cos 4 Á ˜˜ Ë16¯ Ë16¯¯ Ë Ë 16 ¯ Ë 16 ¯¯ Ë Êpˆ Êpˆ Ê 3p ˆ Ê 3p ˆ = 2 - 2 sin 2 Á ˜ ◊ cos 2 Á ˜ - 2 sin 2 Á ˜ ◊ cos 2 Á ˜ Ë16¯ Ë16¯ Ë 16 ¯ Ë 16 ¯ 2 2 1 ÊÊ Ê Êpˆ Ê p ˆˆ Ê 3p ˆ Ê 3p ˆˆ ˆ = 2 - ÁÁ 2 sin Á ˜ cos Á ˜˜ + Á 2 sin Á ˜ cos Á ˜˜ ˜ Ë16¯ Ë16¯¯ Ë 16 ¯ Ë 16 ¯¯ ¯ Ë 2 ËË

=2-

1 Ê 2 Êpˆ Ê 3p ˆ ˆ sin Á ˜ + sin 2 Á ˜˜ Á Ë 8¯ Ë 8 ¯¯ 2Ë

=2-

1 Ê 2 Êpˆ Ê p p ˆˆ sin Á ˜ + sin 2 Á - ˜˜ Á Ë ¯ Ë 2 8 ¯¯ 2Ë 8

=2-

1 Ê 2 Êp ˆ Ê p ˆˆ sin Á ˜ + cos 2 Á ˜˜ Ë 8¯ Ë 8 ¯¯ 2 ÁË

=2-

1 2

=

fi

TR_01a.indd 61

sin x =

5 4 2

…(ii)

Ê p 3p ˆ Êp p ˆ + sin 4 Á - ˜ + sin 4 Á - ˜ Ë 2 16 ¯ Ë 2 16¯

1 (tan 3x – tan x + tan 9x – tan 3x + tan 27x – tan 9x) 2 1 = (tan 27x – tan x) 2 sin x 1 cos x 3 = , = 4. We have sin y 2 cos y 2 tan x 1 = tan y 3 Now tan (x + y)

2 Also, sin y = 2 sin x, cos y = cos x 3 sin2 y + cos2 y 4 = 4 sin 2 x + cos 2 x 9 36 sin 2 x + 4 cos 2 x = 9 2 32 sin x + 4 = 9 2 32 sin x + 4 =1 fi 9 fi 32 sin2 x + 4 = 9 fi 32 sin2 x = 5 5 fi sin 2 x = 32

5 3 3

Êpˆ Ê 3p ˆ = sin 4 Á ˜ + sin 4 Á ˜ Ë16¯ Ë 16 ¯

=

tan x + tan y 1 - tan x tan y tan x + 3 tan x = 1 - tan x ◊ 3 tan x 4 tan x = 1 - 3 tan 2 x

tan x =

…(i)

3 2 6. We have =

3 2 2(cos (a – b) + cos (b – g) + cos (g – a) + 3 = 0 2(cos a cos b + cos b cos g + cos g cos a) + 2(sin a sin b + sin b sin g + sin g sin a) + 3 = 0 (cos2 a + sin2 a) + (cos2 b + sin2 a) + (cos2 g + sin2 g) 2(cos a cos b + cos b cos g + cos g cos a) + 2(sin a sin b + sin b sin g + sin g sin a) = 0 cos2 a + cos2 b + cos2 g + 2 cos a cos b + 2 cos b cos g + 2 cos g cos a + sin2 a + sin2 b + sin2 g

cos (a – b ) + cos (b - g ) + cos (g - a ) = fi fi fi

fi

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+ 2 sin a sin b + 2 sin b sin g + 2 sin g sin a = 0 fi (cos a + cos b + cos g)2 + (sin a + sin b + sin g)2 = 0 fi (cos a + cos b + cos g) = 0 and (sin a + sin b + sin g) = 0 7. We have sin a sin b – cos a cos b + 1 = 0 fi cos a cos b sin a sin b = 1 fi cos (a + b) = 1 Therefore, sin (a + b ) = 1 - cos (a + b ) = 1 - 1 = 0 2

Now, 1 + cot a ◊ tan b cos a ◊ sin b =1+ sin a ◊ cos b sin a ◊ cos b + cos a ◊ sin b = sin a ◊ cos b sin (a + b ) = sin a ◊ cos b =0 8. Given, b + g = a fi tan (b + g) = tan a fi tan (b + g) = tan (90° – a) fi tan (b + g) = cot a tan b + tan g fi = cot a 1 - tan b tan g fi

tan b + tan g = cot b 1 - tan b tan g

fi fi fi fi fi

tan b + tan g = cot b – tan b ◊ cot b tan g tan b + tan g = cot b – tan g tan b + tan g = cot (90° – a) – tan g tan b + tan g = tan a – tan g tan a = tan b + 2 tan g

Ê p ˆ sin (p /24) 9. We have tan Á ˜ = Ë 24 ¯ cos (p /24) = = = = =

TR_01a.indd 62

2 sin (p /24) cos (p /24) 2 cos 2 (p /24) sin (p /12) 1 + cos (p /12) 3 -1 2 2 + 3 +1 3 -1 (2 2 + ( 3 + 1))

¥

(2 2 - ( 3 + 1)) (2 2 - ( 3 + 1))

( 3 - 1)(2 2 - ( 3 + 1)) (8 – ( 3 + 1) 2 )

= = =

2 6 - 3 - 3 - 2 2 + 3 +1 (4 – 2 3) 2 6 -2-2 2 (4 – 2 3) 6 -1- 2 (2 – 3)

= ( 6 - 2 - 1)(2 + 3) = 2 6 - 2 2 - 2 + 18 - 6 - 3 = 6+ 2- 4- 3 = 6 -2- 3+ 2 = ( 3 - 2)( 2 - 1) Thus, a = 3, b = 2, c = 2 and d = 1 Hence, the value of (a + b + c + d + 2) =3+2+2+1+2 = 10 10. Given, x y z = = = m(say) 2p ˆ 2p ˆ cos q Ê Ê cos Á q + cos q ˜ ÁË ˜ Ë 3 ¯ 3 ¯ Now, x + y + z Ê 2p ˆ 2p ˆ ˆ Ê Ê + cos Á q = m ¥ Á cos q + cos Á q + ˜ ˜ Ë ¯ Ë Ë 3 3 ¯ ˜¯ Ê Ê 2p ˆ ˆ = m ¥ Á cos q + 2 cos q ◊ cos Á ˜ ˜ Ë 3 ¯¯ Ë Ê Ê 1ˆˆ = m ¥ Á cos q + 2 cos q ◊ Á - ˜ ˜ Ë 2¯ ¯ Ë = m ¥ (cos q – cos q) =0 Hence, the value of x + y + z is zero. 11. We have sin (25°) sin (35°) sin (85°) = sin (25°) sin (35°) sin (85°) 1 = (4 sin (60° - 25) sin (25°) sin (60° + 25°)) 4 1 = ¥ sin (3 ¥ 25°) 4 1 = ¥ sin (75°) 4 1 = ¥ cos (15°) 4 1 Ê 3 + 1ˆ = ¥Á 4 Ë 2 2 ˜¯ Ê 3 + 1ˆ =Á Ë 8 2 ˜¯ Ê 3 + 1ˆ =Á Ë 128 ˜¯

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Thus, a = 3, b = 1 and c = 128 Hence, the value of (a + b + c + 2) = 3 + 1 + 128 – 2 = 130 12. We have 3 cot (20∞) - 4 cos (20∞) =

3 cos (20°) - 4 sin (20°) cos (20°) sin (20°)

Ê 3 ˆ 2Á cos (20°) - 2 sin (20°) cos (20°)˜ Ë 2 ¯ = sin (20°) 2(sin (60°) cos (20°) - sin (40°)) = sin (20°) (2 sin (60°) cos (20°) - 2 sin (40°)) = sin (20°) (sin (80∞) + sin (40°) - 2 sin (40°)) = sin (20°) (sin (80∞) - sin (40°)) = sin (20°) 2 cos (60∞) sin (20∞) = sin (20°) =1 13. We have sin (2°) + sin (4°) + sin (6°) + sin (8°) + … + sin (180°) = sin (2°) + sin (2° + 2°) + sin (2° + 2.2°) + sin (2° + 3.2°) + … + (2° + (90 – 1)2°) Ê 90∞ ◊ 2° ˆ sin Á Ë 2 ˜¯ 2° ˆ Ê = ¥ sin Á 2∞ + (90∞ - 1) ˜ Ë 2¯ Ê 2° ˆ sin Á ˜ Ë 2¯ 1 ¥ sin (91∞) sin (1∞) cos (1°) = sin (1∞) =

= cot (1°) 14. We have 2p ˆ 2.2p ˆ Ê p ˆ Ê p Ê p sin Á + sin Á + + sin Á + Ë 2013˜¯ Ë 2013 2013˜¯ Ë 2013 2013˜¯ 1006.2p ˆ Ê p + ...... + sin Á + ˜ Ë 2013 2013 ¯ Ê Ê 2p ˆˆ Á Á ˜˜ Ê Ê 2p ˆ ˆ sin Á 2013 ◊ Á 2013˜˜ Ë 2 ¯¯ Ë Á p Á ˜˜ = ¥ sin Á + (2012) Á 2013˜˜ Ë Ë 2013 2 ¯¯ Ê 2p ˆ Á ˜ sin Á 2013˜ Ë 2 ¯

15. Put y = sin (4q) Then tan ( y ) =

1 + 1 + sin (4q ) 1 + 1 - sin (4q ) 1 + (cos (2q ) + sin (2q )) 2

fi

tan ( y ) =

fi

tan ( y ) =

1 + cos (2q ) + sin (2q ) 1 + (cos (2q ) - sin (2q ))

fi

tan ( y ) =

2 cos 2 (q ) + sin (2q ) 2 cos 2 (q ) - sin (2q )

fi

tan ( y ) =

2 cos 2 (q ) + 2 sin (q ) cos (q ) 2 cos 2 (q ) - 2 sin (q ) cos (q )

fi

tan ( y ) =

cos (q ) + sin (q ) cos (q ) - sin (q )

fi fi

1 + (cos (2q ) - sin (2q )) 2

Êp ˆ tan ( y ) = tan Á - q ˜ Ë4 ¯ p y = -q 4

fi 4y = p – 4q fi sin (4y) = sin (p – 4q) = sin (4q) = y Hence, the result. 16. We have Êp ˆ 1 + tan a cos a + sin a = tan Á + a ˜ = Ë4 ¯ 1 - tan a cos a - sin a fi

Êp ˆ 1 + sin 2a tan 2 Á + a ˜ = Ë4 ¯ 1 - sin 2a

Ê p yˆ Ê p xˆ Thus, tan Á + ˜ = tan 3 Á + ˜ , Ë 4 2¯ Ë 4 2¯ fi

1 + sin y (1 + sin x)3 = 1 - sin y (1 – sin x)3

Applying componendo and dividendo, we get, fi

2 sin y 2(3 sin x + sin 3 x) = 2 2(1 + 3 sin 2 x)

fi

sin y =

(3 sin x + sin 3 x) (1 + 3 sin 2 x)

Hence, the result. tan a + tan g 1 + tan a tan g sin a cos g + cos a sin g = cos a cos g + sin a sin g sin (a + g ) = cos (a - y )

17. Given, tan b =

=0

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Thus, 2 tan b sin (2b ) = 1 + tan 2 b =

fi

Ê sin (a + g ) ˆ 2Á Ë cos (a – g ) ˜¯

fi

sin 2 (a + g ) cos 2 (a – g ) 2 sin (a + g ) cos (a – g ) = cos 2 (a – g ) + sin 2 (a + g ) 1+

fi

sin ((a + g ) + (a – g )) + sin ((a + g ) - (a – g )) = 1 + sin 2 (a + g ) - sin 2 (a – g ) sin (2a ) + sin (2g ) = 1 + sin (a + g + a - g ) sin (a + g - a + g ) =

Now, (1 – sin q)(1 – cos q) = 1 – sin q – cos q + sin q cos q = 1 – (sin q + cos q) + sin q cos q Ê 10 ˆ 1 Ê 10 ˆ = 1 - Á –1 + ˜¯ + Á - 10 ˜ Ë Ë ¯ 2 2 4 Ê = Á2 + Ë

sin (2a ) + sin (2g ) 1 + sin (2a ) sin (2g )

Ê 10 - 2 5 ˆ = 8 ¥ Á1 ˜¯ Ë 4 = (8 - 2 10 - 2 5 ) = (8 - 2 (5 + 5)(3 - 5)) = ((5 + 5) + (3 - 5) - 2 (5 + 5) (3 - 5)) = ( (5 + 5) - (3 - 5)) 2

20. We have 3 sin x + 4 cos x = 5 Let y = 3 cos x – 4 sin x Now, y2 + 52 = (3 cos x – 4 sin x)2 + (3 sin x + 4 cos x)2 = 9 cos2 x + 16 sin2 x – 24 sin x cos x + 9 sin2 x + 16 cos2 x + 24 sin x cos x 2 fi y + 25 = 25 fi y2 = 0 fi y=0 fi 3 cos x – 4 sin x = 0 fi 3 cos x = 4 sin x fi tan x = 3/4 Hence, the value of 2 sin x + cos x + 4 tan x Ê 3ˆ Ê 4 ˆ Ê 3ˆ = 2Á ˜ + Á ˜ + 4Á ˜ = 2 + 3 = 5 Ë 5¯ Ë 5 ¯ Ë 4¯

fi 4 sin (27°) = ( (5 + 5) - (3 - 5)) Thus, a = 5, b = 5, c = 3 and d = 5 Now, a + b + c + d + 2 =5+5+3+5+2 = 20. 5 19. We have (1 + sin q )(1 + cos q ) = 4 1 + sin q + cos q + sin q cos q =

fi

Ê t 2 - 1ˆ 5 1+ t + Á = Ë 2 ˜¯ 4

fi fi fi

TR_01a.indd 64

(sin q + cos q = t, say) Ê t 2 - 1ˆ 1 t+Á = Ë 2 ˜¯ 4 1 t 2 + 2t - 1 = 2 2 2t + 4t – 3 = 0

5ˆ ˜ - 10 4¯

Ê 13 ˆ = Á - 10 ˜ Ë4 ¯

Hence, the result. 18. We have 16 sin2 (27°) = 8 ¥ 2 sin2 (27°) = 8 ¥ (1 – cos (54°))

fi

- 4 ± 16 + 24 4 - 4 ± 2 10 1 = = -1 ± 10 4 2 1 t = -1 + 10 2 1 sin q + cos q = -1 + 10 2 t=

5 4

21. We have cos A = tan B fi cos2 A = tan2 B fi cos2 A = sec2 B – 1 fi 1 + cos2 A = sec2 B fi 1 + cos2 A = sec2 B = cot2 C fi 1 + cos2 A = cot2 C fi

2 – sin 2 A =

cos 2C cos 2C = sin 2C 1 - cos 2C

fi

2 – sin 2 A =

tan 2 A 1 - tan 2 A

fi

2 – sin 2 A =

sin 2 A cos 2 A - sin 2 A

fi

2 – sin 2 A =

sin 2 A 1 – 2 sin 2 A

2/10/2017 4:08:04 PM

1.65

The Ratios and Identities

fi fi fi fi

2 – 4 sin2 A – sin2 A + 2 sin4 A = sin2 A 2 sin4 A – 6 sin2 A + 2 = 0 sin4 A – 3 sin2 A + 1 = 0 3± 9- 4 sin 2 A = 2

fi

sin 2 A =

3± 5 2

fi

sin 2 A =

6±2 5 4

fi

Ê 5 - 1ˆ sin A = Á Ë 2 ˜¯

fi

Ê 5 - 1ˆ sin A = Á Ë 2 ˜¯

fi

Ê 5 - 1ˆ sin A = 2 Á = 2 sin (18°) Ë 4 ˜¯

2

4 4 + -4 p Ê ˆ Ê 2 2 3p ˆ sin Á ˜ sin Á ˜ Ë 8¯ Ë 8 ¯ 4 4 = + -4 Êpˆ Êp pˆ sin 2 Á ˜ sin 2 Á - ˜ Ë 8¯ Ë 2 8¯ 4 4 + -4 = p Ê ˆ Êpˆ sin 2 Á ˜ cos 2 Á ˜ Ë 8¯ Ë 8¯ =

=4

2

=

=

Similarly, we can prove that, sin B = 2 sin (18°) = sin C Êpˆ Ê 3p ˆ Ê 5p ˆ Ê 7p ˆ tan 2 Á ˜ + tan 2 Á ˜ + tan 2 Á ˜ + tan 2 Á ˜ Ë 16 ¯ Ë 16 ¯ Ë 16 ¯ Ë 16 ¯ Êpˆ Ê 3p ˆ = tan 2 Á ˜ + tan 2 Á ˜ Ë 16 ¯ Ë 16 ¯ Ê p 3p ˆ Êp p ˆ + tan 2 Á + tan 2 Á - ˜ Ë 2 16 ˜¯ Ë 2 16 ¯ Êpˆ Êpˆ Ê 3p ˆ Ê 3p ˆ = tan 2 Á ˜ + cot 2 Á ˜ + tan 2 Á ˜ + cot 2 Á ˜ Ë 16 ¯ Ë 16 ¯ Ë 16 ¯ Ë 16 ¯ 2

Ê Êpˆ Ê p ˆˆ = Á tan Á ˜ + cot Á ˜ ˜ + Ë 16 ¯ Ë 16 ¯ ¯ Ë 2

Ê Ê 3p ˆ Ê 3p ˆ ˆ ÁË tan ÁË 16 ˜¯ + cot ÁË 16 ˜¯ ˜¯ - 4 1 + Êpˆ 2Ê p ˆ sin Á ˜ cos Á ˜ Ë 16 ¯ Ë 16 ¯ 2

1 Ê 3p ˆ Ê 3p ˆ sin 2 Á ˜ cos 2 Á ˜ Ë 16 ¯ Ë 16 ¯ 4 + = 2 Ê Êpˆ Ê p ˆˆ ÁË 2 sin ÁË 16 ˜¯ cos ÁË 16 ˜¯ ˜¯

Êpˆ Êpˆ sin 2 Á ˜ cos 2 Á ˜ Ë 8¯ Ë 8¯ 8

-4

Ê Êpˆ Ê p ˆˆ ÁË 2 sin ÁË 8 ˜¯ cos ÁË 8 ˜¯ ˜¯

2

-4

8 -4 2Êpˆ sin Á ˜ Ë 4¯ 8 = - 4 = 16 - 4 = 12 1 2 23. We have sin (1°) ◊ sin (2°) ◊ sin (3°) … sin (89°) = sin (1°) ◊ sin (2°) ◊ sin (3°) … sin (44°) ◊ sin (45°) sin (46°) ◊ sin (47°) ◊ sin (48°) … sin (89°) = sin (1°) ◊ sin (2°) ◊ sin (3°) … sin (44°) ◊ sin (45°) cos (44°) ◊ cos (43°) ◊ cos (42°) … cos (1°) = sin (1°) ◊ sin (2°) ◊ sin (3°) … sin (44°) ◊ sin (45°) cos (1°) ◊ cos (2°) ◊ cos (3°) … cos (44°)

Ê Ê 3p ˆ Ê 3p ˆ ˆ ÁË 2 sin ÁË 16 ˜¯ cos ÁË 16 ˜¯ ˜¯

1 1 ¥ 44 (sin (2°) sin (4°) sin (6°) ... sin (88°)) 2 2 1 = 89/2 (sin (2°) sin (4°) sin (6°) ... sin (88°)) 2 =

Thus, sin (1°) ◊ sin (3°) ◊ sin (5°) … sin (89°) 1 6 = 89/2 2

-4

Therefore, n =

4

TR_01a.indd 65

4

=

22. We have

=

1 1 Ê ˆ + -4 Á 2 Êpˆ ˜ p Ê ˆ 2 Á sin Á ˜ cos Á ˜ ˜ Ë 8¯ Ë 8¯¯ Ë

2

-4

89 2

24. We have (1 + tan (1°))(1 + tan (2°))(1 + tan (3°)) … (1 + tan (45°)) = 2n fi (1 + tan (1°))(1 + tan (44°))(1 + tan (2°)) (1 + tan (43°)) … (1 + tan (22°))(1 + tan (23°)) ¥ (1 + tan (45°)) = 2n fi 222 ¥ (1 + 1) = 2n

2/10/2017 4:08:05 PM

1.66

Trigonometry Booster

fi 2n = 23 fi n = 23 Hence, the value of n is 23. 25. We have 3p ˆ Ê 7p ˆ Ê Ê 3p ˆ cos Á ˜ = cos Á p ˜ = - cos ÁË ˜¯ Ë 10 ¯ Ë 10 ¯ 10 pˆ Ê 9p ˆ Ê Êpˆ cos Á ˜ = cos Á p - ˜ = - cos Á ˜ Ë 10 ¯ Ë Ë 10 ¯ 10 ¯ Therefore, Ê Ê p ˆˆ Ê Ê 3p ˆ ˆ Ê Ê 7p ˆ ˆ ÁË1 + cos ÁË 10 ˜¯ ˜¯ ÁË1 + cos ÁË 10 ˜¯ ˜¯ ÁË1 + cos ÁË 10 ˜¯ ˜¯ Ê Ê 9p ˆ ˆ ÁË1 + cos ÁË 10 ˜¯ ˜¯ Ê Ê p ˆˆ Ê Ê 3p ˆ ˆ = Á1 + cos Á ˜ ˜ Á1 + cos Á ˜ ˜ Ë ¯ Ë 10 ¯ ¯ Ë 10 ¯ Ë Ê Ê 3p ˆ ˆ Ê Ê p ˆˆ ÁË1 - cos ÁË 10 ˜¯ ˜¯ ÁË1 - cos ÁË 10 ˜¯ ˜¯ Ê Ê p ˆˆ Ê Ê 3p ˆ ˆ = Á1 - cos 2 Á ˜ ˜ Á1 - cos 2 Á ˜ ˜ Ë 10 ¯ ¯ Ë Ë 10 ¯ ¯ Ë Êpˆ Ê 3p ˆ = sin 2 Á ˜ sin 2 Á ˜ Ë 10 ¯ Ë 10 ¯ = sin 2 (18∞) sin 2 (45∞) 2

Ê 5 - 1ˆ Ê 5 + 1ˆ =Á Ë 4 ˜¯ ÁË 4 ˜¯

2

2

1 Ê 5 – 1ˆ =Á = Ë 16 ˜¯ 16 26. We have cos (60°) cos (36°) cos (42°) cos (78°) =

1 Ê 5 + 1ˆ 1 Á ˜ ◊ (2 cos 78° cos 42°) 2Ë 4 ¯ 2

=

1 Ê 5 + 1ˆ Á ˜ (cos 120° + cos 36°) 4Ë 4 ¯

=

1 Ê 5 + 1ˆ Ê 1 5 + 1ˆ Á ˜Á– + ˜ 4Ë 4 ¯Ë 2 4 ¯

1 Ê 5 + 1ˆ Ê 5 – 1ˆ Á ˜Á ˜ 4Ë 4 ¯Ë 4 ¯ (5 - 1) 4 1 = = = 64 64 16 27. We have f6(q) = sin6 q + cos6 q = 1 – 3 sin2 q cos2 q Also, f4(q) = sin4 q + cos4 q = 1 – 2 sin2 q cos2 q 1 1 Now, f 6 (q ) - f 4 (q ) 6 4 =

TR_01a.indd 66

1 1 (1 - 3 sin 2q cos 2q ) - (1 - 2 sin 2q cos 2q ) 6 4 1 1 1 1 = - sin 2q cos 2q - + sin 2q cos 2q 6 2 4 2 1 1 = 6 4 1 =12 28. We have sin2 (sin q) + cos2 (cos q) = sin2 (cos q) + cos2 (cos q) + sin2 (sin q) – sin2 (cos q) = (sin2 (cos q) + cos2 (cos q)) + sin2 (sin q) – sin2 (cos q) = 1 + (sin2 (sin q) – sin2 (cos q)) Maximum value of f (q) =

Ê Ê Ê Ê p ˆˆ Ê p ˆˆˆ = 1 + Á sin 2 Á sin Á ˜ ˜ - sin 2 Á cos Á ˜ ˜ ˜ Ë 2¯¯ Ë 2¯¯¯ Ë Ë Ë 2 =1 + sin (1) Minimum value of f (q) = 1 + (sin2(sin (q)) – sin2(cos (q))) = 1 – sin2(1) 29. We have f (q) = (3 sin (q) – 4 cos (q) – 10) (3 sin (q) + 4 cos (q) – 10) = (9 sin2(q) – 16 cos2(q)) –10(3 sin q + 4 cos q) – 10(3 sin q – 4 cos q) = (9 sin2(q) – 16 cos2(q)) –10(3 sin q + 4 cos q) + 3 sin q – 4 cos q) = (9 sin2(q) – 16 cos2(q)) – 60 sin (q) = 25 sin2 q – 60 sin (q) – 16 = (5 sin q – 6)2 – 36 – 16 = (5 sin q – 6)2 – 52 Hence, the minimum value of f (q) = 121 – 52 = 69 30. Now, 0 < sin2010 q £ sin2 q …(i) and 0 < cos2014 q £ cos2 q …(ii) Adding (i) and (ii), we get, 0 < sin2010 q + cos2014 q £ sin2 q + cos2 q fi 0 6 (b) 2 £ a £ 6 (c) a > 2 (d) None 34. The number of all possible triplets (a1, a2, a3) such that a1 + a2 cos 2x + a3 sin2 x = 0 for all x, the number of possible 5-tuplets is (a) 0 (b) 1 (c) 2 (d) None 35. If a1 + a2 sin x + a3 cos x + a4 sin 2x + a5 cos 2x = 0 holds for all x, then the number of possible 5-tuplets is (a) 0 (b) 1 (c) 2 (d) infinity 36. The number of solution of the equation x 1 + sin x ◊ sin 2 = 0 in [–p, p] is 2 (a) 0 (b) 1 (c) 2 (d) 3 37. The solution of sin4 x + cos4 x + sin 2x + a = 0 is solvable for 1 1 (b) –3 £ a £ 1 (a) - £ a £ 2 2 3 1 (c) - £ a £ (d) –1 £ a £ 1 2 2 38. The equation sin4 x – 2 cos2 x + a2 = 0 is solvable for (a) - 3 £ a £ 3

(b) - 2 £ a £ 2

(c) –1 £ a £ 1 (d) None 39. The number of pairs (x, y) satisfying the equations sin x + sin y = sin (x + y) and |x| + |y| = 1, is (a) 2 (b) 4 (c) 6 (d) infinity 40. The value of ‘a’ for which the equation 4 cosec2[p (a + x)] + a2 – 4a = 0, has a real solution, if (a) a = 1 (b) a = 2 (c) a = 3 (d) None 41. If sin x + cos x = (a) x =

p ,y=1 4

y+

1 , x Π[0, p], then y (b) y = 0

3p 4 42. |tan x + sec x| = |tan x| + |sec x|, x Π[0, 2 p], if x belongs to that interval (c) y = 2

(d) x =

(a) [0, p]

p p (b) ÈÍ0, ˜ˆ » ÁÊ , p ˙˘ ¯ Ë Î 2 2 ˚

È 3p ˆ Ê 3p ˘ (c) Í0, ˜ » Á , 2p ˙ (d) (p, 2p] ¯ Ë Î 2 2 ˚

TR_02.indd 8

5

43. The number of solutions of

 cos (rx) = 5 in the inter-

r =1

val [0, 2p] are (a) 0 (b) 1 (c) 5 (d) 10 44. If f(x) = max {tan x, cot x}, the number of roots of the 1 equation f ( x) = in (0, 2p) are 2+ 3 (a) 0 (b) 2 (c) 4 (d) 45. If sin x + cos x + tan x + cot x + sec x + cosec x = 7 and sin 2x = a - b c , then a – b + 2c is (a) 0

(b) 14

(c) 2

(d)

3 2

46. If sin4 x + cos4 x + 2 = 4 sin x cos y and 0 < x, y < then sin x + cos y is

p , 2

3 2 4 2 47. The equation cos x – (l + 2) cos x – (l + 3) = 0 possesses a solution if (a) l > – 3 (b) l < – 2 (c) –3 < l < – 2 (d) l Œ z+ 48. If 0 < q < 2p and 2 sin2 q – 5 sin q + 2 > 0, then the range of q is (a) –2

(b) 0

(c) 2

(d)

p 5p (a) ÊÁ 0, ˆ˜ » ÊÁ , 2p ˆ˜ Ë 6¯ Ë 6 ¯ 5p (b) ÊÁ 0, ˆ˜ » (p, 2p) Ë 6¯ p (c) ÊÁ 0, ˆ˜ » (p, 2p) Ë 6¯ (d) None 49. The number of values of x for which sin 2x + cos 4x = 2 are (a) 0 (b) 1 (c) 2 (d) 50. The number of solutions of the equation x3 + x2 + 4x + 2 sin x = 0 in 0 < x < 2p are (a) 0 (b) 1 (c) 2 (d) 4 51. The number of solutions of the equation tan x + sec x = 2cos x lying in the interval [0, 2p] are (a) 0 (b) 1 (c) 2 (d) 3 52. The number of solutions of the equation 2(sin4 2x + cos4 2x) + 3 sin2 x cos2 x = 0 are (a) 0 (b) 1 (c) 2 (d) 3 53. cos 2x + a sin x = 2a – 7 possesses a solution for (a) a || a (b) a > 6 (c) a < 2 (d) a Œ [2, 6] 54. If 0 < x 2

Ê px ˆ = 61. The number of solutions of the equation sin Á Ë 2 3 ˜¯ 2 x – 2 3 x + 4 are (a) it forms an empty set (b) is only one (c) is only two (d) is greater then two 62. Number of real roots of the equation sec q + cosec q = 15 lying between 0 and 2 p are (a) 8 (b) 4 (c) 2 (d) 0 63. The general solution of the equation sin100 x – cos100 x = 1, is p ,nŒz 3 p (b) np + , n Œ z 2 (a) 2np +

p (c) np + p, n Πz 4 p (d) 2np , n Πz 3

TR_02.indd 9

64. The number of solutions of the equation 2cosx = |sin x| in [–2p, 2p] are (a) 1 (b) 2 (c) 3 (d) 4 65. The general solution of the equation 2 cos2x + 1 = 3.2 –sin2x is (a) np. n Œ z (b) (n + 1) p, n Œ z (c) (n – 1) p, n Œ z (d) None 66. If x Œ (0, 1), the greatest root of the equation sin2 p x = 2 cos p x is 1 4

1 2

(b)

3 4

(c)

(d) None

67. The number of solutions of tan (5 p cos a) = cot (5 p sin a) for a Œ (0, 2 p) are (a) 7 (b) 14 (c) 21 (d) 3 68. The number of solution of the equation 1 + sin x ◊ sin2 Ê xˆ ÁË ˜¯ = 0 in [–p, p] is 2 (a) 0 (b) 1 (c) 2 (d) 3 69. The number of solution of the equation |cot x| = cot x + 1 , " x Œ [0, 2 p] are sin x (a) 0 (b) 1 (c) 2 (d) 3 70. The real roots of the equation cos7 x + sin4 x = 1 in (–p, p) are p p (b) - , 0, 2 2

p (a) - , 0 2 (c)

p ,0 2

LEVEL III

(d) 0,

p p , 4 2

(Probems for JEE Advanced)

4 . 3 Solve for x: sin 2x + 12 = 12(sin x – cos x) Solve for x: |sec x + tan x| = |sec x| + |tan x| in [0, 2p]. Let n be a positive integer such that n , find n. Êpˆ Êpˆ sin Á ˜ + cos Á ˜ = Ë 2n ¯ Ë 2n ¯ 2 If cos 2x + a sin x = 2a – 7 possesses a solution then find a. Solve for x: sin100 x – cos100 x = 1.

1. Solve for x: sec x - cosec x = 2. 3. 4.

5. 6.

7. Solve for x: sin10 x + cos10 x = sin 2 x - 3 sin x + 1

8. Solve for x: |cos x |

2

2

29 cos 4 2x 16 =1

9. Find the number of solutions of cos (p x - 4) ◊ cos (p x ) = 1 .

2/10/2017 4:07:36 PM

2.10

Trigonometry Booster

Êpˆ 10. Find the number of solution of x 4 - 2x 2sin 2 Á ˜ x + Ë 2¯ 1=0 11. If cos x + a cos x + 1 = 0 has atleast one real solution, then find the value of a. 12. If the equation tan4 x – 2 sec2 x + b2 = 0 has at least one real solution then find the value of b. 13. If a, b Œ [0, 2p] and the equation x2 + 4 + 3 sin (ax + b) = 2x has at-least one solution, then find (a + b). 14. Find the number of ordered pairs (a, b) satisfying the Ê p x2 ˆ equations |x| + |y| = 4 and sin ÁË ˜ =1 2 ¯ 15. Find the number of values of x in (–2p, 2p) and satisfying log|cos x||sin x| + log|sin x||cos x| = 2. 16 The number of solutions of tan x + sec x = 2 cos x in [0, 2p) is (a) 2 (b) 3 (c) 0 (d) 1 [JEE Main, 2002] 4

17. The number of values of x in the interval [0, 3p] satisfying the equation 2 sin2 x + 5 sin x – 3 = 0 are (a) 4 (b) 6 (c) 1 (d) 2 [JEE Main, 2006] Note No questions asked in between 2007 to 2015. 18. Find all the angles q between p and –p that satisfy the equation Êqˆ 5 cos(2q ) + 2 cos 2 Á ˜ + 1 = 0 Ë 2¯ [Roorkee, 1984] Note No questions asked between 1985 to 1986. 19. Find the general solution to the following equation 2 (sin x – cos 2x) – sin 2x (1 + 2 sin x) + 2 cos x = 0 [Roorkee, 1987] 20. Solve for x and y; x cos3 y + 3x cos y sin2 y = 14, x sin3 y + 3x cos2 y sin y = 13 [Roorkee, 1988] 21. Solve for: x; 4 sin4 x + cos4 x = 1 [Roorkee, 1989] 22. Find all the values of ‘a’ for which the equation sin4 x + cos4 x + sin 2x + a = 0 is valid. Also find the general solution of the equation. [Roorkee, 1990] Note No questions asked in 1991.

the

equation

[Roorkee, 1992]

2

Note No questions asked between 2003 to 2005.

TR_02.indd 10

23. Find the general solution of ( 3 - 1) sin q + ( 3 + 1) cos q = 2 . Note No questions asked in 1993. 24. Solve for q sec q - cosec q =

4 3

[Roorkee, 1994]

Note No questions asked in 1995. 25. If 32 tan8 q = 2 cos2 a – 3 cos a and 3 cos 2q =1, then find the general values of a. [Roorkee, 1996] Note No questions asked in 1997. 26. Find the general values of x and y and satisfying the equations 5 sin x cos y = 1, 4 tan x = tan y. [Roorkee, 1998] Note No questions asked in 1999. 27. Find the smallest positive value of x and y satisfying p ( x - y ) = , cot x + cot y = 2. 4 [Roorkee, 2000] 28. Solve the following equations for x and y: (i) 5(cosec (ii) 2

2

x - sec2 y )

=1

(2cosec x + 3|sec y |)

= 64 [Roorkee, 2001]

LEVEL IV

(Tougher Problems for JEE Advanced)

Solve the following trigonometric equations: Ê xˆ Ê xˆ 1. cot Á ˜ - cosec Á ˜ = cot x Ë 2¯ Ë 2¯ 2. 8 cos x ◊ cos 2x ◊ cos 4x = 3.

sin 6x sin x

tan x tan 2x + +2=0 tan 2x tan x

cos x cos (6x) = –1 cos (4x) + sin (5x) = 2 sin 2x + 5 cos x + 5 sin x + 1 = 0 sin x + sin 2x + sin 3x = cos x + cos 2x + cos 3x in the interval 0 £ x £ 2p 8. sin2 x tan x + cos2 x cot x – sin 2x = 1 + tan x cot x. 9. sin2 4x + cos2 x = 2 sin 4x cos4 x 4. 5. 6. 7.

2/10/2017 4:07:36 PM

2.11

Trigonometric Equations

7 10. sin 4 x + cos 4 x = sin x cos x 2 4 4 11. sin x + cos x pˆ pˆ Ê Ê = 2 cos Á 2x + ˜ cos Á 2x - ˜ Ë Ë 6¯ 6¯ pˆ 1 4 4Ê 12. sin x + sin ÁË x + ˜¯ = 4 4 pˆ Ê 13. If cos Á x + ˜ + cos x = a , then find all values of a so Ë 3¯ that the equation has a real solution. 1 14. Find the number of roots of cos x - x + = 0 lies in 2 Ê pˆ ÁË 0, ˜¯ 2 15. Find the number of integral ordered pairs satisfy the Ïcos( xy ) = x equations Ì Ó tan( xy ) = y 16 Find the number of real solutions of sin2016 x – cos2016 x = 1 in [0, 2p] 17. Find the number of ordered pairs which satisfy the equation x2 + 2x sin (xy) + 1 = 0 for y Œ [0, 2p] 18. Find the number of solutions of the equation sin 5x ◊ cos 3x = sin 6x ◊ cos 2x in [0, p] 19. Find the number of solution of the equation cos 3x ◊ tan p 5x = sin 7x lying in ÈÍ0, ˘˙ Î 2˚ 20. The angles B and C (B > C) of a triangle satisfying the equation 2 tan x – l(1 + tan2 x) = 0, then find the angle A, if 0 < l < 1 21. Determine all values of ‘a’ for which the equation cos4 x – (a + 2) cos2 x – (a + 3) = 0 has a solution and find those. 22. Find all the solution of the equation p È 5p 7p ˘ 2 2 sin x + sin ( (1 - cos x) + sin x ) = 0 in Í , Î 2 2 ˙˚ 8 23. If the equation sin4 x – (k + 2) sin2 x – (k + 3) = 0 has a solution, then find the value of k. 24. Find the number of principal solutions of the equation 4.16sin2x = 26sinx 25. Find the general solution of sec x = 1 + cos x + cos2 x + cos3 x + …

Integer Type Questions 1. Find the number of values of x in (0, 5p) satisfying the equation 3 sin2x – 7 sinx + 2 = 0 2. Find the number of integral values of k, for which the equation 2 cosx + 3 sinx = k + 1 has a solution 3. Find the number of distinct real roots of sin x cos x cos x p p cos x sin x cos x = 0 in ÈÍ- , ˘˙ Î 4 4˚ cos x cos x sin x

TR_02.indd 11

4. Find the number of pairs (x, y) satisfying the equations sin x + sin y = sin (x + y) and |x| + |y| = 1 5. Find the maximum value of f ( x) =

1 + sin 2 x

cos 2 x

4 sin 2x

sin 2 x

1 + cos 2 x

4 sin 2x

2

sin x

2

cos x

1 + 4 sin 2x

6. Find the number of solutions of | x| 10 7. Find the number of solutions of tan x + cot x = 2 cosec x in [–2p, 2p] 8. Find the number of solutions of sin x =

1 in [0, p ] 4 9. If x, y Œ [0, 2p], then find the number of ordered pairs (x, y) satisfying the equation sin x ◊ cos y = 1 10. If x Œ [0, 2p], then find the number of values of x satisfying the equation cos x ◊ cos 2x ◊ cos 3x =

1 sin x Find the number of solutions of tan x tan (4x) = 1, for 0 1 (c) 0 (d) None. 3. The number of real solutions of the equation sin (ex) = 5x + 5–x is (a) 0 (b) 1 (c) 2 (d) infinitely many y2 - y + 1 £ 2 , then the 2 number of ordered pairs of (x, y) is (a) 1 (b) 2 (c) 3 (d) infinitely many

Column I Column II If cos( A + B) - sin( A + B) cos(2B) sin A cos A sin B = 0 (P) np (A) - cos A sin A cos B then B is cos q sin q cos q p (B) If - sin q cos q sin q = 0 , then q is (Q) (2n + 1) 2 - cos q -sin q cos q 1 + sin 2q (C)

If

cos 2q

sin q 1 + cos q 2

sin q 2

2

4 sin 4q 4 sin 4q

sin q 1 + 4 sin 4q 2

=0,

(R) (2n - 1)

p 2

then q is (S)

7p 24

3. Match the following columns: Column I Column II 4 4 (A) If 4 sin x + cos x = 1, then p (P) x is 4 (B) If sec x ◊ cos (5x) + 1 = 0, (Q) where 0 < x < 2p , then x is

p 6

2

4. If 0 £ x £ 2p and 2cosec x ¥

TR_02.indd 13

(C)

If 81sin

2

x

+ 81cos

2

x

= 30 ,

(R)

where 0 < x < 2p, then x is (D) If 2 sin2 x + sin2 2x = 2 where (S) 0 < x < 2p, then x is

-

p 4

np, n ΠI

2/10/2017 4:07:36 PM

2.14

Trigonometry Booster

4. Match the following columns: Column I

7. Match the following columns: If a and b are the roots of a cos q + b sin q = c, then

Column II

(A) If cos q + cos 3q + cos 5q + cos 7q = 0, then (P) q is (B) If sin x – 3 sin 2x + sin 3x = cos x – 3 cos 2x + (Q) cos 3x then x is (C) If sin 4q – sec 2q = 2, (R) then q is (D) If tan (x + 100°) = tan (x + 50°) tan x ◊ tan (S) (x – 50°) then x is (T)

Column I

np p , n ΠI + 2 6 np +

(A)

p , n ŒI 2

(B)

c-a c+a

np , n ŒI 5

2bc a + b3

(D)

Êaˆ Ê bˆ tan Á ˜ ◊ tan Á ˜ is Ë 2¯ Ë 2¯

(S)

c2 - a2 a 2 + b2

Êpˆ ±Á ˜ Ë 3¯

3

8. Match the following columns:

(P)

8

(Q)

4

(R)

3

(S)

2

(T)

7

2 sin2 q – cos 2q = 0 and 2 cos2 q – 3 sin q = 0 in [0, 2p] is 6. Match the following columns:

TR_02.indd 14

(Q)

sin a ◊ sin b is

(R)

cos7 x + sin4 x = 1 in (–p, p), is (B) The number of real roots of cosec x = 1 + cot x in (–2p, 2p) is (C) The number of integral values of k for which, the equation 7 cos x + 5 sin x = 2 k + 1 has a solution is (D) The number of solutions of the pair of equations

Column II

(A) If cos 3x ◊ cos x + sin 3x ◊ sin3 x = 0, then x (P) is (B) If sin 3a = 4 sin a sin (x + a) sin (x – a), (Q) then a is, where a π np (C) If |2 tan x –1| + |2 cot (R) x – 1| = 2, then x is

pˆ Ê ÁË np ± ˜¯ , n Œ I 3

(D)

Ê np p ˆ ± ˜ , n ŒI ÁË 2 4¯

(S)

2b a+c

Êaˆ Ê bˆ tan Á ˜ + tan Á ˜ is Ë 2¯ Ë 2¯

Column II

If sin10 x + cos10 x 29 = cos 4 (2x) , 16 then x is

(P)

(C)

Column I (A) The number of real roots of

3

sin a + sin b is

np p + , n ŒI 5 10

Column I Column II 5 (A) If sin 5x = 16 sin x, p (P) (2n + 1) , n ΠI then x is 4

5. Match the following columns:

Column I

Column II

pˆ Ê ÁË np + ˜¯ , n Œ I 4

Ê np p ˆ + ˜ , n ŒI ÁË 4 8¯

(B) If 4 cos2 x ◊ sin x – 2 (Q) np, n Œ I sin2 x = 3 sin x, then x is (C) If tan2(2x) + cot2(2x) + (R) p np + , n Œ I 2 tan (2x) + 2 cot (2x) 6 =0 (D) If tan2 x ◊ tan2 3x ◊ tan (S) Ê n pˆ ÁË np + (-1) ˜¯ 4x = tan2 x – tan2 3x + 8 tan 4x then x is 9. Observe the following columns: Column I

Column II

(A) If cos (6q) + cos (4q) + cos (2q) (P) 2np, n Œ I + 1 = 0, then q is (B) If 3 – 2 cos q – 4 sin q – cos 2q np (Q) , n ŒI + sin 2q = 0, then q is 3 (C)

If cos q ◊ cos 2q ◊ cos 3q =

1 , 4

(R) (4n + 1) p , n ΠI 2

then q is (D) If sin (5q) + sin (q) = sin (3q), p (S) (2n + 1) then q is 12

Assertion and Reason Codes: (A) Both A and R are individually true and R is the correct explanation of A (B) Both A and R are individually true and R is not the correct explanation of A.

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2.15

Trigonometric Equations

(C) A is true but R is false. (D) A is false but R is true. 1. Assertion (A): The number of real solutions of sin x = x2 + x + 1 is 1 Reason (R): since |sin x| £ 1 (a) A (b) B (c) C (d) D 2. Assertion (A): The number of real solutions of cos x = 3x + 3–x Reason (R): since |cos x| £ 1 (a) A (b) B (c) C (d) D 3. Assertion (A): The maximum value of 3 sin x + 4 cos x + 10 is 15 Reason (R): The least value of 2 sin2 x + 4 is 4 (a) A (b) B (c) C (d) D 4. Assertion (A): The greatest value of sin4 x + cos2 x is 1 Reason (R): The range of the function f(x) = sin2 x + cos2 x is 1 (a) A (b) B (c) C (d) D 5. Assertion (A): a cos x + b cos 3x £ 1 for every x in R Reason (R): since |b| £ 1 (a) A (b) B (c) C (d) D 6. Assertion (A): The set of values of x for which tan 3x - tan 2x = 1 is f 1 + tan 3x ◊ tan 2x Reason (R): Since tan x is not defined at p x = (2n + 1) , n Œ I 2 (a) A (b) B (c) C (d) D 7. Assertion (A): The number of solutions of the equation tan x + sec x = 2 cos x lying in the interval [0, 2p] is 2. Reason (R): The number of solutions of the equation 3 sin2 x – 7 sin x + 2 = 0 in [0, 5p] is 6 (a) A (b) B (c) C (d) D 8. Assertion (A): The number of solutions of tan x ◊ tan 4x = 1 in (0, p) is 5 Reason (R): The number of solutions of |cos x| = sin x in [0, 4p] is 4 (a) A (b) B (c) C (d) D Êp ˆ Êp ˆ 9. Assertion (A): If tan Á sin q ˜ = cot Á cos q ˜ , then Ë2 ¯ Ë2 ¯ sin q + cos q = ± 2 Reason (R): - 2 £ sin q + cos q £ 2 (a) A (b) B (c) C (d) D 10. Assertion (A): sin A = sin B = sin C = 2 sin (18°) Reason (R): If cos A = tan B, cos B = tan C, cos C = tan A (a) A (b) B (c) C (d) D

Questions Asked In Previous Years’ JEE-Advanced Examinations 1. The general solution of the trigonometric equation sin x + cos x = 1 is given by (a) x = 2n p, n Œ I

TR_02.indd 15

pˆ Ê (b) x = Á 2np + ˜ , n Œ I Ë 2¯ p (c) x = ÊÁ np + (-1) n ˆ˜ , n Œ I Ë 4¯ (d) None of these [IIT-JEE, 1981] 2. Find the point of intersections of the curves y = cos x È p p˘ and y = sin 3x where x Œ Í- , ˙ Î 2 2˚ [IIT-JEE, 1982] 3. Find all solutions of 4 cos2 x sin x – 2 sin2 x = 3 sin x [IIT-JEE, 1983] 4. There exist a value of q between 0 and 2p which satisfies the equation sin4 q – 2 sin2 q – 1 = 0? [IIT-JEE, 1984] 5. No questions asked in 1985. 2p 3 , cos x + cos y = , 3 2 where x and y are real. [IIT-JEE, 1986] 7. Find the set of all x in the interval [0, p] for which 2 sin2 x – 3 sin x + 1 ≥ 0. [IIT-JEE, 1987] 8. The smallest +ve root of the equation tan x – x = 0 lies in 6. Find the solution set of x + y =

p (a) ÊÁ 0, ˆ˜ Ë 2¯

p (b) ÊÁ , p ˆ˜ Ë2 ¯

3p ˆ (c) ÊÁ p , ˜ Ë 2¯

3p ˆ (d) ÊÁ , 2p ˜ Ë 2 ¯

[IIT-JEE, 1987] 9. The general solutions of sin x – 3 sin 2x + sin 3x = cos x – 3 cos 2x + cos 3x is (a) np +

p , n ŒI 8

np p + , n ŒI 2 8 np p (c) (-1) n + , n ŒI 2 8 2 (d) 2np + cos -1 ÊÁ ˆ˜ , n Œ I Ë 3¯ (b)

[IIT-JEE, 1989]

10. No questions asked between 1990 to 1992. 11. The number of solutions of the equation tan x + sec x = 2 cos x lying in the interval [0, 2p] is (a) 0 (b) 1 (c) 2 (d) 3 [IIT-JEE, 1993] 12. Determine the smallest +ve value of x (in degrees) for which tan (x + 100°) = tan (x + 50°) tan x tan (x – 50°) [IIT-JEE, 1993]

2/10/2017 4:07:37 PM

2.16

Trigonometry Booster

13. Let n be a +ve integer such that n . Then, Êpˆ Êpˆ sin Á ˜ + cos Á ˜ = Ë 2n ¯ Ë 2n ¯ 2

È p p˘ 25. Find the values of t Œ Í- , ˙ so that Î 2 2˚

(a) 6 £ n £ 8 (c) 4 £ n £ 8

(b) 4 < n £ 8 (d) 4 < n < 8 [IIT-JEE, 1994] 14. Let 2 sin2 x + 3 sin x – 2 ≥ 0 and x2 – x – 2 < 0 (x is measured in radians). Then x lies in the interval Ê p 5p ˆ (a) Á , Ë 6 6 ˜¯

5p ˆ Ê (b) Á -1, ˜ Ë 6¯

(c) (–1, 2)

Êp ˆ (d) Á , 2˜ Ë6 ¯

20. Let n be an odd integer. If sin (nq ) = each value of q, then (a) b0 = 1, b1 = 3 (b) b0 = 0, b1 = n (c) b0 = –1, b1 = n (d) b0 = –1, b1 = n2 – 3n + 3 21. 22.

23. 24.

 br sin q r

for

r =0

[IIT-JEE, 1998] No questions asked between 1999 to 2001. The number of values of k for which the equation 7 cos x + 5 sin x = 2k + 1 has a solution is (a) 4 (b) 8 (c) 10 (d) 12 [IIT-JEE, 2002] No questions asked between 2003 to 2004. Let (a, b) Œ [–p, p] be such that cos (a – b) = 1 and 1 . The number of pairs of a, b satisfying e the system of equations is (a) 0 (b) 1 (c) 2 (d) 4 [IIT-JEE, 2005] cos (a + b) =

{ }

5x 2 - 2x + 1 1 , "x ΠR - 1, 2 3 3x - 2x - 1

[IIT-JEE, 2005] 26. If 0 £ q £ 2p, 2 sin q – 5 sin q + 2 > 0 then the range of q is 2

p 5p ˆ (a) ÊÁ 0, ˆ˜ » ÊÁ , 2p ˜ Ë 6¯ Ë 6 ¯

[IIT-JEE, 1994] 15. Find the smallest +ve value of p for which the equation cos (p sin x) = sin (p cos x) has a solution for x Œ [0, 2p]. [IIT-JEE, 1995] p p˘ È 16. Find all values of q in the interval Í- , ˙ satisfying Î 2 2˚ the equation (1 – tan q)(1 + tan q) sec2 q + 2tan2q = 0 [IIT-JEE, 1996] 17. Find the general value of q satisfying the equation tan2 q + sec 2q = 1 [IIT-JEE, 1997] 18. Find the real roots of the equation cos7 x + sin4 x = 1 in the interval (–p, p) [IIT-JEE, 1997] 19. The number of values of x in the interval [0, 5p] satisfying the equation 3 sin2 x – 7 sin x = 2 = 0 is (a) 0 (b) 5 (c) 6 (d) 10 [IIT-JEE, 1998] n

2 sin t =

5p ˆ (b) ÊÁ 0, » (p , 2p ) Ë 6 ˜¯ p (c) ÊÁ 0, ˆ˜ » (p , 2p ) Ë 6¯ (d) None of these [IIT-JEE, 2006] 27. The number of solutions of the pair of equations 2 sin2 q – cos 2q = 0 and 2 cos2 q – 3 sin q = 0 in the interval [0, 2p]is (a) 0 (b) 1 (c) 2 (d) 4 [IIT-JEE, 2007] 28. If sin q = cos j, then the possible values of 1Ê pˆ [IIT-JEE, 2008] ÁË q ± j - ˜¯ are..... p 2 29. For 0 < q < 6

Ê

p , the solutions of 2

 cosec ÁË q +

m =1

(m - 1)p ˆ mp ˆ Ê ˜¯ cosec ÁË q + ˜ =4 2 4 4 ¯

is (are) (a)

p 4

(b)

p 6

(c)

p 12

(d)

5p 12 [IIT-JEE, 2009]

Ê p pˆ 30. The number of values of q in the interval Á - , ˜ Ë 2 2¯ np such that j π for n Œ I and tan q = cot 5q as well 2 as sin (2q) = cos (4q) is... [IIT-JEE, 2010] 31. The +ve integer value of n > 3 satisfying the equation 1 1 1 is... [IIT-JEE, 2011] = + Êpˆ Ê 2p ˆ Ê 3p ˆ sin Á ˜ sin Á ˜ sin Á ˜ Ë n¯ Ë n ¯ Ë n¯ 32. No questions asked between 2012 to 2013.

TR_02.indd 16

2/10/2017 4:07:37 PM

2.17

Trigonometric Equations

33. For x Π(0, p), the equation sin x + 2 sin 2x Рsin 3x = 3 has (a) infinitely many solutions (b) three solutions (c) one solution (d) no solution [IIT-JEE, 2014]

34. The number of distinct solutions of the equation 5 cos 2 2x + cos 4 x + sin 4 x + cos6 x + sin 6 x = 2 4 in the interval [0, 2p] is ______ [IIT-JEE, 2015]

A NSWERS

LEVEL II 1. 6. 11. 16. 21. 26. 31. 36. 41. 46. 51. 56. 61. 66.

(b) (a) (a) (b) (b) (b) (a) (a) (a) (c) (c) (c) (b) (c)

11. x = 2. 7. 12. 17. 22. 27. 32. 37. 42. 47. 52. 57. 62. 67.

(a, b) (c) (a,b) (b) (c) (a) (c) (c) (b) (c) (a) (c) (b) (b)

3. 8. 13. 18. 23. 28. 33. 38. 43. 48. 53. 58. 63. 68.

(a, d) (d) (b) (c) (a) (b) (b) (b) (b) (a) (d) (a) (b) (a)

4. 9. 14. 19. 24. 29. 34. 39. 44. 49. 54. 59. 64. 69.

LEVEL IV p 1. x = (4n + 2) , n Œ I 3 p 2. x = (2n + 1) , x = np , n Œ I 14 np 3. x = , n ŒI 3 4. x = 2np + p = (2n + 1)p, n Œ I pˆ p Ê 5. x = ÁË 2np + ˜¯ = (4n + 1) , n Œ I 2 2 p 6. x = np - , n Œ I 4 p 5p 7p 13p 2p 4p , , , , , 8 8 8 8 3 3 np a + (-1) n 8. x = 2 2 p 9. x = (2n + 1) 2

7. x =

10. x =

TR_02.indd 17

np Ê pˆ + (-1) n Á - ˜ , n Œ I Ë 12 ¯ 2

(c) (a,b,c) (d) (c) (b) (a) (d) (c) (a) (a) (a) (c) (d) (c)

5. 10. 15. 20. 25. 30. 35. 40. 45. 50. 55. 60. 65. 70.

(a) (b) (c) (b) (c) (c) (b) (b) (c) (b) (d) (b) (a) (b)

np a Ê 1ˆ ± , n Œ I , a = sin -1 Á ˜ Ë 3¯ 2 2

12. x = nð, x = np -

p ,n ŒI 4

13. - 3 £ a £ 3 14. 1 15. 1 16. 1 3p ˆ Ê pˆ Ê 17. Á1, ˜ , Á -1, ˜ Ë 2¯ Ë 2¯ 18. 19. 20. 21.

5 2 90° –3 £ a £ –2

22. x =

13p . 4

23. [–3, –2] 24. 3 25. x = 2np ±

p , n ŒI 3

INTEGER TYPE QUESTIONS

1. 6 6. 6 11. 5

2. 7 7. 4 12. 2

3. 1 8. 6 13. 6

4. 4 9. 3 14. 1

5. 6 10. 2 15. 4

COMPREHENSIVE LINK PASSAGES

Passage-I: Passage-II: Passage-III: Passage-IV: Passage-V:

1. 1. 1. 1. 1.

(d) 2. (a, b) 2. (b) 2. (b) 2. (a) 2.

(a) 3. (a, c) 3. (b) 3. (c) 3. (c) 3.

(c) 4. (b) (c) (a, b, c, d) (b) 4. (b) (a) 4. (b)

5. (a) 4. (d) 5. (b) 5. (a)

MATRIX MATCH

1. (A) Æ (S), (B) Æ (S), (C) Æ (S), (D) Æ (S) 2. (A) Æ (Q), (B) Æ (R), (C) Æ (S).

2/10/2017 4:07:37 PM

2.18 3. 4. 5. 6. 7.

Trigonometry Booster

(A) Æ (S), (B) Æ (P), (C) Æ (Q),(D) Æ (P, Q) (A) Æ (S), (B) Æ (P), (C) Æ (Q, R),(D)Æ (T) (A) Æ (R), (B) Æ (S), (C) Æ (P), (D) Æ (S) (A) Æ (S), (B) Æ (P), (C) Æ (Q), (D) Æ (R) (A) Æ (R), (B) Æ (S), (C) Æ (P), (D) Æ (Q)

H INTS

LEVEL 1 np , where n ΠI 3 2. We have, cos2 (5q) = 0 q=

1. (a) 6. (a)

AND

2. (a) 7. (b)

3. (b) 8. (b)

4. (b) 9. (d)

1 2

sin 2q =

fi

Êpˆ cos 2 (5q ) = cos 2 Á ˜ Ë 2¯

fi

Ê 1 ˆ Êpˆ sin 2q = Á = sin 2 Á ˜ Ë 4¯ Ë 2 ˜¯

fi

Êpˆ (5q ) = np ± Á ˜ Ë 2¯

fi

Êpˆ q = np ± Á ˜ , where n Œ I Ë 4¯

fi

1Ê Ê p ˆˆ q = Á np ± Á ˜ ˜ , where n Œ I Ë 2¯¯ 5Ë

2

7. We have, tan (q – 15°) = tan (q + 15°) fi

tan(q - 15°) 3 = tan(q + 15°) 1

fi

Êpˆ tan q = tan Á ˜ Ë 3¯

fi

tan(q - 15°) + tan(q + 15°) 3 + 1 = tan(q - 15°) - tan(q + 15°) 3 - 1

fi

Êpˆ q = np + Á ˜ , where n Œ I Ë 3¯

fi

sin(q + 15° + q - 15°) 3 + 1 = sin(q + 15° - q + 15°) 3 - 1

fi fi

2 sin (2q) = 2 sin (2q) = 1

4. We have, sin 2q = sin q fi 2 sin q cos q = sin q fi sin q (2 cos q – 1) = 0 fi sin q = 0 and (2 cos q – 1) = 0 fi

1 sin q = 0 and cos q = 2

fi

q = np and q = 2np ±

p , where n ΠI 3

5. We have, sin (9 q) = sin q fi sin (9q) – sin q = 0 fi fi fi fi fi

5. (a) 10. (a)

S OLUTIONS

fi

3. We have, tan q = 3

TR_02.indd 18

ASSERTION AND REASON

where n ΠI 6. We have, 5 sin2 q + 3 cos2 q = 4 fi 2 sin 2 q + 3(sin2 q + cos2 q) = 4 fi 2 sin2 q + 3 = 4 fi 2 sin2 q = 1

1. We have, sin 3q = 0 fi 3q = np fi

8. (A) Æ (Q), (B) Æ (R), (C) Æ (S), (D) Æ (P) 9. (A) Æ (S), (B) Æ (R), (C) Æ (Q), (D) Æ (P).

Ê 9q + q ˆ Ê 9q - q ˆ 2 cos Á sin Á =0 Ë 2 ˜¯ Ë 2 ˜¯ 2 cos (5q) sin (4q) = 0 cos (5q) = 0 and sin (4q) = 0 p (5q ) = (2n + 1) and (4q ) = np 2 p Ê np ˆ q = (2n + 1) and q = Á ˜ Ë 4¯ 10

p q = (4n + 1) , n ΠI 4 8. We have, tan2 (q) + cot2 (q) = 2 fi

1 =2 tan 2 (q )

fi

tan 2 (q ) +

fi fi fi

tan4 (q) – 2 tan2 (q) + 1 = 0 (tan2 (q) – 1)2 = 0 (tan2 (q) – 1) = 0

fi

Êpˆ q = np ± Á ˜ , n Œ I Ë 4¯

9. We have, (cos (3q) + cos (q)) + cos (2q) = 0 fi (cos (3q) + cos (q)) + cos (2q) = 0 fi 2 cos (2q) cos (q) + cos (2q) = 0 fi cos (2q) (2 cos (q) + 1) = 0 fi cos (2q) = 0 and (2 cos (q) + 1) = 0

2/10/2017 4:07:37 PM

2.19

Trigonometric Equations

1 2

fi

cos (2q ) = 0 and cos (q ) = -

fi

Êpˆ Ê 2p ˆ q = (2n + 1) Á ˜ and q = np ± Á ˜ Ë 4¯ Ë 3 ¯

10. We have sin (2q) + sin (4q) + sin (6q) = 0 sin (6q) + sin (2q) + sin (4q) = 0 2 sin (4q), cos (2q) + sin (4q) = 0 sin (4q) (2 cos (2q) + 1) = 0 sin (4q) = 0 and (2 cos (2q) + 1) = 0 1 2 Ê np ˆ Ê 2p ˆ fi q = Á ˜ and (2q ) = np ± Á ˜ Ë 4¯ Ë 3 ¯ Ê np ˆ Ê np ˆ Ê p ˆ fi q = Á ˜ and q = Á ˜ ± Á ˜ , n Œ I Ë 4¯ Ë 2 ¯ Ë 3¯ 11. We have tan (q) + tan (2q) + tan (q) tan (2q) = 1 fi tan (2q) + tan (q) = 1 – tan (q) tan (2q) fi

(4q ) = np and cos (2q ) = -

fi

tan (2q ) + tan (q ) =1 1 - tan (q ) tan (2q )

fi

tan (3q) = 1

fi

Êpˆ tan (3q ) = tan Á ˜ Ë 4¯

fi

Êpˆ (3q ) = np + Á ˜ Ë 4¯

Ê np ˆ Ê p ˆ q = Á ˜ + Á ˜ , n ŒI Ë 3 ¯ Ë 12 ¯ 12. We have tan (q) + tan (2q) + tan (3q) = tan (q), tan (2q), tan (3q) fi tan (q) + tan (2q) = – tan (3q) + tan (q) tan (2q) tan (3q) fi tna (q) + tan (2q) = – tan (3q) (1 – tan (q), tan (2q)) fi

(cosec q – 1 + 3) (1 + cosec q) = 0 (cosec q + 2) (1 + cosec q) = 0 cosec q = –1, –2

fi

sin q = -1, sin q = -

1 2

p Ê pˆ q = (4n - 1) , q = np + (-1) n Á - ˜ , n Œ I Ë 6¯ 2 14. Given equation is 2 tan q – cot q = –1 fi 2 tan q = cot q – 1 fi

fi fi fi fi fi fi fi fi

1 -1 tan q 2 tan2 q + tan q – 1 = 0 2 tan2 q + 2 tan q – tan q – 1 = 0 2 tan q (tan q + 1) – (tan q + 1) = 0 (2 tan q – 1)(tan q + 1) = 0 (2 tan q – 1) = 0, (tan q + 1) = 0 2 tan q =

1 2 pˆ Ê Ê 1ˆ q = Á np - ˜ , q = np + a , a = tan -1 Á ˜ Ë ¯ Ë 2¯ 4 tan q = -1,

15. Given equation is tan 2q + (1 - 3) tan q - 3 = 0 fi

tan 2q + tan q - 3 (tan q + 1) = 0

fi

tan q (tan q + 1) - 3 (tan q + 1) = 0

fi

(tan q - 3)(tan q + 1) = 0

fi

tan q = 3 , tan q = -1

p p , q = np - , n ΠI 3 4 16. Given equation is fi

q = np +

pˆ 2p ˆ Ê Ê tan q + tan Á q + ˜ + tan Á q + ˜ =3 Ë ¯ Ë 3 3 ¯

fi

Ê tan (q ) + tan (2q ) ˆ ÁË (1 - tan (q ) ◊ tan (2q )) ˜¯ = - tan (3q )

fi

Ê Êp ˆ Êp ˆˆ tan q + tan Á + q ˜ + tan Á p - Á - q ˜ ˜ = 3 Ë3 ¯ Ë3 ¯¯ Ë

fi fi fi

tan (3q) = – tan (3q) 2 tan (3q) = 0 (3q) = np

fi

Êp ˆ Êp ˆ tan q + tan Á + q ˜ - tan Á - q ˜ = 3 Ë3 ¯ Ë3 ¯ tan q +

fi

Ê np ˆ q = Á ˜ , n ŒI Ë 3¯

fi

3 + tan q 3 - tan q =3 1 - 3 tan q 1 + 3 tan q

fi

tan q +

8 tan q =3 1 - 3 tan 2q

fi

9 tan q - 3 tan 3q =3 1 - 3 tan 2q

13. Given equation is cot 2q + fi fi

TR_02.indd 19

fi fi fi

3 +3=0 sin q

cot2 q + 3 (1 + cosec q) = 0 (cosec2 q – 1) + 3 (1 + cosec q) = 0

2/10/2017 4:07:37 PM

2.20

Trigonometry Booster

3 tan q - tan 3q fi =1 1 - 3 tan 2q fi tan (3q) = 1 p fi 3q = np + 4 np p fi q= + , n ŒI 3 12 17. Given equation is 3 tan (q – 60°) = tan (q + 60°) tan (q + 60°) tan (q - 60°)

fi

3=

fi

tan (q + 60°) =3 tan (q - 60°)

fi

tan (q + 60°) 3 = tan (q - 60°) 1

fi

tan (q + 60°) + tan (q - 60°) 3 + 1 = tan (q + 60°) - tan (q - 60°) 3 - 1

fi

sin (q + 60° + q - 60°) =2 sin (q + 60° - q + 60°)

fi

sin (2q ) =2 sin (120°)

fi

sin (2q) = 2 sin (120°)

fi

sin (2q ) = 2 ¥

3 = 3 2

It is not possible. Hence, the equation has no solution. 18. Given equation is tan q + tan 2q + tan 3q = 0 fi tan q + tan 2q + tan (2q + q) = 0 tan (2q ) + tan (q ) fi tan q + tan 2q + =0 1 - tan (2q )tan (q ) fi

1 Ê ˆ (tan q + tan 2q ) Á1 + =0 Ë 1 - tan (2q ) tan (q ) ˜¯

fi

1 Ê ˆ (tan q + tan 2q ) = 0, Á1 + =0 Ë 1 - tan (2q ) tan (q ) ˜¯

when (tan q + tan 2q) = 0 2 tan q fi tan q + =0 1 - tan 2q

TR_02.indd 20

fi

2 Ê ˆ tan q Á1 + =0 2 ˜ Ë 1 - tan q ¯

fi

2 Ê ˆ tan q = 0, Á1 + =0 Ë è1 - tan 2 ˜¯

fi

tan q = 0,

2 = -1 1 - tan 2q

fi fi

tan q = 0, 1 – tan2 q = –2 tan q = 0, tan2 q = 3

fi

q = np , q = np ±

p , n ŒI 3

1 ˆ when ÊÁ1 + =0 Ë 1 - tan (2q ) tan (q ) ˜¯ fi

1 = -1 1 - tan q tan 2q

fi fi

1 – tan q tan 2 q = –1 tan q tan 2q = 2

fi

Ê 2 tan q ˆ tan q Á =2 Ë 1 - tan 2q ˜¯

fi

tan2 q = 1 – tan2 q

fi

tan 2q =

1 Ê 1 ˆ = tan 2a , a = tan -1 Á Ë 2 ˜¯ 2

fi q = np ± a, n Œ I 19. Given equation is fi

cos 2q cos 4q =

1 2

fi 2 cos (4q) cos (2q) = 1 fi cos (6q) + cos (2q) = 1 fi cos (6q) = 1 – cos (2q) 20. Given equation is cot q – tan q = cos q – sin q fi

Ê (cos q + sin q ) ˆ (cos q - sin q ) Á - 1˜ = 0 Ë sin q cos q ¯

fi (cos q – sin q) = 0, (cos q + sin q) = sin q cos q fi tan q = 1, (cos q + sin q ) = sin q cos q when tan q = 1 fi

q = np +

p , n ŒI 4

when (cos q + sin q) = sin q cos q No real value of q satisfies the given equation. 21. Given equation is (1 – tan q)(1 + sin 2q) = 1 + tan q fi (cos q – sin q)(cos q + sin q)2 = (cos q + sin q) fi (cos q + sin q)(cos 2q – 1) = 0 fi tan (q) = –1, sin2 q = 0 fi tan (q) = –1, sin (q) = 0 fi

q = np -

p , q = np , n ΠI 4

22. Given equation is 2 sin2 q + sin2 2q = 2 fi 2 sin2 q + 4 sin2 q cos2 q = 2 fi sin2 q + 2 sin2 q cos2 q = 1 fi 2 sin2 q cos2 q = 1 – sin2 q

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2.21

Trigonometric Equations

fi fi fi

2 sin2 q cos2 q = cos2 q (2 sin2 q – 1) cos2 q = 0 (2 sin2 q – 1) = 0, cos2 q = 0

fi

1 sin q = , cos q = 0 2 2

p p q = (2n + 1) , q = np ± , n Œ I 2 4 23. Given equation is sin (3a) = 4 sin q sin (q + a) sin (q + a) fi sin (3a) = 4 sin q (sin2 q – sin2 a) fi 3 sin a – 4 sin2 a = 4 sin a (sin2 q – sin2 a) It is possible only when 3 sin 2 q = 4 fi

fi

Ê 3ˆ sin 2q = Á ˜ Ë 2 ¯

2

p , n ŒI 3 24. Given equation is 4 sin q sin 2q sin 4q = sin 3 q fi 4 sin q sin (3q – q) sin (3q + q) = sin 3q fi 4 sin q [sin2 (3q) – sin2 (q)] = sin 3q fi 4 sin q [sin2 (3q) – sin2 (q)] = 3 sin q – 4 sin3 q fi sin q [4 sin2 (3q) – 4 sin2 (q) + 4 sin2 q – 3] = 0 fi sin q [4 sin2 (3q) – 3] = 0 fi sin q = 0, [4 sin2 (3q) – 3] = 0 fi

fi

q = np ±

sin q = 0, sin 2 (3q ) =

3 4

p , n ŒI 3 25. We have sin (q) + cos (q) = 1 fi

fi

q = np , q = np ±

1 Ê 1 ˆ 2Á sin (q ) + cos (q )˜ = 1 Ë 2 ¯ 2

pˆ Ê sin Á q + ˜ = 1 Ë 6¯

fi

pˆ Ê Êpˆ sin Á q + ˜ = sin Á ˜ Ë Ë ¯ 6 2¯

fi

pˆ Ê nÊpˆ ÁË q + ˜¯ = np + (-1) ÁË ˜¯ 6 2

Êpˆ p q = np + (-1) n Á ˜ Ë 2¯ 6 27. We have sin (2q) + cos (2q) + sin (q) + cos (q) + 1 = 0 fi (sin (q) + cos (q)) + (1 + sin (2q)) + cos (2q) = 0 fi (sin (q) + cos (q)) + (sin (q) + cos (q))2 + (cos2 q – sin2 q) = 0. fi (sin (q) + cos (q)) + (sin (q) + cos (q))2 + (cos q + sin q) (cos q – sin q) = 0 fi (sin (q) + cos (q)) (1 + (sin (q) + cos (q)) + (cos q – sin q)) = 0 fi (sin (q) + cos (q)) (1 + 2 cos q) = 0 fi (sin (q) + cos (q)) = 0 and (1 + 2 cos q) = 0 fi

fi

Ê 1 Êp ˆˆ ÁË sin ÁË + q ˜¯ ˜¯ = 0 and cos q = 4 2

fi

Êp ˆ Ê 2p ˆ ÁË + q ˜¯ = np and q = 2np ± ÁË ˜¯ 4 3

fi

q = np -

p Ê 2p ˆ and q = 2np ± Á ˜ , n Œ I Ë 3 ¯ 4

28. We have sin2 q + sin q cos q + cos2 q = 1 fi (sin3 q + cos3 q) + sin q cos q = 1 fi (sin q + cos q)(1 – sin q cos q) + sin q cos q = 1 fi (sin q + cos q)(1 – sin q cos q) = (1 – sin q cos q) fi (sin q + cos q – 1)(1 – sin q cos q) = 0 fi (sin q + cos q – 1) = 0 and (1 – sin q cos q) = 0 fi

(sin q + cos q ) = 1 and sin (2q ) =

fi

1 1 Ê 1 ˆ sin (q ) + cos (q )˜ = ÁË ¯ 2 2 2

fi

Ê p ˆˆ Ê Êpˆ ÁË sin ÁË q + ˜¯ ˜¯ = sin ÁË ˜¯ 4 2

fi

pˆ 1 Ê sin Á q + ˜ = Ë 4¯ 2

and

Êpˆ sin (2q ) = sin Á ˜ Ë 6¯

fi

pˆ Ê Ê Ê p ˆˆ sin Á q + ˜ = Á sin Á ˜ ˜ Ë Ë 4¯¯ 4¯ Ë

fi

Êpˆ Êpˆ q = np + (-1) n Á ˜ - Á ˜ Ë 2¯ Ë 4¯

fi

pˆ Ê Ê n Ê p ˆˆ ÁË q + ˜¯ = ÁË np + (-1) ÁË ˜¯ ˜¯ 4 4

fi

Ê Êpˆ pˆ q = Á np + (-1) n Á ˜ - ˜ , n Œ I Ë 4¯ 4¯ Ë

26. We have fi

TR_02.indd 21

fi

3 sin (q ) + cos (q ) = 2

3 1 sin q + cos q = 1 2 2

1 2

1Ê Ê p ˆˆ and q = Á np + (-1) n Á ˜ ˜ , where n Œ I Ë 6 ¯¯ Ë 2 29. Given equation is sin q + 3 cos q = 2 fi

1 3 1 sin q + cos q = 2 2 2

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2.22

Trigonometry Booster

fi fi fi 30. 31. 32. 33. 34. 35. 36. 37.

pˆ 1 Ê sin Á q + ˜ = Ë 3¯ 2 pˆ Ê nÊpˆ ÁË q + ˜¯ = np + (-1) ÁË ˜¯ 3 4 Êpˆ p q = np + (-1) n Á ˜ - , n Œ I Ë 4¯ 3

Do yourself. Do yourself. Do yourself. Do yourself. Do yourself. Do yourself. Do yourself. Given equation is

Êp ˆ 5q = (2k + 1)p - Á - q ˜ Ë3 ¯ fi

4q = (2k + 1)p -

p 3

p p - , k ŒI 4 12 1 39. We have sin (q ) = 2 fi

q = (2k + 1)

fi

q =-

p 6

Ê pˆ Hence, the principal value of q is ÁË - ˜¯ 6

cos q + 3 sin q = 2 cos 2q

40. We have sin (q ) =

1 2

fi

1 3 cos q + sin q = cos 2q 2 2

fi

fi

pˆ Ê cos Á q - ˜ = cos 2q Ë 3¯

Hence the principal value of q is

pˆ Ê ÁË q - ˜¯ = 2np ± 2q 3 Taking positive one, we get fi

pˆ Ê q = - Á 2np + ˜ Ë 3¯ Taking negative one, we get, 2np p fi q= + , n ŒI 3 9 38. Given equation is 3(cos q - 3 sin q ) = 4 sin 2q ◊ cos 3q fi

3 cos q - 3 sin q = 2(sin 5q - sin q )

fi

3 cos q - sin q = 2(sin 5q )

fi

3 1 cos q - sin q = (sin 5q ) 2 2 Êp ˆ sin Á - q ˜ = sin 5q Ë3 ¯

fi fi

Êp ˆ 5q = np + (-1) n Á - q ˜ Ë3 ¯

when n is even Êp ˆ 5q = 2kp + Á - q ˜ Ë3 ¯

TR_02.indd 22

when n is odd

p 3

fi

6q = 2kp +

fi

kp p q= + , k ŒI 3 18

q=

p 3p , 4 4 p 4

41. We have tan (q ) = - 3 fi

(q ) = -

p 3

Hence, the principal value of q is -

p . 3

42. Given, tan q = –1 fi

q=

3p p ,4 4

Hence, the principal value of q is -

p . 4

1 2 p p q = ,3 3

43. Given, cos q = fi

Hence, the principal value of q is 44. Given, cos q = fi

q=

p . 3

1 2

2p 2p ,3 3

Hence, the principal value of q is

2p . 3

45. Given, tan q = - 3 fi

q=

2p p ,3 3

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2.23

Trigonometric Equations

Hence, the principal value of q is - p . 3 46. Given, sec q = 2 . fi fi

cos q = q=

1 2

p p ,4 4

Hence, the principal value of q is 47. Now sin (q ) = fi

q=

q=

1 2

3p 4 3p . 4

Hence, the general values of q is 3p ˆ Ê ÁË 2np + ˜ 4¯

q=

fi

3p 7p q= , 4 4 3p . 4

3p ˆ Hence, the general values of q is ÊÁ 2np + ˜ , where Ë 4¯ n Œ I. 49. Given, cos q =

p 4p , 3 3 Hence, the general solution is fi

q=

p , n ŒI 3

fi

Ê tan A + tan B ˆ ÁË 1 - tan A ◊ tan B ˜¯ = 1

fi

tan (A + B) = 1

fi

Êpˆ tan ( A + B ) = tan Á ˜ Ë 4¯

fi

Êpˆ ( A + B) = np + Á ˜ , where n Œ I Ë 4¯

fi

( A - B) =

fi

3 2 p 11p ( A + B) = , 6 6

fi

p 7p , 4 4 and tan q = –1

cos ( A + B) =

Here, we observe that A – B is positive So, A > B fi A+B>A–B 11p Ï 11p Ï ÔÔ A + B = 6 ÔÔ A + B = 6 or Ì Ì Ô A- B= p Ô A - B = 5ð ÔÓ ÔÓ 4 4

1 2

q=

On solving, we get,

3p 7p , 4 4 Hence, the general solution is q=

q = 2np +

1 3

p 5p , 4 4 2 Also, sec ( A + B ) = 3

1 2

Thus, the common value of q is

TR_02.indd 23

tan q =

52. Given, tan (A – B) = 1

p 3p , 4 4 Also, tan q = –1

fi

fi

and

51. We have, (1 + tan A) (1 + tab B) = 2 fi 1 + tan A + tan B + tan A ◊ tan B = 2 fi tan A + tan B = 1 – tan A ◊ tan B

Thus, the common value of q is

fi

q=

q = 2np +

p 3p , 4 4

48. We have sin(q ) =

1 2

p 2p , 3 3

fi

1 2

and cos (q ) = fi

p . 4

50. Given, sin q =

7p , n ŒI 4

25p Ï 19p Ï ÔÔ A = 24 ÔÔ A = 24 or Ì Ì Ô B = 19p Ô B = 7p ÔÓ ÔÓ 24 24 General values of tan tan (A – B) = 1 is

( A - B) = np +

p , n ŒI 4

…(i)

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2.24

Trigonometry Booster

General values of sec ( A + B ) = is

( A + B) = 2np +

2 3

p , n ŒI 6

…(ii)

On solving (i) and (ii), we get p p Ï ÔÔ A = (2n + m) 2 + 24 Ì Ô B = (2n - m) p - 5p ÔÓ 2 24 53. We have sin (p cos q) = cos (p sin q)

56. Given equations are 3 sin2 A + 2 sin2 B = 1 and 3 sin 2A – 2 sin 2B = 0 From (ii), we get, 3 sin 2A = 2 sin 2B fi

sin 2A sin 2B = 2 3

fi

sin 2B 3 = sin 2A 2

From (i), we get 3 (2 sin 2 A) + (2 sin 2 B) = 1 2

fi

Êp ˆ sin (p cos q ) = sin Á - p sin q ˜ Ë2 ¯

fi

Êp ˆ (p cos q ) = Á - p sin q ˜ Ë2 ¯

fi

3 (1 - cos 2A) + (1 - cos 2B) = 1 2

fi

Ê1 ˆ cos q = Á - sin q ˜ Ë2 ¯

fi

3 3 cos 2A + cos 2B = 2 2

fi

cos q + sin q =

fi

sin 2B sin 2B cos 2A + cos 2B = sin 2A sin 2A

fi fi fi fi fi

sin 2B cos 2A + sin 2A cos 2B = sin 2B sin (2A + 2B) = sin 2B sin (2A + 2B) = sin (p – 2B) (2A + 2) = (p = 2B) (2A + 4B) = p

fi

( A + 2B ) =

fi

1 2

1 1 1 cos q + sin q = 2 2 2 2

pˆ 1 Ê cos Á q + ˜ = Ë 4¯ 2 2 Similarly, we can prove that,

fi

pˆ 1 Ê cos Á q - ˜ = Ë 4¯ 2 2

57. Given, x + y =

54. We have tan (p cos q) = cot (p sin q)

p and tan x + tan y = 1 4

Êp ˆ tan (p cos q ) = tan Á - p sin q ˜ Ë2 ¯

fi

Êpˆ tan ( x + y ) = tan Á ˜ Ë 4¯

fi

Êp ˆ (p cos q ) = Á - p sin q ˜ Ë2 ¯

fi

tan x + tan y =1 1 - tan x ◊ tan y

fi

cos (q ) + sin (q ) =

1 2

1 1 1 cos (q ) + sin (q ) = 2 2 2 2

pˆ 1 Ê cos Á q - ˜ = Ë 4¯ 2 2 55. Given, sin A = sin B and cos A = cos B Dividing (i) and (ii), we get, fi

fi fi

TR_02.indd 24

p 2

fi

fi

…(i) …(ii)

…(i) …(ii)

fi 1 – tan x ◊ tan y = 1 fi tan x ◊ tan y = 0 fi tan x = 0 & tan y = 0 fi x = np = y Thus, no values of x and y satisfy the given equations. Therefore, the given equations have no solutions. 58. Given, sin x + sin y = 1 …(i) and cos 2x – cos 2y = 1 …(ii) From (ii), we get, cos 2x – cos 2y = 1 fi 1 – 2 sin2 x – 1 + sin2 y = 1 1 2

sin A sin B = cos A cos B

fi

sin 2 x - sin 2 y = -

tan A = tan B A = np + B, where n ΠI

fi

Ê 3ˆ y = np + (-1) n sin -1 Á ˜ , Ë 4¯

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2.25

Trigonometric Equations

where n ΠI 1 2 Adding (i) and (iii), we get, 1 2 sin x = 2 1 sin x = fi 4 fi

sin x - sin y = -

fi

Ê 1ˆ x = np + (-1) n sin -1 Á ˜ , n Œ I , Ë 4¯

Subtracting (i) and (iii), we get

fi

fi

3 2 sin y = 2 3 sin y = 4

fi

3ˆ fi y = np + (-1) sin Á ˜ , n Œ I Ë 4¯ 59. Given, sin x = 2 sin y n

-1 Ê

fi

Ê 2p ˆ sin x = 2 sin Á - x˜ Ë 3 ¯

fi

Ê 3 ˆ 1 sin x = 2 Á cos x + sin x˜ Ë 2 ¯ 2

fi

sin x = 3 cos x + sin x

fi fi

3 cos x = 0 cos x = 0 p x = (2n + 1) 2

fi

p p when x = (2n + 1) , then y = np 6 2 Hence, the solutions are p Ï ÔÔ x = (2n + 1) 2 , n ŒI Ì Ô y = np - p ÔÓ 6 3 2p and cos x + cos y = 60. Given, x + y = 2 3 3 Now cos x + cos y = 2

TR_02.indd 25

…(iii)

It is not possible, since the maximum value of LHS is 2. So, the given system of equations has no solutions. 61. Given equations are r sin q = 3 …(i) and r = 4(1 + sin q) …(ii) Eliminating (i) and (ii), we get 4(1 + sin q) sin q = 3 fi 4 sin2 q + 4 sin q – 3 = 0 fi 4 sin2 q + 6 sin q – 2 sin q – 3 = 0 fi 2 sin q(2 sin q + 3) –1(2 sin q + 3) = 0 fi (2 sin q + 3)(2 sin q – 1) = 0 3 1 sin q = - , 2 2 1 sin q = 2

p 5p , 6 6 62. Given equations are sin x + sin y = 1 and cos 2x – cos 2y = 1 Now, cos 2x – cos 2y = 1 fi 1 – 2 sin2 x –1 + 2 sin2 y = 1 fi –2 sin2 x – 1 + 2 sin2 y = 0 fi 2(sin2 x – sin2 y) = –1 fi

q=

fi

(sin x + sin y )(sin x - sin y ) = -

fi

(sin x - sin y ) = -

…(i)

1 2

1 2

…(ii)

On solving, we get sin x = 0, sin y = 1 fi

p x = np , y = (4n + 1) , n ΠI 2

Hence, the solutions are x = np Ï Ô Ì p , n ŒI ÔÓ y = (4n + 1) 2 63. Given curves are y = cos x and y = sin 2x Thus, sin 2x = cos x fi 2 isn x cos x = cos x fi (2 sin x – 1) cos x = 0 fi (2 sin x – 1) = 0, cos x = 0

fi

Ê 2p ˆ 3 cos x + cos Á - x˜ = Ë 3 ¯ 2

fi

1 3 3 cos x - cos x + sin x = 2 2 2

fi

fi

1 3 3 cos x + sin x = 2 2 2

then y =

fi

cos x + 3 sin x = 3

fi

1 sin x = , cos x = 0 2 p 5p p 3p x= , , , 6 6 2 2 3 3 ,,0 2 2

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2.26

Trigonometry Booster

Hence, the solutions are Ê p 3 ˆ Ê 5p 3 ˆ Ê p ˆ Ê 3p ˆ , - ˜ , Á , 0˜ , Á , 0˜ ÁË , ˜¯ , ÁË 6 2 6 2 ¯ Ë2 ¯ Ë 2 ¯ 64. Given equation is fi fi

3 2 2(cos x + cos y ) + 2 cos (x + y) + 2 = 1 cos x + cos y + cos ( x + y ) = -

fi

Êx+ 4 cos Á Ë 2

yˆ Êx˜¯ cos ÁË 2

yˆ 2Ê x + ˜¯ + 4 cos ÁË 2

fi

Ê x+ yˆ Ê x- yˆ Ê x+ yˆ 4 cos 2 Á + 4cos Á cos Á +1= 0 Ë 2 ˜¯ Ë 2 ˜¯ Ë 2 ˜¯

yˆ ˜¯ = 1

For real x and y, Ê x - yˆ 16 cos 2 Á - 16 ≥ 0 Ë 2 ˜¯ fi

Ê x - yˆ cos 2 Á ≥1 Ë 2 ˜¯

fi

Ê x - yˆ cos 2 Á =1 Ë 2 ˜¯

fi

Ê x - yˆ ÁË ˜ =0 2 ¯

fi x=y The given equation Ê x + yˆ Ê x - yˆ Ê x + yˆ 4 cos 2 Á + 4 cos Á cos Á +1= 0 Ë 2 ˜¯ Ë 2 ˜¯ Ë 2 ˜¯ reduces to 4 cos2 (x) + cos (x) + 1 = 0 fi (2 cos (x) + 1)2 = 0 1 fi cos ( x) = 2 2p =y 3 65. Given equation is 8 cos q cos j cos (q + j) + 1 = 0 fi

x=

1 4 fi 4[cos (q + j) + cos (q + j)] cos (q + j) + 1 = 0 fi 4 cos2 (q + j) + cos (q + j) cos (q + j) + 1 = 0 For all real 0 < q, j < p, 16 cos2 (q – j) – 16 > 0 fi cos2 (q – j) ≥ 1 fi cos2 (q – j) = 1 fi q–j=0 fi q=j when q = j, then the equation 4 cos2 (q + j) + 4 cos (q – j) cos (q + j) + 1 = 0 reduces to fi 4 cos2 (2q) + 4 cos (2q) + 1 = 0 fi (2 cos (2q) + 1)2 = 0 fi

TR_02.indd 26

2 cos q cos j cos (q + j ) = -

fi

(2 cos (2q) + 1) = 0

fi

cos (2q ) = -

fi

(2q ) =

fi

q=

66. We have

1 2

2p 3

p =j 3

tan 3x - tan 2x =1 1 + tan 3x ◊ tan2x

fi fi

tan (3x – 2x) = 1 tan x = 1 p fi x = np + , where n Œ I 4 But the values of x do not satisfy the given equation. Hence, the set of values of x is f. 67. Given equation is tan x + sec x = 2 cos x fi (1 + sin x) = 2 cos2 x fi (1 + sin x) = 2(1 – sin2 x) fi (1 + sin x ) = 2(1 + sin x) ◊ (1 – sin x) fi (1 + sin x)(1 – 2 + 2 sin x) = 0 fi (1 + sin x)(2 sin x – 1) = 0 fi (1 + sin x) = 0 and (2 sin x – 1) = 0 fi

sin x = - 1 and sin x =

fi

x=

1 2

p p 5p , , 2 6 6

p does not satisfy the given equation. 2 p 5p and . Thus, the values of x are 6 6 Hence, the number of solutions is 2. 68. Given equation is 2 sin2 x + 6 sin x – sin x – 3 = 0 fi 2 sin2 x + 6 sin x – sin x – 3 = 0 fi 2 sin x(sin x + 3) – 1(sin x + 3) = 0 fi (sin x + 3)(2 sin x – 1) = 0 But x =

fi

sin x = - 3,

fi

sin x =

1 2

1 2

p 5p 13p 17p , , , 6 6 6 6 Hence, the number of values of x is 4. 69. Given tan (x + 20°) = tan (x – 10°) tan x ◊ tan (x + 10°) fi

x=

fi

tan ( x + 20°) = tan ( x - 10°) ◊ tan ( x + 10°) tan x

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2.27

Trigonometric Equations

fi

sin ( x + 20°) cos x sin ( x - 10°) sin ( x + 10°) = cos ( x + 20°) sin x cos ( x - 10°) ◊ cos ( x + 10°)

fi sin ( x + 20°) cos x + cos ( x + 20°)sin x sin ( x + 20°) cos x - cos ( x + 20°)sin x sin ( x -10°)sin ( x +10°) + cos ( x -10°)◊cos ( x +10°) = sin ( x -10°)sin ( x +10°) - cos ( x -10°)◊cos ( x +10°) fi

sin ( x + 20° + x) cos ( x + 10° - x + 10°) =sin ( x + 20° - x) cos ( x + 10° + x - 10°)

fi

sin (2x + 20°) cos (20°) =sin (20°) cos (2x)

fi sin (2x + 20°) cos (2x) = –sin (20°) cos (20°) fi 2 sin (2x + 20°) cos (2x) = –2 sin (20°) cos (20°) fi sin (4x + 20°) + sin (20°) = sin (40°) fi sin (4x + 20°) = –sin (40°) – sin (20°) fi sin (4x + 20°) = –2 sin (30°) cos (10°) fi sin (4xz + 20°) = –cos (10°) fi sin (4x + 20°) = –sin (80°) fi sin (4x + 20°) = sin (–80°) fi sin (4x + 20°) = sin (p – (–80°)) fi (4x + 20°) = (p – (–80°)) fi (4x + 20°) = 260° fi 4x = 260° – 20° = 240° fi x = 60° Hence, the smallest positive value of x is 60° 70. Given, sin2 x + cos2 y = 2 sec2 z Here, LHS £ 2 and RHS ≥ 2 It is possible only when sin2 x = 1, cos2 y = 1, sec2 z = 1 fi cos2 x = 0, sin2 y = 0, cos2 z = 1 fi cos2 x = 0, sin2 y = 0, sin2 z = 1 fi cos x = 0, sin y = 0, sin z = 0 p fi x = (2n + 1) , y = mp , z = kp 2 where, n, m, k Œ I. 71. The given equation can be expressed as 5(2 cos2 x – 1) + (1 cos x) + 1 = 0 fi 10 cos2 x + cos x – 3 = 0 fi (5 cos x + 3) (2 cos x – 1) = 0 fi (5 cos x + 3) = 0, (2 cos x – 1) = 0 cos x = -

fi

cos x =

fi

-1 Ê p ˆ x = 2np ± a = 2np ± cos ÁË ˜¯ , 3

fi

TR_02.indd 27

3 = cos a , 5

fi

1 Êpˆ = cos Á ˜ Ë 3¯ 2

p x = 2np ± , n Œ Z 3

72. Given equation is 4 sin4 x + cos4 x = 1 fi 4 sin4 x = 1 – cos4 x fi 4 sin4 x = (1 + cos2 x) sin2 x fi sin2 x (4 sin2 x – cos2 x – 1) = 0 fi sin2 x = 0, (5 sin2 x – 2) = 0 fi

sin x = 0, sin 2 x =

2 5

Ê 2ˆ x = np , x = np ± a , a = sin -1 Á Ë 5 ˜¯ 73. Given equation is 4 cos2 x sin x – 2 sin2 x = 2 sin x fi 4(1 – sin2 x) sin x – 2 sin2 x = 2 sin x fi 2 (1 – sin2 x) sin x – sin2 x = sin x fi 2 sin x – 2sin3 x – sin2 x – sin x = 0 fi sin x – 2 sin3 x – sin2 x = 0 fi 2 sin3 x + sin2 x – sin x = 0 fi sin x (2 sin2 x + sin x – 1) = 0 fi sin x = 0, (2 sin2 x + sin x – 1) = 0 fi

fi fi fi

-1 ± 3 2 sin x = 0, sin x = 1, sin x = –2 sin x = 0, sin x = 1 sin x = 0, sin x =

p x = np , x = (4n + 1) , n Œ I 2 74. Given equation is sin 3x + cos 2 x = 1 fi sin 3x = 1 – cos 2x fi sin x (3 – 4 sin2 x) = 2 sin2 x fi sin x (3 – 4 sin2 x – 2 sin x) = 0 fi sin x = 0, (4 sin2 x + 2 sin x –3) = 0 fi

fi

sin x = 0, sin x =

-2 ± 4 + 48 8

fi

sin x = 0, sin x =

-1 ± 13 4

fi

sin x = 0, sin x =

13 - 1 4

Ê 13 - 1ˆ n x = np , x = np + (-1) a , a = sin -1 Á Ë 4 ˜¯ 75. Given equation is fi

2 cos 2x + 2 sin x = 2 fi

2 sin x = 2(1 - cos 2x)

fi

2 sin x = 4 sin 2 x

fi

sin x = 2 2 sin 2 x

fi

sin x (1 - 2 2 sin 3/2 x) = 0

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Trigonometry Booster

1

fi

sin x = 0, sin 3/2 x =

fi

sin x = 0, sin x =

1

fi

Êpˆ x = np , n = np + (-1) n Á ˜ , n Œ I Ë 4¯

2 2

3 sin 2x 2 1 + (sin3 x + cos3 x) = 3 sin x cos x 1 + (sin3 x + cos3 x) – 3 sin x cos x ◊ 1 = 0 (sin x + cos x + 1)(2 – sinx cos x – sin x – cos x) 1 + sin 3 x + cos3 x =

fi

(sin x + cos x + 1) = 0 sin x + cos x = –1 1 1 1 sin x + cos x = 2 2 2

fi

pˆ 1 Ê sin Á x + ˜ = Ë ¯ 4 2

fi

pˆ Ê nÊ pˆ ÁË x + ˜¯ = np + (-1) ÁË - ˜¯ 4 4

Ê pˆ p x = np + (-1) n Á - ˜ - , n Œ I Ë 4¯ 4 77. Given equation is 7 16

fi

1 - 3 sin 2 x cos 2 x =

7 16

fi

3 sin 2 x cos 2 x = 1 -

7 9 = 16 16

fi

3 16 3 4 sin 2 x cos 2 x = 4 3 sin 2 (2x) = 4

fi fi

sin 2 x cos 2 x =

fi

(2x) = np ±

fi

x=

p 3

78. Given equation is

fi

TR_02.indd 28

17 cos 2 2x 16

(sin 4 x + cos 4 x) 2 - 2 sin 4 x cos 4 x =

-1 ± 5 4

fi

sin 2 (2x) =

5 -1 4

fi

Ê 2x = np ± a , a = sin -1 Á Ë

5 - 1ˆ ˜ 4 ¯

fi

x=

fi

Ê sin 2 x ˆ 2 cos 2 x - 1 = 6 Á - 2 cos 2 x Ë cos 2 x ˜¯

fi fi fi fi fi fi

np p ± , n ŒI 3 6

sin 8 x + cos8 x =

sin 2 (2x) =

(1 - 2 sin 2 x cos 2 x) 2 - 2 sin 4 x cos 4 x =

Ê np a 5 - 1ˆ ± , a = sin -1 Á ˜ Ë 2 2 4 ¯ 79. Given equation is 2 sin2 x + 2 = cos2 3x fi 2 sin3 x + 2 = 1 – sin2 3x fi 2 sin3 x + sin2 3x + 1 = 0 fi 2 sin3 x + (3 sin x – 4 sin3 x)2 + 1 = 0 fi 2 sin3 x + 9 sin2 x – 24 sin4 x + 16 sin6 x + 1 = 0 fi 16 sin6 x – 24 sin4 x + 2 sin3 x + 9 sin2 x + 1 = 0 fi sin x = –1 p x = (4n - 1) , n Œ I 2 fi 80. Given equation is cos 4x = cos2 3x fi 2 cos2 2x – 1 = cos2 3x fi 2 cos2 2x = 1 + cos2 3x It is possible only when cos2 2x = 1, cos2 3x = 1 It is true for x = 0 Hence, the solution is x = np, n Œ I 81. Given equation is cos 2x = 6 tran2 x – 2 cos2 x

fi

sin 6 x + cos6 x =

fi

fi

2

76. Given equation is

fi fi fi =0 fi fi

fi fi fi fi fi

17 cos 2 2x 16 17 (1 - 4 sin 2 x cos 2 x) + 2 sin 4 x cos 4 x = cos 2 2x 16 16(1 – 4 sin2 x cos2 x = 2 sin4 x cos4 x) 17(cos4 x + sin4 x – 2 sin2 x cos2 x) 17(1– 4 sin2 x cos2 x) 32 sn4 x cos4 x + 4 sin2 x cos2 x – 1 = 0 2 sin4 (2x) + sin2 2x – 1 = 0

fi

17 cos 2 2x 16

fi fi

2 cos4 x – cos2 x = 6 – 6 cos2 x – 2 cos4 x 4 cos4 x +_5 cos2 x – 6 = 0 4 cos4 x + 8 cos2 x – 3 cos2 x –6 = 0 4 cos2 x (cos2 x + 2) – 3(cos2 x + 2) = 0 (4 cos2 x – 3)(cos2 x + 2) = 0 (4 cos2 x – 3) = 0 3 cos 2 x = 4 p x = np ± , n Œ I 6

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Trigonometric Equations

82. The given equation can be written as (2 sin x – cos x)(1 + cos x) = (1 – cos x)(1 + cos x) fi (1 + cos x)(2 sin x – cos x – 1 + cos x) = 0 fi (1 + cos x)(2 sin x – 1) = 0 fi cos x = –1, sin x = 1/2 1 Êpˆ fi cos x = –1 = cos p, sin x = = sin ÁË ˜¯ 2 6 p fi x = 2np ± p, x = np + (-1) , n ŒZ 6 83. Given equation is 2 sin2 x + sin x – 1 = 0 n

fi fi fi

p , x = np + a , a = tan -1 (3) 4 87. Given equation is fi

x = np -

2 cos 2 x - 3 sin x + 1 = 0 fi

2 - 2 sin 2 x - 3 sin x + 1 = 0

fi

3 - 2 sin 2 x - 3 sin x = 0

fi

2 sin 2 x + 3 sin x - 3 = 0

fi

sin x =

-1 ± 3 1 = , -1 4 2

fi

sin x =

- 3 ± 27 - 3 ± 3 3 = 4 4

fi

sin x =

1 , sin x = - 1 2

fi

sin x =

-4 3 2 3 , 4 4

fi

sin x = - 3,

fi

sin x =

fi

Êpˆ x = np + (-1) n Á ˜ , n Œ I Ë 3¯

p 5p 3p , , 6 6 2 84. Given equation is 5 sin2 x + 7 sin x – 6 = 0 fi 5 sin2 x + 10 sin x – 3 sin x – 6 = 0 fi 5 sin x (sin x + 2) – 3(sin x + 2) = 0 fi (5 sin x – 3)(sin x + 2) = 0 fi (5 sin x – 3) = 0, (sin x + 2) = 0 fi (5 sin x – 3) = 0 fi

x=

3 5

fi

sin x =

fi

Ê 3ˆ x = np + (-1) n a , a = sin -1 Á ˜ Ë 5¯ Ê 3ˆ Ê 3ˆ x = sin -1 Á ˜ , p - sin -1 Á ˜ Ë 5¯ Ë 5¯

85. Given equation is 1 4 2 4 sin x – 4 cos x – 1 = 0 4 – 4 cos2 x – 4 cos x – 1= 0 3 – 4 cos2 x – 4 cos x = 0 4 cos2 x + 4 cos x – 3 = 0 sin 2 x - cos x =

fi fi fi fi fi fi

-4 ± 8 1 3 cos x = = ,8 2 2 cos x =

1 2

p 5p fi x= , 3 3 86. Given equation is tan2 x – 2 tan x – 3 = 0 fi (tan x – 3)(tan x + 1) = 0

3 2

3 2

88. The given equation can be written as (sin x + sin 5x) + sin 3x = 0 fi 2 sin 3x ◊ cos 2x + sin 3x = 0 fi sin 3x(2 cos 2x + 1) = 0 fi sin 3x = 0, cos 2x = –1/2 fi

Ê 2p ˆ sin 3x = 0, cos 2x = cos Á ˜ Ë 3 ¯

fi

3x = np, 2 x = 2np ±

fi

x=

Hence, the solution is

TR_02.indd 29

(tan x – 3) = 0, (tan x + 1) = 0 (tan x – 3) = 0, (tan x + 1) = 0 tan x = –1, tan x = 3

2p ,nŒZ 3

np p , x = np ± , n Œ Z 3 3

p 3 89. Given equation is cos x – cos 2x = sin 3x fi

x = 0,

fi

Ê 3x ˆ Ê xˆ Ê 3x ˆ Ê 3x ˆ 2 sin Á ˜ sin Á ˜ = 2 sin Á ˜ cos Á ˜ Ë 2¯ Ë 2¯ Ë 2¯ Ë 2¯

fi

Ê 3x ˆ Ê Ê xˆ Ê 3x ˆ ˆ 2 sin Á ˜ Á sin Á ˜ - cos Á ˜ ˜ = 0 Ë 2 ¯Ë Ë 2¯ Ë 2 ¯¯

fi

Ê Ê 3x ˆ Ê xˆ Ê 3x ˆ ˆ sin Á ˜ = 0, Á sin Á ˜ - cos Á ˜ ˜ = 0 Ë 2¯ Ë 2¯ Ë 2 ¯¯ Ë

Ê 3x ˆ when sin Á ˜ = 0 Ë 2¯ Then

3x = np 2

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Trigonometry Booster

fi

x=

2np 3

Ê x 3x ˆ when Á sin ÊÁ ˆ˜ - cos ÊÁ ˆ˜ ˜ = 0 Ë 2¯ Ë 2 ¯¯ Ë

Ê 5x ˆ Ê xˆ cos Á ˜ = sin x, cos Á ˜ = 0 Ë 2¯ Ë 2¯

fi

Ê 5x ˆ Êp ˆ Ê xˆ cos Á ˜ = cos Á - x˜ , cos Á ˜ = 0 Ë 2¯ Ë2 ¯ Ë 2¯

fi

Ê Ê xˆ Ê 3x ˆ ˆ ÁË sin ÁË ˜¯ = cos ÁË ˜¯ ˜¯ 2 2

fi

p Ê 5x ˆ Êp ˆ Ê xˆ ÁË ˜¯ = 2np ± ÁË - x˜¯ , ÁË ˜¯ = (2n + 1) 2 2 2 2

fi

Ê xˆ Ê 3x ˆ sin Á ˜ = cos Á ˜ Ë 2¯ Ë 2¯

fi

x=

fi

Ê 3x ˆ Ê p xˆ cos Á ˜ = cos Á - ˜ Ë 2¯ Ë 2 2¯

Ê 3x ˆ Ê p xˆ ÁË ˜¯ = 2np ± ÁË - ˜¯ 2 2 2 Taking positive sign, we get,

fi

2x = 2np +

p 2

p 4 Taking negative sign, we get,

fi

x = np +

p x = 2np - , n ΠI 2 90. Given equation is sin 7x + sin 4x + sin x = 0 fi (sin 7x + sin x) + sin 4x = 0 fi 2 sin 4x cos 3x = sin 4x = 0 fi sin 4x(2 cos 3x + 1) = 0 fi sin 4x = 0, (2 cos 3x + 1) = 0 fi fi

1 2 2p 4x = np , 3x = 2np ± 3

sin 4x = 0, cos 3x = -

np 2np 2p ,x= ± , n ŒI 4 3 9 Hence, the solutions are 2p p p , , x = 0, 9 4 2 91. Given equation is fi

x=

Ê 3x ˆ Ê xˆ cos 3x + cos 2x = sin Á ˜ + sin Á ˜ Ë 2¯ Ë 2¯ fi fi fi

TR_02.indd 30

fi

Ê 5x ˆ Ê xˆ Ê xˆ 2 cos Á ˜ cos Á ˜ = 2 sin x cos Á ˜ Ë 2¯ Ë 2¯ Ë 2¯ Ê ˆ Ê 5x ˆ Ê xˆ 2 Á cos Á ˜ - sin x˜ cos Á ˜ = 0 Ë 2¯ Ë 2¯ Ë ¯ Ê ˆ Ê 5x ˆ Ê xˆ 2 Á cos Á ˜ - sin x˜ = 0, cos Á ˜ = 0 Ë 2¯ Ë 2¯ Ë ¯

4np ± (p - 2x), x = (2n + 1)p 5

4np ± (p - 2x), x = (2n + 1)p 5 p p 3p 13p 17p 7p 5p 29p fi x= , , , , , , , 3 5 5 15 15 5 3 15 92. Do yourself. 93. Given equation is cos 2x + cos 4x = 2 cos x fi 2 cos 3x cos x = 2 cos x fi (2 cos 3x –1) cos x = 0 fi (2 cos 3x – 1) = 0, cos x = 0 fi

x=

fi

cos 3x =

1 , cos x = 0 2

2np 2p p ± , x = (2n + 1) 3 9 2 94. Given equation is sin 2x + cos 2x + sin x + cos x + 1 = 0 fi (1 + sin 2x) + (sin x + cos x) + cos 2x = 0 fi (sin x + cos x)2 + (sin x + cos x) + (cos2 x – sin2 x) = 0 fi (sin x + cos x)(2 cos x + 1) = 0 fi (sin x + cos x) = 0, (2 cos x + 1) = 0 fi

x=

fi

tan x = - 1, cos x = -

fi

x = np -

1 2

2p p , x = 2np ± , n ŒI 4 3

95. Do yourself. 96. Given equation is tan 3x + tan x = 2 tan 2x fi

sin 4x 2 sin 2x = cos 3x cos x cos 2x

fi

2 sin 2x cos 2x 2 sin 2x = cos 3x cos x cos 2x

fi

1 ˆ Ê cos 2x 2 sin 2x Á =0 Ë cos 3x cos x cos 2x ˜¯

fi

1 ˆ Ê cos 2x 2 sin 2x = 0, Á = Ë cos 3x cos x cos 2x ˜¯

fi fi

sin 2x = 0, 2 cos2 2x = cos 4x + cos 2x sin 2x = 0, 2 cos2 2x = 2 cos2 2x – 1 + cos 2x

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Trigonometric Equations

fi fi

sin 2x = 0, cos 2x = 1 2x = np, 2x = 2np

np , x = np , n ΠI 2 97. Given equation is (1 Рtan x)(1 + sin 2x) = (1 + tan x) fi

x=

fi

2 tan x ˆ Ê (1 - tan x) Á1 + = (1 + tan x) Ë 1 + tan 2 x ¯˜

fi fi fi fi fi fi

(1 – tan x)(1 + tan x)2 = (1 + tan x) (1 + tan2 x) (1 – tan2 x)(1 + tan x) = (1 + tan x) (1 + tan 2 x) ((1 – tan2 x) – (1 + tan2 x))(1 + tan x) = 0 tan2 x(1 + tan x) = 0 tan2 x = 0, (1 + tan x) = 0 tan2 x = 0, tan x = 4

fi

x = np , x = np -

p , n ŒI 4

98. Given equation is sin x – 3 sin 2x + sin 3x = cos x – 3 cos 2x + cos 3x fi (sin 3x + sin x) – 3 sin 2x = (cos 3x + cos x) – 3 cos 2x fi 2 sin 2x cos x – 3 sin 2x = 2 cos 2x cos x – 3 cos 2x fi sin 2x (2 cos x – 3) = (2 cos x – 3) cos 2x fi

sin 2x (2 cos x - 3) = (2 cos x - 3) cos 2x

fi

sin 2x =1 cos 2x

fi

tan 2x = 1

fi

2x = np +

fi

x=

fi

p 4

np p + , n ŒI 2 8

fi

Ê 2p ˆ sin x = 0, cos 6x = cos Á ˜ Ë 3 ¯

fi

x = np , 6x = 2np ±

2p , n ŒZ 3

fi

x = np , x = (3n ± 1)

p , n ŒZ 9

x=

np np ,x= , n ŒI 4 2

103. Given equation is sec x cos 5x + 1 = 0 fi cos 5x + cos x = 0 fi 2 cos (3x) cos (2x) = 0 fi 2 cos (3x) = 0, cos (2x) = 0 fi cos (3x) = 0, cos (2x) = 0 p p , 2x = (2n + 1) , n ΠI 2 2 p p fi x = (2n + 1) , x = (2n + 1) , n ΠI 6 4 Hence, the solutions are fi

3x = (2n + 1)

p p p 3p 5p , , , , 6 4 2 4 6 105. Given equation is cos (6x) cos x = 1 fi 2 cos (6x) cos x = –2 fi cos 7x + cos 5x = –2 It is possible only when cos (7x) = –1, cos (5x) = –1 x=

p p , x = (2n + 1) , n Œ I 7 5 106. The given equation can be written as 2 sin2 x – 5 sin x cos x – 8 cos2 x = –2(sin2 x + cos2 x) fi 2 tan2 x – 5 tan x – 8 = – (tan2 x + 1) fi 4 tan2 x – 5 tan x – 6 = 0 fi (tan x – 2) (4 tan x + 3) = 0 3 fi tan x = - 2, tan x = 4 fi x = np + a, x = np + b, where Ê 3ˆ fi a = tan -1 (2), b = tan -1 Á - ˜ , n Œ Z. Ë 4¯ fi

99. The given equation can be written as (2 sin 2x ◊ sin x) 2 sin 4x – sin 3x = 0 fi 2(cos x – cos 3x) sin 4x – sin 3x = 0 fi 2 sin 4x cos x – 2sin 4x cos 3x – sin 3x = 0 fi (sin 5x + sin 3x) – (sin 7x + sin x) – sin 3x = 0 fi (sin 7x – sin 5x) + sin x = 0 fi sin x (2 cos 6x + 1) = 0 fi sin x = 0, cos 6x = –1/2,

TR_02.indd 31

100. Do yourself. 101. Do yourself. 102. Given equation is sin 4x sin 2x = cos 6x – cos 2x fi sin 4x sin 2x = –2 sin 4x sin 2x fi 3 sin 4x sin 2x = 0 fi sin 4x = 0, sin 2x = 0 fi 4x = np, 2x = np, n Œ I

x = (2n + 1)

107. Given equation is 5 sin2 x – 7 sin x cos x + 10 cos2 x = 4 fi 5 tan2 x – 7 tan x + 10 = 4 sec2 x fi 5 tan2 x – 7 tan x + 10 = 4 + 4 tan2 x fi tan2 x – 7 tan x + 6 = 0 fi (tan x – 1) (tan x – 6) = 0 fi tan x = 1, 6 p fi x = np + , x = np + a , a = tan -1 (5) 4

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Trigonometry Booster

108. Given equation is 2 sin2 x – 5 sin x cos x – 8 cos2 x = –3 fi 2 tan2 x – 5 tan x – 8 = –3 sec2 x fi 2 tan2 x – 5 tan x – 8 = –3 – 3 tan2 x fi 5 tan2 x – 5 tan x – 5 = 0 fi tan2 x – tan x – 1 = 0 1± 5 2

fi

tan x =

fi

Ê1 ± 5ˆ x = np + a , a = tan -1 Á Ë 2 ˜¯

109. Given equation is sin2 x cos x + sin2 x cos2 x + sin x cos3 x = 1 fi sin x cos x [sin2 x + sin x cos x + cos2 x] = 1 fi sin x cos x [1 + sin x cos x] = 1 fi 2 sin x cos x [2 + 2 sin x cos x] = 4 fi sin (2x) (2 + sin (2x)) = 4 fi sin2 (2x) + 2 sin (2x) – 4 = 0 fi

sin (2x) =

- 2 ± 20 2

-2 ± 2 5 fi = -1 ± 5 sin (2x) = 2 It is not possible. So, it has no solution. 110. The given equation can be written as 2 1Ê Ê xˆˆ = - Á1 + tan 2 Á ˜ ˜ 1+ Ë 2¯ ¯ sin x 2Ë fi

Ê 2t ˆ Ê 2t ˆ 2Á , + 2˜ = - (1 + t 2 ) ¥ Á 2 Ë1 + t ¯ Ë 1 + t 2 ˜¯

where t = tan (x/2) fi t3 + 2t2 + 3t + 2 = 0 fi t3 + t2 + t2 + t + 2t + 2 = 0 fi (t + 1)(t2 + t + 2) = 0 fi t + 1 = 0, t2 + t + 2 π 0 fi

Ê xˆ Ê -p ˆ tan Á ˜ = -1 = tan Á Ë 2¯ Ë 4 ˜¯

x p = np - , n ΠZ 2 4 p fi x = 2np - , n ΠZ . 2 111. Given equation is (cos x Рsin x)(2 tan x + sec x) + 2 = 0 fi (cos x Рsin x)(2 sin x + 1) + 2 cos x = 0 fi

TR_02.indd 32

Ê 1- tan 2 ( x /2) 2 tan ( x /2) ˆ Ê 4 tan ( x /2) ˆ +1 Á ˜ Ë 1+ tan 2 ( x /2) 1 + tan 2 ( x /2) ¯ ÁË 1 + tan 2 ( x /2) ˜¯ Ê 1- tan 2 ( x /2) ˆ + 2Á ˜ =0 Ë 1+ tan 2 ( x /2) ¯

Ê xˆ Put tan Á ˜ = t and then solve it. Ë 2¯ 112. Given equation is x x - cos3 2 2 = cos x 2 + sin x 3

sin 3

fi

Ê sin x ˆ Ê xˆ Ê xˆˆ Ê sin Á ˜ - cos Á ˜ ˜ Á1 + ˜ Ë 2¯ Ë 2¯ ¯ Ë ËÁ 2 ¯ 2 + sin x Ê 2 Ê xˆ 2 Ê xˆˆ ÁË cos ÁË ˜¯ - sin ÁË ˜¯ ˜¯ 2 2 = 3

3 Ê xˆ Ê xˆ = Á cos Á ˜ + sin ˜ Ë 2¯ 2 Ë 2¯ It is not possible. So, it has no solution. 113. Given equation is fi

-

fi

Ê xˆ Ê xˆ cot Á ˜ - cosec Á ˜ = cot x Ë 2¯ Ë 2¯ Ê xˆ cos Á ˜ - 1 Ë 2¯ = cot x Ê xˆ sin Á ˜ Ë 2¯

Ê Ê xˆˆ 2 (sin x + 2) = - Á1 + tan 2 Á ˜ ˜ sin x Ë 2¯ ¯ Ë

x Ê ˆ x ˆ Ê 2 tan 2 tan Á ˜ xˆ Á ˜ Ê 2 2 2 fi 2Á + 2˜ = - Á1 + tan ˜ ◊ Á Ë ¯ x˜ 2 x 2 2 Á 1 + tan ˜ Á 1 + tan ˜ Ë ¯ Ë 2 2¯ fi

fi

fi

Ê xˆ Ê xˆ 2 sin 2 Á ˜ = - sin Á ˜ cot x Ë 2¯ Ë 2¯

fi

Ê Ê xˆ ÁË 2 sin ÁË ˜¯ + cot 2

fi

Ê ˆ Ê xˆ Ê xˆ ÁË 2 sin ÁË ˜¯ + cot x˜¯ = 0, sin ÁË ˜¯ = 0 2 2

ˆ Ê xˆ x˜ sin Á ˜ = 0 Ë 2¯ ¯

Ê xˆ when sin Á ˜ = 0 Ë 2¯ fi

x = 2np, n Π1

when ÊÁ 2 sin ÊÁ x ˆ˜ + cot Ë 2¯ Ë

ˆ x˜ = 0 ¯

fi

cos x Ê xˆ 2 sin Á ˜ = Ë 2¯ sin x

fi

Ê 2 Ê xˆ 2 Ê xˆˆ ÁË cos ÁË ˜¯ - sin ÁË ˜¯ ˜¯ x Ê ˆ 2 2 2 sin Á ˜ = Ë 2¯ Ê xˆ Ê xˆ 2 sin Á ˜ cos Á ˜ Ë 2¯ Ë 2¯

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2.33

Trigonometric Equations

fi

Ê xˆ Ê xˆ Ê xˆ Ê xˆ 4 sin 2 Á ˜ cos Á ˜ + cos 2 Á ˜ - sin 2 Á ˜ = 0 Ë 2¯ Ë 2¯ Ë 2¯ Ë 2¯

s -s Êq +q +q +q ˆ Now, tan Á 1 2 3 4 ˜ = 1 3 Ë ¯ 1 - s2 + s4 2

fi

Ê xˆ Ê xˆ Ê xˆ Ê xˆ cos 2 Á ˜ + 4 sin 2 Á ˜ cos Á ˜ - sin 2 Á ˜ = 0 Ë 2¯ Ë 2¯ Ë 2¯ Ë 2¯

fi

Êq +q +q +q ˆ tan Á 1 2 3 4 ˜ Ë ¯ 2

fi

Êq +q +q +q ˆ Êpˆ tan Á 1 2 3 4 ˜ = tan Á ˜ Ë ¯ Ë 2¯ 2

fi

Ê q1 + q 2 + q 3 + q 4 ˆ Êpˆ ÁË ˜¯ = np + ÁË ˜¯ 2 2

For all real x, Ê xˆ Ê xˆ 16 sin 4 Á ˜ + 4 sin 2 Á ˜ ≥ 0 Ë 2¯ Ë 2¯ fi fi

4Ê

xˆ Ê xˆ 4 sin Á ˜ + sin 2 Á ˜ = 0 Ë 2¯ Ë 2¯ Ê xˆ Ê Ê xˆ ˆ sin 2 Á ˜ Á 4 sin 2 Á ˜ + 1˜ = 0 Ë 2¯ Ë Ë 2¯ ¯

fi

Ê Ê xˆ Ê xˆ ˆ sin 2 Á ˜ = 0, Á 4 sin 2 Á ˜ + 1˜ = 0 Ë 2¯ Ë 2¯ ¯ Ë

fi

Ê xˆ sin 2 Á ˜ = 0 Ë 2¯

fi

Ê xˆ sin Á ˜ = 0 Ë 2¯

fi x = 2np, n ΠI 114. Given equation is sin (q + a) = k sin (2q) fi sin q cos a + cos q sin a = k sin (2q) fi

Ê Ê Êqˆ ˆ 2Êqˆ ˆ Á 2 tan ÁË 2 ˜¯ ˜ Á 1 - tan ÁË 2 ˜¯ ˜ Á ˜ cos a + Á ˜ sin a Á 1 + tan 2 ÁÊ q ˜ˆ ˜ Á 1 + tan 2 ÁÊ q ˜ˆ ˜ Ë 2¯ ¯ Ë 2¯ ¯ Ë Ë Ê Êqˆ ˆ Ê 2Êqˆ ˆ Á 2 tan ÁË 2 ˜¯ ˜ Á 1 - tan ÁË 2 ˜¯ ˜ = 2k Á ˜Á ˜ Á 1 + tan 2 ÁÊ q ˜ˆ ˜ Á 1 + tan 2 ÁÊ q ˜ˆ ˜ Ë 2¯ ¯ Ë Ë 2¯ ¯ Ë

fi

Ê1 - t2 ˆ Ê 2t ˆ cos a + Á ˜ sin a ÁË 1 + t 2 ˜¯ Ë1 + t2 ¯ 2 Ê 2t ˆ Ê 1 - t ˆ = 2k ◊ Á ˜ 2˜Á Ë1 + t ¯ Ë1 + t2 ¯

2t(1 + t2) cos a + (1 + t4) sin a = 4k t(1 – t2) (sin a)t4 – (4k + 2 cos a)t3 + (4k – 2 cos a)t = sin a = 0 Let t1, t2, t3 and t4 be its four roots

fi fi

4k + 2 cos a = s1 sin a S t1t2 = 0 = s2 S t1 =

2 cos a - 4k = s3 sin a sin a S t1t2t3t4 = = -1 = s4 sin a S t1t2t3 =

TR_02.indd 33

fi 115. Let

(q1 + q2 + q3 + q4) = (2n + 1)p, n ΠI sin x + cos x = t

t2 - 1 2 So, the given equation can be reduces to fi

sin x ◊ cos x =

Ê t 2 - 1ˆ t - 2 2Á =0 Ë 2 ˜¯ fi

2t 2 - t - 2 = 0

fi

( 2t + 1)(t - 2) = 0

fi

t = 2, -

1 2

When sin x + cos x = fi

1 2

sin x +

1 2

2 cos x = 1

fi

pˆ Ê Êpˆ sin Á x + ˜ = 1 = sin Á ˜ Ë Ë 2¯ 4¯

fi

x+

np p = np + ( - 1) n , n ΠZ 4 2

When sin x + cos x = fi

1 2

1 1 1 sin x + cos x = 2 2 2

fi

pˆ Ê Ê pˆ sin Á x + ˜ = sin Á - ˜ Ë Ë 6¯ 4¯

fi

pˆ Ê nÊ pˆ ÁË x + ˜¯ = np + ( - 1) ÁË - ˜¯ 4 6

fi

x = np - ( - 1) n

p p - , n ŒZ 6 4

117. Given equation is sin x + cos x = 1 – sin x cos x Put sin x + cos x = t fi

sin x ◊ cos x =

…(i)

t2 - 1 2

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Trigonometry Booster

2 sin x cos x – 12(sin x – cos x) + 12 = 0 sin x cos x– 6(sin x – cos x) + 6 = 0 Put sin x + cos x = t

Now, equation (i) becomes t -1 2 2t = 2 – t2 + 1 t2 + 2t – 3 = 0 (t + 3) (t – 1) = 0 (t + 3) = 0, (t – 1) = 0 sin x + cos x = 1, sin x + cos x = –3 sin x + cos x = 1 2

t =1-

fi fi fi fi fi fi fi

1 2

1

sin x +

2

cos x =

t2 - 1 2 Now, equation (i) becomes

1 2

fi

pˆ 1 Ê sin Á x + ˜ = Ë ¯ 4 2

fi

Êpˆ p x = np + (-1) n Á ˜ - , n Œ I Ë 4¯ 4

118. Given equation is 3 1 + sin 3 x + cos3 x = sin 2x 2

…(i)

fi 1 + (sin x + cos x)(1 – sin x cos x) = 3 sin x cos x Put sin x + cos x = t t2 - 1 2 Now, equation (i) becomes

fi

sin x . cos x =

Ê t 2 - 1ˆ 3 2 1 + t Á1 ˜ = (t - 1) Ë 2 ¯ 2 fi fi fi fi fi fi fi

2 + t(3 – t2) = 3(t2 – 1) 2 + 3t – t3 – 3t2 + 3 = 0 3t – t3 – 3t2 + 5 = 0 t3 + 3t2 – 3t – 5 = 0 t3 + t2 + 2t2 + 2t – 5t – 5 = 0 t2(t + 1) + 2t(t + 1) – 5(t + 1) = 0 (t2 + 2t – 5) (t + 1) = 0

fi

t = - 1, t =

fi

sin x + cos x = - 1,

fi

sin x + cos x = –1

fi

1 2

sin x +

- 2 ± 24 2

1 2

cos x = -

sin x ◊ cos x =

fi

t2 - 1 - 6t + 6 = 0 2

fi fi fi fi fi

t2 – 1 – 12t + 12 = 0 t2 – 12t + 1 = 0 (t – 1)(t – 1) = 0 t = 1, 1 sin x + cos x = 1

fi

pˆ 1 Ê sin Á x + ˜ = Ë ¯ 4 2

fi

Êpˆ p x = np + (-1) n Á ˜ - , n Œ I Ë 4¯ 4

Hence, the solution are p , 2p 2 120. The given equation can be written as sin 6x + cos 4x = –2 fi sin 6x = –1 and cos 4x = –1 x = 0,

fi fi fi fi

1 2

fi

pˆ 1 Ê sin Á x + ˜ = Ë ¯ 4 2

fi

Ê pˆ p x = np + (-1) n Á - ˜ - , n Œ I Ë 4¯ 4

3p , cos 4x = cos p 2 3p 6x = 2np + , 4x = 2np + p , n Œ Z 2 np p np p x= + ,x= + , n ŒZ 3 4 2 4

sin 6x = sin

p 5p 7p 11p , , , ,… 4 4 12 12 p 3p 5p 7p …x = , , , ,… 4 4 4 4

x=

p 5p , 4 4 Hence, the general solution will be, x=

5p p , 2np + , n ŒZ 4 4 p p fi x = 2np + , (2n + 1)p + , n Œ Z 4 4 p fi x = mp + , m Œ Z 4 122. Given equation is sin4 x = 1 + tan8 x It is possible only when sin4 x = 1, tan8 x = 0 fi sin2 x = 1, tan x = 0 fi

119. Given equation is sin 2x – 12(sin x – cos x) + 12 = 0

TR_02.indd 34

fi

fi

- 2 ± 24 2

…(i)

x = 2np +

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2.35

Trigonometric Equations

p , x = np , n Œ I 2 There is no common value which satisfies both the above equations. Hence, the equation has no solution. 123. Given sin2 x + cos2 y = 2 sec2 z Here, LHS £ 2 and RHS ≥ 2 It is possible only when sin2 x = 1, cos2 y = 1, sec2 z = 1 fi cos2 x = 0, sin2 y = 0, cos2 z = 1 fi cos2 x = 0, sin2 y = 0, sin2 z = 1 fi cos x = 0, sin y = 0, sin z = 0 fi

fi

x = np ±

x = (2n + 1)

p , y = mp , z = k p 2

3x =

fi

x = 2np ±

When cos x = fi

3p , 2x = p 2

fi

1 =2 1 - |cos x |

fi

1 - |cos x | =

to

= 82

fi 1 + |cos x| + cos2 x + |cos x|3 + cos4 x + |cos x|5 + … to = 2

1 2 1 When cos x = 2

TR_02.indd 35

cos x = ±

1 2 1 1 |cos x | = 1 - = 2 2 1 cos x = ± 2

1 + sin q + sin 2q +

p p x = 2np + = (4n + 1) , n ΠI 2 2 126. The given equation can be written as

fi

=4

p 2p . ,± 3 3

128. Given equation is

p 2 2np p fi x= , x = (4n + 1) 4 10 p Thus, x = satisfies both 2 Hence, the solution is

fi

to

1

Hence the values of x are ±

4x = 2np , 5x = (4n + 1)

1 =2 1 - |cos x |

p 2p . ,± 3 3

|cos x| cos 2 x |cos x|3 cos 4 x |cos x|5

21- |cos x| = 4 = 22

fi

p , n ŒI 2 125. Given equation is cos 4x + sin 5x = 2 It is possible only when cos 4x = 1, sin 5x = 1

|cos x | cos 2 x |cos x |3 cos 4 x |cos x |5

2p , n ŒZ 3

fi

fi

x=

81

1 Ê 2p ˆ = cos Á ˜ Ë 3 ¯ 2

Hence the values of x are ±

x = 2np +

fi

p , n ŒZ 3

x = 2np ±

21

p p ,x= 2 2 Hence, the general solution is

fi

Êpˆ cos x = cos Á ˜ Ë 3¯

127. Given equation is

where, n, m, k Œ I 124. Given equation is sin 3x + cos 2x + 2 = 0 It is possible only when sin 3x = –1, cos 2x = –1 fi

fi

=4+2 3

fi

1 =4+2 3 1 - sin q

fi

1 - sin q =

fi

sin q = 1 -

fi

sin q = 1 -

fi

Êpˆ q = np + (-1) n Á ˜ , n Œ I Ë 3¯

1 4+2 3 1 4+2 3 1 4+2 3

=1-

2- 3 3 = 2 2

129. Given equation is sin 2 x - 3 sin x + 1

|cos x |

2

2

=1

fi

3 1ˆ Ê 2 ÁË sin x - sin x + ˜¯ log |cos x | = 0 2 2

fi fi

(2 sin2 x – 3 sin x + 1) log |cos x| = 0 (sin x – 1)(2 sin x – 1) log |cos x| = 0

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Trigonometry Booster

fi

(sin x – 1) = 0, (2 sin x – 1) = 0, log |cos x| = 0

fi

1 sin x = 1, sin x = ,log |cos x | = 0 2 1 sin x = ,|cos x | = 1 2

fi fi fi

fi

1 sin x = , cos = 1, cos x = -1 2 Êpˆ x = np + (-1) n Á ˜ , x = 2np , x = (2n + 1)p Ë 6¯

130. Given equation is esin x – e–sin x – 4 = 0 1 fi t - - 4 = 0, t = esin x t 2 fi t – 4t – 1 = 0 fi (t – 2)2 = 5

Ê sin x ˆ Ê cos x ˆ log cos x Á + log sin x Á =0 ˜ Ë cos x ¯ Ë sin x ˜¯

fi logcos x (sin x) + logsin x (cos x) = 2 It is possible only when sin x = cos x fi tan x = 1 p 4 133. Given equation is fi

x=

2

2

3sin 2x + 2cos x + 31- sin 2x + 2sin x = 28 2

2

fi

3sin 2x + 2cos x + 31- sin 2x + 2 - 2cos x = 28

fi

3sin 2x + 2cos x +

33

2

3sin 2x + 2cos

fi

t=2± 5

fi

esin x = 2 ± 5

fi

a+

fi

sin x = log e (2 ± 5)

fi

sin x = log e (2 + 5)

fi

sin x = log e (2 + 5) > 1

fi fi fi

a2 – 28a + 27 = 0 (a – 27)(a – 1) = 0 a = 27, 1

e[sin

2

x sin 4 x sin 6 x

=e

=e

2

2

tan2 x = 0 x = np, n ΠI

when a = 8, then 2 fi tan2 x = 3 fi tan x = 3

tan 2 x

=8=2

cos x cos x + sin x 1 1 = = 1 + tan x 3 +1 =

( 3 - 1) 2

2 27 = 28, a = 3sin 2x + 2cos x a

when a = 1, then 3sin 2x + 2cos x = 30 fi sin 2x + 2 cos2 x = 0 fi sin 2x + 1 + cos 2x = 0 fi sin 2x + cos 2x = –1

3

1 2

sin 2x +

1 2

cos 2x = -

1

fi

pˆ 1 Ê sin Á 2x + ˜ = Ë 4¯ 2

fi

pˆ Ê nÊ pˆ ÁË 2x + ˜¯ = np + (-1) ÁË - ˜¯ 4 4

2

fi fi

Now,

=2

tan 2 x

fi

when a = 1, then 2 tan x = 1 = 20 2 tan x = 1 = 20

= 28

2

tan 2 x log e 2

Let a = 2 tan x Thus, a2 – 9a + 8 = 0 fi (a – 1)(a – 8) = 0 fi a = 1, 8

fi

x

2

to ]log e 2

Ê sin 2 x ˆ ÁË 1- cos2 x ˜¯ log e 2

2

when a = 27, then 3sin 2x + 2cos x = 33 fi sin 2x + 2 cos2 x = 3 fi sin 2x + 1 + cos 2x = 3 fi sin 2x + cos 2x = 2 It is not possible.

It is not possible So, it has no solution. 131. We have

TR_02.indd 36

132. Given equation is logcos x tan x + logsin x cot x = 0

2

np Ê pˆ p + (-1) n Á - ˜ - , n Œ I Ë 8¯ 8 2 134. Do yourself. 135. Given equation is fi

x=

1 Ê xˆ 2 cos 2 Á ˜ sin 2 x = x 2 + 2 Ë 2¯ x Here, LHS < 2 for 0 < x <

p 2

and RHS ≥ 2 So, it has no solutions.

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2.37

Trigonometric Equations

fi 1 – 2 sin x cos x = t2 fi 2 sin x cos x = (1 – t2) (i) reduces to (1 – t2) + 12 = 12t fi (1 – t2) = 12(t – 1) fi (t + 1)(t – 1) = –12(t –1) fi (t – 1)(t + 1 + 12) = 0 fi (t – 1)(t + 13) = 0 fi t = 1, –13 when t = 1, then sin x – cos x = 1

136. Given equation is Ê x2 + 2 cos 2 Á Ë 6

xˆ x -x ˜¯ = 2 + 2

It is possible only when x = 0 Hence, the solution is x = 0.

LEVEL III 1. The given equation is sec x - cosec x =

4 3

1 1 4 = cos x sin x 3 fi 3(sin x – cos x) = 4 sin x cos x Put (sin x – cos x) fi 1 – 2 sin x cos x = t2 fi

fi

sin x cos x =

…(i)

1 - t2 2

Ê1 - t2 ˆ (i) reduces to 3t = 4 Á Ë 2 ˜¯ fi fi fi fi fi

3t = 2(1 – t2) 2t2 + 3t – 2 = 0 2t2 + 4t – t – 2 = 0 2t(t + 2) – (t + 2) = 0 (1 + 2)(2t – 1) = 0

fi

t=

1 , -2 2

1 when t = 2 1 2

fi

sin x - cos x =

fi

1 1 Ê 1 ˆ sin x cos x˜ = ÁË ¯ 2 2 2 2

fi

pˆ 1 Ê sin Á x - ˜ = Ë ¯ 4 2 2

fi

pˆ Ê n -1 Ê 1 ˆ ÁË x - ˜¯ = np + (-1) ◊ sin Á Ë 2 2 ˜¯ 4

Ê np Ê 1 ˆˆ x=Á , n ŒI + np + (-1) n ◊ sin -1 Á Ë 2 2 ˜¯ ˜¯ Ë 4 when t = –2 fi sin x – cos x = 2 It is impossible, since the maximum value of (sin x - cos x) is 2 . 2. The given equation is sin 2x + 12 = 12 (sin x – cos x) Put (sin x – cos x) = t

fi

1 1 Ê 1 ˆ sin x cos x˜ = ÁË ¯ 2 2 2

fi

pˆ Ê Êpˆ sin Á x - ˜ = sin Á ˜ Ë ¯ Ë 4¯ 4

fi

pˆ Ê n Êpˆ ÁË x - ˜¯ = np + (-1) ÁË ˜¯ 4 4

fi

x = np +

when t = –13 sin x – cos x = –13 It is impossible, since the maximum value of (sin x - cos x) is 2 3. The given equation is |sec x + tan x| = |sec x| + |tan x| fi sec x ◊ tan x ≥ 0 fi

sin x cos 2 x

≥0

fi

sin x ≥ 0, cos x π 0

fi

x Œ[0, p ] -

fi

È pˆ Êp ˘ x Œ Í0, ˜ » Á , p ˙ Î 2¯ Ë 2 ˚

{} p 2

p p Hence, the solution set is ÈÍ0, ˆ˜ » ÊÁ , p ˘˙ ¯ Ë Î 2 2 ˚ 4. The given equation is fi

n Êpˆ Êpˆ sin Á ˜ + cos Á ˜ = Ë 2n ¯ Ë 2n ¯ 2 1 1 n Êpˆ Êpˆ sin Á ˜ + cos Á ˜ = Ë ¯ Ë ¯ 2 2 n n 2 2 2 2

fi

TR_02.indd 37

p Êpˆ + (-1) n Á ˜ , n Œ I Ë 4¯ 4

fi

n Êp p ˆ cos Á - ˜ = Ë 4 2n ¯ 2 2

It is satisfied for n = 6 only. 5. The given equation is cos 2x + a sin x = 2a – 7 fi 1 – 2 sin2 x + a sin x = 2a – 7 fi 2 sin2 x – a sin x + (2a – 8) = 0

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Trigonometry Booster

8. The given equation is

fi

a ± a 2 - 16(a - 4) sin x = 4

fi

sin x =

a ± a - 16a + 64 4

fi

3 1ˆ Ê 2 ÁË sin x - sin x + ˜¯ log |cos x | = 0 2 2

fi

sin x =

a ± (a - 8) 2 4

fi

3 1ˆ Ê 2 ÁË sin x - sin x + ˜¯ = 0, log |cos x | = 0 2 2

fi

a ± (a - 8) sin x = 4

fi

sin x =

2a - 8 ,2 4

fi

sin x =

a-4 2

fi

Ê a - 4ˆ -1 £ Á £1 Ë 2 ˜¯

sin 2 x - 3 sin x + 1

|cos x |

2

5

29 Ê 1 - cos 2x ˆ Ê 1 - cos 2x ˆ 4 ÁË ˜¯ + ÁË ˜¯ = cos 2x 2 2 16 fi

2 29 (1 + 10 cos 2 2x + 5 cos 4 2x) = cos 4 2x 32 16

fi fi fi fi

(1 + 10 cos2 2x + 5 cos4 2x) = 29 cos4 2x 24 cos4 2x – 10 cos2 2x – 1 = 0 (2 cos2 2x – 1)(12 cos2 2x + 1) = 0 (2 cos2 2x – 1) = 0, since (12 cos2 2x + 1) π 0

fi

cos 2 2x =

1 2

fi fi

2 cos 2x – 1 = 0 cos 4x = 0

fi

4x = (2n + 1)

fi

p x = (2n + 1) , n ΠI 8

TR_02.indd 38

=1

fi

sin x =

1 ,1 2

fi

sin x =

1 , since |cos x | = 0 2

fi

Êpˆ x = np + (-1) n Á ˜ , n Œ I Ë 6¯

when log |cos x| = 0 fi log |cos x| = log 1 fi |cos x| = 1 fi cos x = ±1 when cos x = 1 fi x 2np when cos x = –1 x (2n + 1)p Hence, the solution is Êpˆ x = 2np , (2n + 1)p , np + (-1) n Á ˜ , n Œ I Ë 6¯ 9. It is possible only when cos(p x - 4) = 1 and cos(p x ) = 1 x = 4 and x = 0 x = 0 does not satisfy the equation simultaneously. Hence, the solution is x = 4 Therefore, the number of solution is 1. 10. The given equation is Êpˆ x 4 - 2x 2 sin 2 Á ˜ x + 1 = 0 Ë 2¯

2

2

Ê 2 2Êpˆ ˆ 4Êpˆ ÁË x - sin ÁË ˜¯ x˜¯ + 1 - sin ÁË ˜¯ x = 0 2 2

p , n ŒI 2

Hence, the solution is x = (2n + 1)

2

3 1 when ÊÁ sin 2 x - sin x + ˆ˜ = 0 Ë 2 2¯ 2 fi (2 sin x – 3 sin x + 1) = 0 fi (2 sin2 x – 2 sin x – sin x + 1) = 0 fi 2 sin x (sin x – 1) – (sin x – 1) = 0 fi (2 sin x – 1)(sin x –1) = 0 fi (2 sin x – 1) = 0, (sin x – 1) = 0

fi –2 £ (a – 4) £ 3 fi 2£a£6 fi a Œ [2, 6] 6. The given equation is sin100 x – cos100 x = 1 It is possible only when sin x = 1, cos x = 0 Hence, the general solution is p x = np + , n Œ I 2 7. The given equation is 29 fi sin10 x + cos10 x = cos 4 2x 16 5

2

It is possible only when Ê 2 2Êpˆ ˆ 4Êpˆ ÁË x - sin ÁË ˜¯ x˜¯ = 0, 1 - sin ÁË ˜¯ x = 0 2 2 p , n ŒI . 8

p when 1 - sin 4 ÊÁ ˆ˜ x = 0 Ë 2¯

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2.39

Trigonometric Equations

fi

Êpˆ sin 4 Á ˜ x = 1 Ë 2¯

Ê p x2 ˆ =1 Also, sin Á Ë 2 ˜¯

fi

Êpˆ Êpˆ sin 4 Á ˜ x = sin 2 Á ˜ Ë 2¯ Ë 2¯

fi

fi

Êpˆ Êpˆ ÁË ˜¯ x = np ± ÁË ˜¯ 2 2

fi

x = (2n ± 1), n Œ I

Ê p ˆ when Á x 2 - sin 2 ÁÊ ˜ˆ x˜ = 0 Ë 2¯ ¯ Ë fi

Êpˆ x 2 = sin 2 Á ˜ x Ë 2¯

fi x2 = 1 fi x ±1 Hence, the number of solutions is 2. 11. The given equation is cos4 x + a cos2 x + 1 = 0 …(i) Let cos2 x = 1 Then t Œ [0, 1] (i) reduces to t2 + at + 1 = 0. since it has at-least one real root in [0, 1], so a2 – 4 ≥ 0 and 1 + a + 1 £ 0 fi |a| ≥ 2, a £ –2 fi a ≥ 2, a £ –2; a £ –2 fi a £ –2 fi a Œ (– , –2] 12. The given equation is tan4 x – 2 sec2 x + b2 = 0 fi tan2 x = 2 (1 + tan2 x) + b2 = 0 fi tan4 x – 2 tan2 x + 1 = 3 – b2 fi (tan2 x – 1)2 = 3 – b2 fi (3 – b2) = (tan2 x – 1)2 ≥ 0 fi (3 – b2) ≥ 0 fi b2 £ 3 fi |b| £ 3 13. The given equation is x2 + 4 + 3 sin (ax + b) = 2x fi (x2 – 2x + 1) + 3 + 3 sin (ax + b) = 0 fi (x – 1)2 + 3(1 + sin (ax + b)) = 0 It is possible only when (x – 1) = 0, (1 + sin (ax + b)) = 0 fi x = 1, sin (ax + b) = –1 fi sin (a + b) = –1 fi

(a + b) = (4n - 1)

3p 7p , 2 2 14. The given equation is |x| + |y| = 4 fi –4 £ x, y £ 4 fi |x| £ 4, |y| £ 4 fi

TR_02.indd 39

( a + b) =

p , n ŒI 2

Ê p x2 ˆ p ÁË ˜ = (4n + 1) , n Œ I 2 ¯ 2

fi fi fi Then

x2 = (4n + 1) x2 = 1 x = ±1 |y| = 4 – 1 = 3 y = ±3 Thus, the possible ordered pairs are (1, 3), (1, –3), (–1, 3), (–1, –3). 15. The given equation is log|cos x| |sin x| + log|sin x| |cos x| = 2 It is possible only when |sin x| = |cos x| π 1 fi |tan x| = 1 3p 5p 7p p ,± ,± ,± 4 4 4 4 Hence, the number of values of x is 8. 16. The given equation is tan x + sec x = 2 cos x sin x + 1 fi = 2 cos x cos x fi

x=±

fi fi fi fi

sin x + 1 = 2 cos2 x 1 + sin x = 2 (1 – sin x) (1 + sin x) (1 + sin x) (1 – 2 + 2 sin x) = 0 (1 + sin x) = 0, (2 sin x – 1) = 0

fi

sin x = - 1, sin x =

fi

x=

3p p 5p ;x= , 2 6 6

fi

x=

p 5p , 6 6

Hence, the solutions are

1 2

{ }

p 5p . , 6 6

17. The given equation is 2 sin2 x + 5 sin x –3 = 0 fi 2 sin2 x + 6 sin x – sin x – 3 = 0 fi 2 sin x (sin x + 3) – (sin x + 3) = 0 fi (sin x + 3) (2 sin x – 1) = 0 fi (sin x + 3) = 0, (2 sin x – 1) = 0 fi 2(sin x – 1) = 0 1 sin x = fi 2 p 5p 11p 17p , , , 6 6 6 6 Hence, the number of solutions is 4. fi

x=

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2.40

Trigonometry Booster

18. The given equation is Êqˆ 5 cos(2q ) + 2 cos Á ˜ + 1 = 0 Ë 2¯ 2

fi fi fi fi fi

5(2 cos2 q – 1) + (1 + cos q) + 1 = 0 10 cos2 q + cos q – 3 = 0 10 cos2 q + 6 cos q – 5 cos q – 3 = 0 2 cos q (5 cos q + 3 – 1)(5 cos q + 3) = 0 (2 cos q – 1)(5 cos q + 3) = 0

fi

1 3 cos q = , cos q = 2 5

When cos q =

1 2

3 5

p Ê 3ˆ - - cos -1 Á - ˜ Ë 5¯ 2

Hence, the solutions are p p Ê 3ˆ p Ê 3ˆ , + cos -1 Á - ˜ , - - cos -1 Á - ˜ Ë 5¯ Ë 5¯ 3 2 2 19. The given equation is 2(sin x – cos 2x) – sin 2x (1 + 2 sin x) + 2 cos x = 0 fi 2 sin x – 2 cos 2x – 2 sin x cos x – sin 2x + 2 cos x = 0 fi 2 sin x (1 – cos x) + 4 cos3 x – 4 cos2 x – 2 cos x + 2 = 0 fi 2 sin x (1 – cos x) + 4 cos2 x (cos x – 1) –2 (cos x – 1) = 0 fi (cos x – 1)(4 cos2 x – 2 – 2 sin x) = 0 fi (cos x – 1)(2 sin2 x + sin x – 1) = 0 fi (cos x – 1)(sin x + 1) (2 sin x – 1) = 0 1 fi cos x = 1, sin x = - 1, sin x = 2 p Êpˆ fi x = 2np , (4n - 1) , np + (-1) n Á ˜ Ë 6¯ 2 20. The given equations are x cos3 y + 3x cos y sin2 y = 14, 3 x sin y + 3x cos2 y sin y = 13 Adding and subtracting, we get, x(cos y + sin y)3 = 27 …(i) x (cos y – sin y)3 = 1 …(ii) Dividing (ii) by (i), we get, q=±

(cos y + sin y )3 fi

TR_02.indd 40

fi

tan y =

fi

Ê 1ˆ y = tan -1 Á ˜ Ë 2¯

1 2

-1 Ê 1 ˆ Put the value of y = tan ÁË ˜¯ into (ii), we get 2 3

fi

1 ˆ Ê 2 xÁ ˜ =1 Ë 5 5¯ 3

p Ê 3ˆ Then q = + cos -1 Á - ˜ Ë 5¯ 2 and

1 + tan y =3 1 - tan y

Ê 1 ˆ xÁ ˜ = 1 Ë 5¯

p p Then q = - , 3 3 When cos q = -

fi

= 27 (cos y - sin y )3 (cos y + sin y ) =3 (cos y - sin y )

x=5 5 fi Hence, the solutions are Ê 1ˆ x = 5 5 and y = tan -1 Á ˜ Ë 2¯ 21. The given equation is 4 sin4 x + cos4 x = 1 fi 5 sin4 x + cos4 x = (sin2 x + cos2 x)2 fi 4 sin4 x + cos4 x = sin4 x + cos4 x + 2 sin2 x cos2 x fi 3 sin4 x – 2 sin2 x cos2 x = 0 fi 3 sin4 x – 2 sin2 x + 2 sin4 x = 0 fi 5 sin4 x – 2 sin2 x = 0 fi sin2 x (5 sin2 x – 2) = 0 fi sin2 x = 0, (5 sin2 x – 2) = 0 2 5

fi

sin x = 0, sin 2 x =

fi

Ê 2ˆ x = np , np ± a , where a = sin -1 Á Ë 5 ˜¯

22. The given equation is sin4 x + cos4 x + sin 2x + a = 0 fi 1 – 2 sin2 x cos2 x + sin 2x + a = 0 1 fi 1 - (4 sin 2 x cos 2 x) + sin 2x + a = 0 2 fi 1 - 1 (sin 2x) 2 + sin 2x + a = 0 2 fi 2 – (sin 2x)2 + 2 sin 2x + 2a = 0 fi (sin 2x)2 + 2 sin 2x – 2a = 2a fi (sin 2x – 1)2 – 3 = 2a fi (sin 2x – 1)2 = 2a + 3 fi 2a + 3 = (sin 2x – 1)2 ≥ 0 3 2 Also, (sin 2x – 1)2 ≥ 0 fi (sin 2x – 1) ≥ 0 fi sin 2x ≥ 1 fi sin 2x = 1 fi

a≥-

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2.41

Trigonometric Equations

fi

25. Given 3 cos 2q = 1 1 fi cos 2q = 3

Êpˆ 2x = np + (-1) n Á ˜ Ë 2¯

np Êpˆ + (-1) n Á ˜ , n Œ I Ë 4¯ 2 23. The given equation is fi

x=

fi

Ê 3 - 1ˆ Ê 3 + 1ˆ 1 ÁË ˜¯ sin q + ÁË ˜ cos q = 2 2 2 2 ¯ 2 1 sin q cos (75°) + cos q sin (75°) = 2

2

1 3

4

fi

5p ˆ 1 Ê sin Á q + ˜= Ë 12 ¯ 2

fi

Ê 1ˆ 32 Á ˜ = 2 cos 2a - 3 cos a Ë 2¯

fi

5p ˆ 1 Ê Êpˆ sin Á q + = sin Á ˜ ˜¯ = Ë Ë 4¯ 12 2

fi

5p ˆ Ê nÊpˆ ÁË q + ˜¯ = np + (-1) ÁË ˜¯ , n Œ I 12 4

fi

Ê Ê p ˆ 5p ˆ , n ŒI q = Á np + (-1) n Á ˜ Ë 4 ¯ 12 ˜¯ Ë

fi fi fi fi fi fi

2 cos2 a – 3 cos a = 2 2 cos2 a – 3 cos a – 2 = 0 2 cos2 a – 4 cos a + cos a – 2 = 0 2 cos a (cos a – 2) + 1 (cos a – 2) = 0 (2 cos a + 1) (cos a – 2) = 0 (2 cos a + 1) = 0, (cos a – 2) = 0

fi

1 cos a = - , 2 2 cos a = 2 is not possible.

24. The given equation is sec q - cosec q = 3(sin q – cos q) = 4 sin q cos q Let sin q – cos q = t’

4 3

fi …(i)

1 - t2 2 Equation (i) reduces to

Then sin q cos q =

fi

Ê1 - t2 ˆ 3t = 4 ¥ Á = 2(1 - t 2 ) Ë 2 ˜¯

fi fi

2t2 + 3t – 2 = 0 (2t – 1)(t + 2) = 0

1 fi t = , -2 2 when t = –2, (sin q – cos q) = –2 It is not possible. 1 1 when t = , sin q - cos q = 2 2 pˆ 1 Ê fi sin Á q - ˜ = Ë 4¯ 2 2

TR_02.indd 41

=

1 + tan q fi 1 + tan2 q = 3 – 3 tan2 q fi 4 tan2 q = 2 1 fi tan 2q = 2 Also, it is given that, fi 32 tan8 q = 2 cos2 a – 3 cos a

( 3 - 1)sin q + ( 3 + 1)cos q = 2 fi

1 - tan 2q

fi

pˆ 1 Ê sin Á q - ˜ = = sin a Ë ¯ 4 2 2

fi

pˆ Ê n -1 Ê 1 ˆ ÁË q - ˜¯ = np + (-1) a , where a = sin Á Ë 2 2 ˜¯ 4

fi

pˆ Ê q = Á np + (-1) n a + ˜ , n Œ I Ë 4¯

Also, when cos a = -

1 2

2p 3 26. Given equations are 5 sin x cos y = 1 and 4 tan x = tan y fi 5 sin x cos y = 1 and 4 sin x cos y = cos x sin y Dividing (i) by (ii), we get, fi

x = 2np ±

4 5 Adding (i) and (iii), we get, sin (x + y) = 1 cos x sin y =

p , n ŒI 2 Subtracting (iii) from (i), we get,

fi

( x + y ) = (4n + 1)

3 sin ( x - y ) = - = sin a 5 fi (x – y) = mp + (–1)n a, m Œ I From (iv) and (v), we get, 2x = (2n + m)p + fi

x = (2n + m)

…(i) …(ii) …(iii)

…(iv)

…(v)

p + (-1) m a 2

p p a Ê 3ˆ + + (-1) m , a = sin -1 Á - ˜ Ë 5¯ 2 4 2

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Trigonometry Booster

and fi

p Ê 3ˆ 2y = (2n - m)p + - (-1) m a , a = sin -1 Á - ˜ Ë 5¯ 2 p p a y = (2n - m) + - (-1) m 2 4 2

27. Given equations are ( x - y ) =

p , 4

and cot x + cot y = 2 fi cos x sin y + sin x cos y = 2 sin x sin y fi sin (x + y) = cos (x – y) – cos (x + y) fi sin (x + y) + cos (x + y) = cos (x – y) 1 1 1 sin ( x + y ) + cos ( x + y ) = cos ( x - y ) 2 2 2

fi

1 1 1 1 sin ( x + y ) + cos ( x + y ) = ◊ 2 2 2 2 pˆ 1 Ê sin Á x + y + ˜ = Ë 4¯ 2

fi

p ˆ 5p Ê ÁË x + y + ˜¯ = 4 6

fi

( x + y) =

Also, ( x - y ) = Thus, x =

p 4

=1

2(2cosec x + 3|sec y |) = 64 From (i), we get cosec2 x – 3 sec2 y = 0 fi cosec2 x = 3 sec2 y cosec x = 3 |sec y | fi Also, from (ii), we get 2(2cosec x + fi fi fi fi fi fi

= 64 = 26

2 cosec x + 3 |sec y | = 6 2 cosec x + cosec x = 6, from (iii) 3 cosec x = 6 cosec x = 2 1 sin x = 2 Êpˆ x = np + (-1) n Á ˜ , n Œ I Ë 6¯

Again, |sec y | =

TR_02.indd 42

3|sec y|)

2 3

fi

Êpˆ y = np ± Á ˜ , n Œ I . Ë 6¯

3 Êpˆ = cos 2 Á ˜ Ë 6¯ 4

Ï n Êpˆ ÔÔ x = np + (-1) ÁË 6 ˜¯ , n Œ I Ì Ô y = np ± Ê p ˆ , n Œ I ÁË ˜¯ ÔÓ 6

Ê xˆ Ê xˆ cot Á ˜ - cosec Á ˜ = cot x Ë 2¯ Ë 2¯

5p p 7p - = 6 4 12

x - sec2 y )

cos 2 y =

1. Given equation is

28. The given equations are 2

fi

LEVEL IV

5p p and y = 12 6

5(cosec

|cos y | =

Hence, the solutions are

fi

fi

3 2

fi

…(i) …(ii)

…(iii)

fi

cos ( x /2) - 1 = cot x sin ( x /2)

fi

-

2 sin 2 ( x /4) = cot x sin ( x /2)

fi

-

2 sin 2 ( x /4) = cot x 2 sin ( x /4) cos ( x /4)

fi

tan (x/4) + cot x = 0

fi

sin ( x /4) cos x + =0 cos ( x /4) sin x

fi

xˆ Ê cos Á x - ˜ = 0 Ë 4¯

fi

Ê 3x ˆ cos Á ˜ = 0 Ë 4¯

fi

3x p = (2n + 1) 4 2

fi

x = (4n + 2)

p , n ŒI 3

2. Given equation is 8 cos x ◊ cos 2x ◊ cos 4x = fi fi fi fi fi

sin 6x sin x

4 sin 2x cos 2x cos 4x = sin 6x 2 sin 4x cos 4x = sin 6x sin 8x – sin 6x = 0 2 cos (7x) sin x = 0 cos (7x) = 0, sin x = 0

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2.43

Trigonometric Equations

fi

(7x) = (2n + 1)

fi

x = (2n + 1)

p , x = np 2

p , x = np , n ΠI 14

2. Given equation is tan x tan 2x + +2=0 fi tan 2x tan x fi fi fi fi fi

(tan x + tan 2x)2 = 0 (tan x + tan 2x) = 0 sin (2x + x) = 0 sin (3x) = 0 3x = np

np , n ŒI 3 4. Given equation is cos x cos (6x) = –1 fi 2 cos (6x) cos x = –2 fi cos (7x) + cos (5x) = –2 It is possible only when cos (7x) = –1, cos (5x) = –1 fi 7x = (2k + 1)p, 5x = (2m + 1)p fi

x=

p p , x = (2m + 1) 7 5 when k = 3 and m = 2, then common value of x is p Hence, the general solution is x = 2np + p = (2n + 1)p, n ΠI 5. Given equation is cos (4x) + sin (5x) = 2 It is possible only when cos (4x) = 1, sin (5x) = 1 fi

fi

x = (2k + 1)

4x = 2k p , 5x = (4m + 1)

p 2

kp p , x = (4m + 1) , k , m ΠI 2 10 when k = 1 and m = 1, then the common value of x is p . 2 Hence, the general solution is

fi

x=

pˆ p Ê x = Á 2np + ˜ = (4n + 1) , n Œ I Ë ¯ 2 2 6. Given equation is (1 + sin 2x) + 5(sin x + cos x) = 0 fi (sin x + cos x)2 + 5(sin x + cos x) = 0 fi ((sin x + cos x) + 5)(sin x + cos x) = 0 fi ((sin x + cos x) + 5) = 0, (sin x + cos x) = 0 fi (sin x + cos x) = 0 fi tan x = 1 p x = np - , n Œ I 4

TR_02.indd 43

7. Given equation is sin x + sin 2x + sin 3x = cos x +cos 2x + cos 3x fi (sin 3x + sin x) + sin 2x = (cos 3x + cos x) + cos 2x fi 2 sin 2x cos x + sin 2x = 2 cos 2x cos x + cos 2x fi sin 2x(2 cos x +1) = cos 2x(2 cos x + 1) fi (sin 2x – cos 2x)(2 cos x + 1) = 0 fi (sin 2x – cos 2x) = 0, (2 cos x + 1) = 0 when (sin 2x – cos 2x) = 0 fi tan 2x = 1 p 4 np p fi x= + 2 8 when 2 cos x + 1 = 0 fi

2x = np +

fi

cos x = -

fi

1 2 2p x = 2np ± 3

Hence, the solution is p 5p 9p 13p 2p 4p , , , , , 8 8 8 8 3 3 8. Given equation is x=

sin 3 x cos3 x + - 2 sin x cos x = 2 cos x sin x fi

sin 4 x + cos 4 x = 2 sin x cos x + 2 sin x cos x

fi fi fi

1 – 2 sin2 x cos2 x = 2 sin2 x cos2 x + sin (2x) 4 sin2 x cos2 x + sin (2x) – 1 = 0 sin2 2x + sin (2x) – 1 = 0

fi

sin (2x) =

-1 ± 1 + 4 -1 ± 5 = 2 2

fi

sin (2x) =

-1 + 5 2

fi

Ê -1 + 5 ˆ sin(2x) = sin a , a = sin -1 Á ˜ Ë 2 ¯

fi

(2x) = np + (–1)n a

np a + (-1) n 2 2 9. Given equation is sin2 4x + cos2 x = 2 sin 4x cos4 x fi sin2 4x – 2 sin 4x cos4 x + cos2 x = 0 fi (sin2 4x – cos4 x)2 + cos2 x – cos8 x = 0 fi (sin2 4x – cos4 x)2 + cos2 x (1 – cos6 x) = 0 It is possible only when (sin2 4x – cos4 x) = 0, cos2 x (1 – cos6 x) = 0 Now, cos x = 0, cos2 x = 1 fi

x=

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2.44

Trigonometry Booster

when cos x = 0 then x = (2n + 1) So,

Ê sin 4 Á n + Ë

12. Given equation is

p 2

1ˆ ˜p =0 2¯

fi

which is true when cos2 x = 1, then x = np which is not satisfied the equation sin (4x) – cos4 x = 0 Hence, the solution is x = (2n + 1)

2

p . 2

10. Given equation is 7 sin x cos x 2 7 1 - 2 sin 2 x cos 2 x = sin(2x) 4 sin 4 x + cos 4 x =

fi fi fi fi fi fi fi fi

4 – 8 sin x cos x = 7 sin (2x) 4 – 2 sin2 2x – 7 sin (2x) = 0 2 sin2 2x + 7 sin (2x) – 4 = 0 2 sin2 2x + 8 sin (2x) – sin (2x) – 3 = 0 (2 sin (2x) +1)(sin (2x) + 4) = 0 (2 sin (2x) + 1) = 0, (sin (2x) + 4) = 0 (2 sin (2x) + 1) = 0

fi

1 sin (2x) = 2

fi

Ê pˆ sin (2x) = sin Á - ˜ Ë 6¯

fi

Ê pˆ 2x = np + (-1) Á - ˜ Ë 6¯

2

fi

Ê p ˆˆ Ê (2 sin x) + Á 2 sin 2 Á x + ˜ ˜ = 1 Ë Ë 4¯¯

fi

Ê p ˆˆ Ê (1 - cos (2x)) 2 + Á1 - cos Á 2x + ˜ ˜ = 1 Ë Ë 2¯¯

fi fi fi fi

(1 – cos (2x))2 + (1 + sin (2x))2 = 1 1 – 2 cos (2x) + 1 + 2 sin (2x) = 0 2 – 2 (cos (2x) – sin (2x)) = 0 (cos (2x) – sin (2x)) = 1

fi

1 1 Ê 1 ˆ cos (2x) sin (2x)˜ = ÁË ¯ 2 2 2

fi

pˆ 1 Ê cos Á 2x + ˜ = Ë ¯ 4 2

fi

pˆ p Ê ÁË 2x + ˜¯ = 2np ± 4 4

2

x = np , x = np -

3 sin x + cos x 2 3 sin x 2

The equation will provide us a real solutions if 1 2

fi fi fi fi fi

2 – 4 sin2 x cos2 x = 2 cos (4x) + 1 2 – sin2 2x = 2 cos (4x) + 1 2 cos (4x) + sin2 2x = 1 2(1 – 2 sin2 2x) + sin2 (2x) = 1 3 sin2 (2x) = 1

fi

1 3 1 sin 2 (2x) = = sin 2a 3 2x = np ± a np a Ê 1ˆ x= ± , n Œ I , a = sin -1 Á ˜ Ë 3¯ 2 2

fi

TR_02.indd 44

1 2

1 - 2 sin 2 x cos 2 x = cos(4x) +

fi

2

1 = cos x 2 3 = cos x 2

fi

fi

2

Êpˆ Êpˆ = cos x cos Á ˜ - sin x sin Á ˜ + cos x Ë 3¯ Ë 3¯

np Ê pˆ fi x= + (-1) n Á - ˜ , n Œ I Ë 12 ¯ 2 11. Given equation is

sin 2 (2x) =

2

p , n ŒI 4 pˆ Ê 13. We have a = cos Á x + ˜ + cos x Ë 3¯ fi

n

sin 4 x + cos 4 x = cos(4x) +

pˆ 1 Ê sin 4 x + sin 4 Á x + ˜ = Ë 4¯ 4 pˆ Ê 4 sin 4 x + 4 sin 4 Á x + ˜ = 1 Ë 4¯

fi

9 3 + £a£ 4 4

9 3 + 4 4

- 3£a£ 3

14. Let f ( x) = cos x - x + Now, f (0) = 1 +

1 2

1 3 = >0 2 2

p 1 1 p Êpˆ f Á ˜ =0- + = - 1 and (1 - 2) < 0 , so there is no value of q satisfying the given equation. 5. No questions asked in 1985. 3 6. We have cos x + cos y = 2 Ê x + yˆ Ê x - yˆ 3 2 cos Á cos Á = Ë 2 ˜¯ Ë 2 ˜¯ 2 Êpˆ Ê x - yˆ 3 2 cos Á ˜ cos Á = fi Ë 2 ˜¯ 2 Ë 3¯ fi fi

1 Ê x - yˆ 3 ¥ cos Á = Ë 2 ˜¯ 2 2 Ê x - yˆ 3 cos Á = Ë 2 ˜¯ 2 2¥

It is not possible, so the solution set is x=j

TR_02.indd 49

7. The given inequation is 2 sin2 x – 3 sin x + 1 ≥ 0 fi (2 sin x – 1)(sin x – 1) ≥ 0 fi

sin x £

fi

sin x £

1 and sin x ≥ 1 2

1 and sin x = 1 2 p È p ˘ È 5p ˘ x Œ Í0, ˙ » Í , p ˙ and x = 2 Î 6˚ Î 6 ˚

Hence, the solution set is

{}

p È p ˘ È 5p ˘ x Œ Í0, ˙ » Í , p ˙ » 2 Î 6˚ Î 6 ˚ 8. As we know that tan x ≥ x

Ê pˆ So there is no root between Á 0, ˜ Ë 2¯ Êp ˆ Ê 3p ˆ , 2p ˜ ÁË , p ˜¯ and ÁË ¯ 2 2 Ê 3p ˆ But there is a root in Á p , ˜ Ë 2¯ 9. The given equation is sin x – 3 sin 2x + sin 3x = cos x – 3 cos 2x = cos 3x fi (sin x + sin 3x) – 3 sin 2x (cos x + cos 3x) – 3 cos 2x fi 2 sin 2x cos x – 3 sin 2x = 2 cos 2x cos x – 3 cos 2x fi (2 cos x – 3) (sin 2x – cos 2x) = 0 fi (sin 2x – cos 2x) = 0, ({ 2 cos x – 3 π 0) tan 2x = 1 fi

Êpˆ tan 2x = 1 = tan Á ˜ Ë 4¯

fi

pˆ Ê 2x = Á np + ˜ , n Œ I Ë 4¯

fi

Ê np p ˆ x=Á + ˜ , n ŒI Ë 2 8¯

10. No questions asked in between 1990 to 1992. 11. The given equation is tan x + sec x = 2 cos x fi sin x + 1 = 2 cos2 x fi sin x + 1 = 2(1 – sin2 x) fi (sin x + 1) = 2(1 – sin x)(1 + sin x) fi (sin x + 1)(1 – 2(1 – sin x)) = 0 fi

sin x = - 1, sin x =

fi

x=

1 2

p 5p 3p , , 6 6 2

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2.50

Trigonometry Booster

3p But x = does not satisfy the given equation 2 Hence, the number of solutions is 2. 12. The given equation can be written as tan (x + 100°) cot x = tan (x + 50°) tan (x – 50°) fi

sin ( x + 100°) cos x sin ( x + 50°) sin ( x - 50°) = cos ( x + 100°) sin x cos ( x + 50°) cos ( x - 50°)

Applying compenendo and dividendo, we get, fi sin ( x +100°) cos x + cos ( x +100°)sin x sin ( x +100°) cos x - cos ( x +100°)sin x sin ( x + 50°)sin ( x - 50°) + cos ( x + 50°) cos ( x - 50°) = sin ( x + 50°)sin ( x - 50°) - cos ( x + 50°) cos ( x - 50°) fi

sin ( x + 100° + x) cos ( x + 50° - x + 50°) = sin ( x + 100° - x) - cos ( x + 50° + x - 50°)

fi sin (2x + 100°) cos 2x = –sin (100°) cos (100°) fi 2 sin (2x + 100°) cos 2x = –2 sin (100°) cos (100°) fi sin (4x + 100°) + sin (100°) = –sin (200°) fi sin (4x + 100°) = –(sin (200°) + sin (100°)) fi sin (4x + 100°) = –2 sin (150°) cos (50°) 1 ¥ sin (40°) 2 fi sin (4x + 100°) = –sin (40°) = sin (220°) fi (4x + 100°) = (220°) fi 4x = 120° fi x = 30° Hence, the result.

fi

sin (4x + 100°) = - 2 ¥

p 2n The given equation reduces to

13. Let q =

n 2 n 2 (sin q + cos q ) = 4

sin q + cos q = fi fi

1 + 2 sin q cos q =

fi

sin 2q =

n 4

n Ê n - 4ˆ -1= Á Ë 4 ˜¯ 4

As per choices, n ≥ 4 p fi 0 < 2q < 2 fi 0 < sin 2q < 1 Ê n - 4ˆ fi 0 k or f (x) < k When we solve the inequation, we often use the graphs of the functions y = f (x) and y = k. Then, the solution of the inequality f (x) > k is the value of x, for which the point (x, f (x)) of the graph of y = f (x) lies above the straight line y = k. Y

Type II: An in-equation is of the form sin x < k Rule: Find the smallest values of x satisfies the given inequation and then add 2n p with that values of x. Y

y=k X¢

X

y=k X¢

Y¢

X

O

Y¢

Similarly, when we solve f (x) < k, then the solution of the inequalition f (x) < k is the values of x for which the point (x, f (x)) of the graph of y = f (x) lies below the straight line y = k.

Type III: An in-equation is of the form cos x > k Rule: First we find the smallest interval for which x satisfies the given in-equation and then add 2n p with each values of x. Y

Y

X¢

y=k X

O

y=k X¢

O

x

Y¢

Y¢

Type I: An in-equation is of the form sin x > k Rule: Find the smallest values of x satisfies the given inequation and then add 2n p with that values of x

Type IV: An in-equation is of the form cos x < k Rule: First we find the smallest interval for which x satisfies the given inequation and then add 2n p with each values of x. Y

Y

X¢

O Y¢

TR_03.indd 1

y=k y=k X

X¢

O cos–1 (k)

X 2p – cos–1(k)

Y¢

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3.2

Trigonometry Booster

Type V: An in-equation is of the form tan x > k Rule: First we find the smallest interval for which x satisfies the given in-equation and then add n p with each values of x.

Type VI: An in-equation is of the form tan x < k Rule: First we find the smallest interval for which x satisfies the given in equation and then add n p with each values of x.

Y

Y

y=k X¢

–p O 2

p 2

y=k

X

X¢

–p O 2

Y¢

p 2

X

Y¢

E XERCISES

LEVEL I

Type 6

(Problems Based on Fundamentals)

Type 1 1. 2. 3. 4.

Solve sin x > 1/2 Solve sin x > 1/3 Solve sin x ≥ 1 Solve sin x > 0

Type 2 5. Solve sin x < 1/2 6. Solve sin x < 1/5 3 7. Solve sin x £ 2

Type 3 1 2

8. Solve cos x > 9. Solve cos x ≥

1 2

Type 4 10. Solve cos x <

1 3

11. Solve cos x £

3 2

12. Solve tan x > 1

14. Solve tan x > 2

TR_03.indd 2

Mixed Problems 17. Solve sin x > cos x 18. Solve cos x > sin x 1 1 19. Solve - £ cos x < 2 2 20. Solve |sin x + cos x| = |sin x| + |cos x| 21. Solve sin x sin 2x < sin 3x sin 4x, Ê pˆ " x Œ Á 0, ˜ Ë 2¯ 22. Solve cos x – sin x – cos 2 x > 0, " x Œ (0, 2p) 5 1 23. Solve sin 2 x + sin 2 2x > cos2x 4 4 24. Solve 6 sin2x – sin x cos x – cos2x > 2 13 6 6 25. Solve sin x + cos x > 16 5 8 13 27. Solve the inequality sin 6 x + cos6 x > 16 5 2 1 2 28. Solve sin x + sin 2 x > cos 2 x 4 4 3 3 26. Solve cos x ◊ cos 3x - sin x sin 3x >

Type 5 13. Solve tan x ≥

15. Solve tan x < 1 16. Solve tan x £ 3

1 3

Ê pˆ 29. Solve sin 3x sin 4x > sin x sin 2x " x Œ Á 0, ˜ Ë 2¯ 30. Solve |sin x + cos x| = |sin x| + |cos x 31. Solve |sec x + tan x| = |sec x| + |tan x 32. Solve for x: sin2 x + sin x – 2 < 0 and x2 – 3x + 2 < 0

2/10/2017 4:05:41 PM

3.3

Trigonometric In-Equation

33. Solve for x: 2 sin 2 x + sin x – 1 < 0 and x2 + x – 2 < 0 34. Solve for x: tan 2 x – 5 tan x + 6 > 0 and x 2 – 16 £ 0 x -1 35. Solve for x: < 0 and tan2 x + tan x – 6 < 0 5- x 36. Solve for x: [sin x] = 0, where [,] = GIF

LEVEL II

8. 2( 2 - 1)sin x - 2 cos 2x + 2( 2 - 1) < 0

(Mixed Problems)

1. sin (3x – 1) > 0 2. cos (2x – 3) < 0 1 3. sinx £ 2 4. cos x £ 5. 6. 7. 8. 9. 10. 11. 12.

Passage I If x1, x2, x3 ΠR, then

1

3 sin x + cos x > 1

14. sin x - 3 cos x < 1 15. 16. 17. 18.

sin2 x + sin x – 2 < 0 sin2 x + 3 sin x + 2 < 0 cos2 x – cos x > 0 sin x + cos x – cos 2 x > 0

19. x2 + x – 2 < 0 and sin x > 20. x2 –1 £ 0 and cos x <

1 2

1 2

21. 4x2 – 1 £ 0 and tan x ≥

1

3 22. x – 3x + 2 < 0 and (sin x)2 – sin x > 0 2

LEVEL III

(Problems for JEE Advanced)

Q. Solve for x: sin x + cos x > 3 sin x - cos x 2. |sin x| > |cos x| sin x 3. cot x + ≥0 cos x - 2

1.

4. sin x + cos x > 2 cos 2 x

TR_03.indd 3

9. sin 2x > 2 sin 2 x + (2 - 2) cos 2 x , x Π(0, 2p) 10. 1 + log4 sin x + 2 log16 cos x > 0

Comprehensive Link Passages

2 |sin 2x + cos 2x| = |sin 2x| + |cos 2x| 2 cos2 x + cos x < 1 4 sin2 x – 1 £ 0 4 cos2 x – 3 ≥ 0 |sin x| > |cos x| |cos x| > |sin x| sin x + cos x > 1 sin x + cos x < 1

13.

5. 4 sin x sin 2x sin 3x > sin 4x cos 2 2x ≥ 3 tan x 6. cos 2 x cos x + 2 cos 2 x + cos 3x 7. >1 cos x + 2 cos 2 x - 1

f ( x1 ) + f ( x2 ) + f ( x3 ) Êx +x +x ˆ £ f Á 1 2 3˜ Ë ¯ 3 3 Find 1. The value of sin a + sin b + sin g is, where a + b + g = 180° 3 3 3 (a) £ 1 (b) £ 3 (c) £ (d) £ 2 2 2. The value of cos a + cos b + cos g is, where a + b + g = 180° 3 3 (a) £ 2 (b) £ (c) £ 3 (d) £ 2 2 3. The value of cot a + cot b + cot g is, where a + b + g = 180° 3 (a) ≥ 1 (b) ≥ 3 (c) ≥ 3/2 (d) ≥ 2 4. The value of cot a cot b cot g is, where a + b + g = 180° 1 1 1 3 (c) £ (d) £ (a) £ (b) £ 3 3 2 3 3 2 5. The value of sin2 a + sin2 b + sin2 g is, where a + b + g = 180° (a) £ 9/4 (b) £ 3/4 (c) £ 3/2 (d) £ 1/2 6. The value of sin a sin b . sing is, where a + b + g = 180° 1 1 3 3 3 (d) £ (b) £ (c) £ (a) £ 2 3 3 2 4 8 7. The value of cot2 a + cot2 b + cot2 g is, where a + b + g = 180° (a) ≥ 1 (b) ≥ 3 (c) ≥ 3 (d) ≥ 3/2 Passage II If | f (x) + g (x)| = |f (x)| + |g (x)|, then f (x) . g (x) ≥ 0 On the basis of the above information answer the following questions. 1. If |sec x + tan x| = |sec x| + |tan x| " x Œ [0, 2p], then x does not satisfy the equation is p (d) 2p (a) 0 (b) p (c) 2 2. If |x – 1| + |x – 3| = 2, then x is (a) x > 1 (b) x > 3 (c) x < 1 (d) 1 < x < 3

2/10/2017 4:05:42 PM

3.4

Trigonometry Booster

3. If |sin x + cos x| = |sin x| + |cos x|, " x Π[0, 2p] then the solution set is

(C) If |tan x| £ 1, (R) x Œ [–p, p] then x is (D) If cos x – sin (S) x ≥ 1, x Œ [0, 2p] then x is

È p˘ (a) Í0, ˙ Î 2˚ È p ˘ È 3p ˘ (b) Í0, ˙ » Íp , » {2p } 2 ˙˚ Î 2˚ Î È 3p ˘ (c) Íp , » {2p } 2 ˙˚ Î (d) [0, 2p]

1. Match the following columns: Column I Column II (A) The number of solutions of (P) 6 1 sin x > in (0, 2p) is 2 (B) The number of solutions of |tan x| (Q) 0 £ 1 in (–p, p) is (C) The number of solutions of |cos x| (R) 4 > 1 in (0, 2013p) is (D) The number of solutions of (S) 2 sin x + cos x = |sin x| + |cos x| in (0, 2p) is 2. Match the following columns:

(B) If 4 sin2 x – 8 (Q) sin x + 3 £ 0, x Œ [0, 2p] then x is

È p 5p ˘ Í6 , 6 ˙ Î ˚

Assertion and Reason

Matrix Match (For JEE-Advanced Examination Only)

Column I (A) If sin x . cos3 (P) x > cos x . sin3 x, x Π[0, 2p] Then x is

Ê pˆ ÁË 0, ˜¯ 4

Column II 3p ˘ È Í-p , - 4 ˙ Î ˚ È 3p ˘ Í 4 , p˙ Î ˚

È p p˘ Í- 4 , 4 ˙ » Î ˚

È 3p ˘ Í 2 , 2p ˙ » {0} Î ˚

Codes: (A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true but R is not the correct explanation of A. (C) A is true but R is false. (D) A is false but R is true. 1. Assertion (A): The value of tan 3a . cot a cannot lie between 3 and 1/3. Reason (R): In a triangle ABC, the maximum value of Ê Aˆ Ê Bˆ Ê Cˆ 1 sin Á ˜ sin Á ˜ sin Á ˜ is . Ë 2¯ Ë 2¯ Ë 2¯ 8 2. Assertion (A): The minimum value of a2 tan2 q + b2 cot2 q is 2ab. Reason (R): For positive real numbers AM ≥ GM a2 b2 + 3. Assertion (A): The minimum value of cos 2 x sin 2 x is (a + b)2 Reason (R): The maximum value of 3 sin2 x + 4 cos2 x is 4 È p˘ 4. Assertion (A): For all q Œ Í0, ˙ , Î 2˚ cos (sin q) > sin (cos q) Reason (R): In a triangle ABC, the maximum value of sin A + sin B + sin C 3 is . cot A + cot B + cot C 2 5. Assertion (A): cot–1 x ≥ 2 fi x Œ (– , 2] Reason (R): cot–1 x is a decreasing function.

A NSWERS

LEVEL II

TR_03.indd 4

3p pˆ Ê , np - ˜ , n Œ I 4. x Œ Á np Ë 4 4¯

Ê 2np + 1 (2np + 1)p + 1ˆ , 1. x Œ Á ˜¯ , n Œ I Ë 3 3

pˆ Ê 6. x Œ Á np - p , np - ˜ , n Œ I Ë 3¯

p 3 p 3ˆ Ê 2. x ŒÁ (4n + 1) + , (4n + 3) + ˜ , n Œ I Ë 4 2 4 2¯

p pˆ Ê 7. x Œ Á np - , np + ˜ , n Œ I Ë 6 6¯

p pˆ Ê 3. x Œ Á np - , np + ˜ , n Œ I Ë 6 6¯

p pˆ Ê 8. x Œ Á np - , np + ˜ , n Œ I Ë 6 6¯

2/10/2017 4:05:42 PM

3.5

Trigonometric In-Equation

9. x Œ

ÏÊ p pˆ Ê p pˆ¸ ÌÁ 2np - , np - ˜ » Á np + , np + ˜ ˝ Ë ¯ Ë 2 4 4 2¯˛ n ŒI Ó

p pˆ Ê 10. x Œ Á np - , np + ˜ , n Œ I Ë 4 4¯ 11. x Œ (2np , (2n + 1)p ), n Œ I pˆ Ê 12. x Œ Á (2n - 1)p , 2np + ˜ , n Œ I Ë 2¯ 2p ˆ Ê 13. x Œ Á 2 np , 2np + ˜ , n ŒI Ë 3 ¯ 5p pˆ Ê , 2 np + ˜ , n Œ I 14. x Œ Á 2 np Ë 6 2¯ 3p pˆ 15. x Œ ÊÁ 2 np , 2np + ˜ , n Œ I Ë 2 2¯ 16. x = j p 5p ˆ Ê 17. x Œ Á 2 np + , 2 np + ˜ , n Œ I Ë 12 12 ¯ 3 p 3 p 7 p Ê ˆ Ê ˆ 18. x Œ Á 0, , ˜ »Á ˜ Ë 4¯ Ë 4 4¯ p 19. < x 0

8. Here we should draw the graphs of y = cos x and 1 . y= 2

Y X¢

+

+

O

+

p

y=1 x

Y

y = –1

Y¢

fi

x=

X¢

(2 np , (2n + 1) p ) n ŒI

5. Here we should draw the graph of y = sin x and y = Y

X¢

– 7p 6

p – 4p O 3 3

X

5p 2

y= 3 2 X

O p 6

–p 4

1 . 2

y = 1/2 x

O

y= 1 2 X

p 4

Y¢

Hence, the solution set is p pˆ Ê x= ÁË 2np - , 2np + ˜¯ 4 4 n ŒI 9. Here we should draw the graphs of y = cos x and y = Y

Y¢

Hence, the solution set is x= n ŒI

X¢

7p pˆ Ê , 2np + ˜ ÁË 2 np 6 6¯

6. Here we should draw the graphs of y = sin x and 1 y= . 5

O

sin–1 1 5 Y¢

Hence, the solution set is Ê -1 Ê 1 ˆ ˆ Á (2n - 1)p - sin ÁË 5 ˜¯ ,˜ Á ˜ x= ˜ -1 Ê 1 ˆ n ŒI Á + n p 2 sin ÁË ˜¯ ÁË ˜¯ 5

p 3

Hence, the solution set is x=

y = 1/5 X

O

y= 1 2 X

Y¢

Y

X¢ –p – sin–1 1 5

–p 3

1 2

p p˘ È Í2np - 3 , 2np + 3 ˙ ˚ n ŒI Î

10. Here we should draw the graphs of y = cos x and 1 y= . 3 Y

X¢

O

1 cos–1 3

1 2p – cos–1 3

y = 1/3 X

Y¢

TR_03.indd 6

2/10/2017 4:05:46 PM

3.7

Trigonometric In-Equation

14. Here we should draw the graphs of y = tan x and y = 2.

Hence, the solution set is

Y

Ê ˆ -1 Ê 1 ˆ Á 2np + cos ÁË 3 ˜¯ , ˜ Á ˜ x= -1 Ê 1 ˆ ˜ n ŒI Á ÁË 2(n + 1)p - cos ÁË 3 ˜¯ ˜¯ 11. Here we should draw the graphs of y = cos x and y= 3. 2

p 2 – p O Y = tan–1(2) 2

X¢

Y

X¢

X

Y¢

p 6

O

y=2

2p – p 6

y= 3 2 X

Hence, the solution set is x= n ŒI

pˆ Ê -1 ÁË np + tan (2), np + ˜¯ 2

15. Here we should draw the graphs of y = tan x and y = 1. Y

Y¢

Hence, the solution set is x=

p p˘ È Í2np + 6 , 2(n + 1)p - 6 ˙ ˚ n ŒI Î

y=1

12. Here we should draw the graphs of y = tan x and y = 1.

X¢

p p –p O 4 2 2

X

Y Y¢

y=1 X¢

–p O p p 4 2 2

X

Hence, the solution set is p pˆ Ê x= ÁË np - , np + ˜¯ 2 4 n ŒI 16. Here we should draw the graphs of y = tan x and y= 3.

Y¢

Y

Hence, the solution set is x= n ŒI

p pˆ Ê ÁË np + , np + ˜¯ 4 2

y= 3

13. Here we should draw the graphs of y = tan x and 1 . y= 3

X¢

p p –p O 3 2 2

X

Y Y¢

X¢

–p O p p 6 2 2

Y¢

Hence, the solution set is x=

TR_03.indd 7

p pˆ È Ínp + 6 , np + 2 ˜¯ n ŒI Î

y=1 3 X

Hence, the solution set is p p˘ È x= Ínp - 2 , np + 3 ˙ ˚ n ŒI Î 17. We have sin x > cos x fi sin x – cos x > 0 1 1 fi sin x cos x > 0 2 2 fi

pˆ Ê sin Á x - ˜ > 0 Ë 4¯

2/10/2017 4:05:48 PM

3.8

Trigonometry Booster

fi

fi

p pˆ Ê x Œ Á 2np + , (2n + 1)p + ˜ Ë 4 4¯

Hence, the solution set is x=

p pˆ Ê ÁË 2np + , (2n + 1)p + ˜¯ 4 4 n ŒI

18. We have cos x > sin x fi cos x – sin x > 0 1 1 fi cos x sin x > 0 2 2 fi

pˆ Ê cos Á x + ˜ > 0 Ë 4¯

fi

3p pˆ Ê x Œ Á 2np , 2np + ˜ Ë 4 4¯

fi

1 2

and cos x ≥ -

1 2

p 7p ˆ Ê x Œ Á 2np + , 2np + ˜ Ë 4 4¯

p p˘ È x Œ Í2np - , 2np + ˙ 3 3˚ Î Hence, the solution set is

and

x=

p 7p ˆ Ê ÁË 2np + , 2np + ˜¯ 4 4 nŒI

p p˘ È ÍÎ2np + 3 , 2np + 3 ˙˚ 20. We have |sin x + cos x| = |sin x| + |cos x| As we know that, if |f (x) + g (x)| = |f (x)| + |g (x)| then f (x)g (x) ≥ 0 Thus, sin x cos x ≥ 0 fi sin 2 x ≥ 0 fi

p˘ È x Œ Ínp , np + ˙ 2˚ Î

Hence, the solution set is x=

p˘ È Ínp , np + 2 ˙ ˚ n ŒI Î

21. We have sin x sin 2x < sin 3x sin 4x fi 2 sin x sin 2x < 2 sin 3x sin 4x fi cos x – cos 3x < cos x – cos 7x fi cos 3x > cos 7x fi cos 3x – cos 7x > 0 fi 2 sin 5x sin 2x > 0 fi sin 5x > 0 (since sin 2x is +ve for 0 < x < p/2)

TR_03.indd 8

Ê pˆ Ê pˆ x = Á 0, ˜ » Á 0, ˜ Ë 5¯ Ë 2¯ 22. We have cos x – sin x – cos 2x > 0 fi (cos x – sin x) – (cos2 x – sin2 x) > 0 fi (cos x – sin x)(1 – cos x – sin x) > 0. fi (sin x – cos x)(sin x + cos x – 1) > 0 pˆ Ê sin Á x - ˜ ◊ (sin x + cos x - 1) > 0 fi Ë 4¯

Hence, the solution set is 3p pˆ Ê , 2np + ˜ x= ÁË 2np 4 4¯ n ŒI 19. We have cosx <

0 8 cos 2x fi 5(1 – cos 2x) + 2(1 – cos2 2x) > 8 cos 2x fi 5 – 5 cos 2x + 2 – 2 cos2 2x – 8 cos 2x > 0 fi 2 cos2 2x + 13 cos 2x – 7 < 0 fi 2 cos2 2x + 14 cos 2x – cos 2x – 7 < 0 fi 2 cos 2x (cos 2x + 7) – (cos 2x +7) < 0 fi (cos 2x + 7)(2 cos 2x – 1) < 0 fi 2 cos 2x – 1 < 0 fi cos 2x < 1/2 p 5p ˆ Ê x Œ Á np + , np + ˜ fi Ë 6 6¯ Hence, the solution set is 5p ˆ p Ê x= ÁË np + , np + ˜¯ 6 6 n ŒI

23. We have

24. We have 6 sin2 x – sin x cos x – cos2 x > 2 fi 6 sin2 x – sin x cos x – cos2 x > 2 (sin2 x + cos2 x) fi 4 sin2 x – sin x cos x – 3 cos2 x > 0 fi 4 tan2 x – tan x – 3 > 0 fi 4 tan2 x – 4 tan x + 3 tan x – 3 > 0 fi 4 tan x (tan x – 1) + 3(tan x – 1) > 0 fi (tan x – 1)(4 tan x + 3) > 0 fi tan x < –3/4 and tan x > 1 Ê p Ê 3ˆ ˆ x Œ Á np - , np - tan -1 Á ˜ ˜ fi Ë 4¯ ¯ Ë 2 p pˆ Ê and x Œ Á np + , np + ˜ Ë 4 2¯ Hence, the solution set is Ê p Ê 3ˆ ˆ x = Á np - , np - tan -1 Á ˜ ˜ Ë 4¯ ¯ Ë 2 p pˆ Ê » Á np + , np + ˜ , n Œ I Ë 4 2¯

2/10/2017 4:05:49 PM

3.9

Trigonometric In-Equation

25. We have sin 6 x + cos6 x > fi

13 16

(1 - 3 sin 2 x cos 2 x) >

fi

Hence, the solution set is

13 16

x=

Ê Ê np p np p ˆ ˆ ÁË ÁË 2 - 12 , 2 + 12 ˜¯ ˜¯

fi

13 Ê 3 2 2 ˆ ÁË 1 - (2 sin x) (2 cos x)˜¯ > 4 16

fi

Ê 3 ˆ 13 ÁË1 - (1 - cos 2x) (1 + cos 2x)˜¯ > 4 16

fi

Ê 3 ˆ 13 2 ÁË1 - (1 - cos 2x)˜¯ > 4 16

fi

Ê 3 2 ˆ 13 ÁË1 - sin 2x˜¯ > 4 16

fi

1 - 3sin 2 x cos 2 x >

fi

Ê 3 ˆ 13 2 ÁË1 - (2 sin 2x)˜¯ > 8 16

fi

fi

Ê 3 ˆ 13 ÁË1 - (1 - cos 4x)˜¯ > 8 16

3 13 1 - (sin 2 2x) > 4 16

fi

fi

Ê5 3 ˆ 13 ÁË + cos 4x˜¯ > 8 8 16

3 13 1 - (2 sin 2 4x) > 8 16

fi

3 13 1 - (1 - cos 4x) > 8 16

fi

5 3 13 + cos 4x > 8 8 16

fi

3 3 cos 4x > 8 16

fi

cos 4x >

fi

2 np -

fi fi fi

nŒI

27. The given in-equation is

3 3 cos 4x > 8 16 cos 4x > 1/2 p pˆ Ê 4x Œ Á 2np - , 2np + ˜ Ë 3 3¯

cos3 x ◊ cos 3x - sin 3 x sin 3x >

fi

5 8

5 2

sin2 3x + cos2 3x + 3(cos 3x cos x - sin 3x sin x) > 3 cos 4x + 1 >

fi

cos 4x >

fi

2np -

13 16

13 16

1 2

p p < 4x < 2np + , n ΠI 3 3

np p np p -

13 16

-3sin 2 x cos 2 x(sin 2 x + cos 2 x) >

Ê np p np p ˆ x ŒÁ - , + Ë 2 12 2 12 ˜¯ Hence, the solution set is Ê np p np p ˆ x= - , + ˜ ÁË 2 12 2 12 ¯ nŒI

fi

sin 6 x + cos6 x > fi

fi

TR_03.indd 9

np p np p - cos 2x 4 4 2 fi 5 sin x + sin2 2x > 4 cos 2x fi 5 (2 sin2 x) + 2 (sin2 2x) > 4 cos 2x fi 5 (1 – cos 2x) + 2(1 – cos2 2x) > 8 cos 2x fi 2 cos2 2x + 13 cos 2x – 7 < 0 fi 2 cos2 2x + 14 cos 2x – cos 2x – 7 < 0 fi 2 cos 2x (cos 2x + 7) – 1(cos 2x + 7) < 0 fi (2 cos 2x – 1) (cos 2x + 7) < 0 fi

-7 < cos 2x <

1 2

2/10/2017 4:05:50 PM

3.10

Trigonometry Booster

32. We have sin 2 x + sin x – 2 < 0 fi (sin x + 2) (sin x – 1) < 0 fi –2 < sin x < 1 fi sin x < 1 3p p fi cos x – cos 3x fi –cos 7x > –cos 3x fi cos 7x < cos 3x fi cos 7x – cos 3x < 0 fi –2 sin 5x sin 2x < 0 fi sin 5x sin 2 x > 0 fi fi fi

33. We have 2 sin 2 x + sin x – 1 < 0 fi (2 sin x – 1) (sin x + 1) < 0 fi – 1 < sin x < 2 p p fi - 0, since sin 2x is positive in Á 0, ˜ Ë 2¯ 0 < 5x < p p 0 tan–1(3) Also, x 2 – 16 £ 0 fi (x + 4)(x – 4) £ 0 fi –4 £ x £ 4 Hence, the solution set is x Œ (–4, tan–1(2)) » (tan–1(3), 4) 35. We have tan2 x + tan x – 6 < 0 fi (tan x + 3) (tan x – 2) < 0 fi – 3 < tan x < 2 fi tan–1(–3) < x < tan–1(2) x -1 0 x-5 fi x < 1 and x > 5 Hence, the solution set is x Œ (1, tan–1(2)) 36. We have [sin x] = 0 fi 0 £ sin x < 1 Case-I: When sin x ≥ 0 fi 2n p £ x £ (2n + 1) p, n Œ I Case-II: When sin x < 1 3p p fi 2np < x < 2np + 2 2 fi

fi

Ê p Ê p ˆˆ x Œ Á (4n - 3) , Á (4n + 1) ˜ ˜ Ë 2 Ë 2 ¯¯

Hence, the solution set is

2/10/2017 4:05:51 PM

3.11

Trigonometric In-Equation

Ê [2np, (2n + 1) p ] ˆ Á ˜ x= p pˆ˜ Á Ê nŒI Á » Á (4n - 3) , (4n + 1) ˜ ˜ Ë Ë 2 2¯¯

8. 4 cos2 x – 3 ≥ 0 3 cos 2 x ≥ fi 4 fi

LEVEL II

2. cos (2x – 3) < 0 3p p 2np + < (2x - 3) < 2np + fi 2 2 fi

2n p +

3p 3 p 3 + < x < np + + 4 2 4 2

1 3. sin x £ 2 fi

4. cos x £ fi

2 3p p £ x £ np + , n Œ I 4 4

5. |sin 2x + cos 2x| = |sin 2x| + |cos 2x| fi sin (2x) cos (2x) ≥ 0 fi 2 sin (2x) cos (2x) ≥ 0 fi sin (4x) ≥ 0 fi 2n p £ 4x £ (2n + 1)p, 2np (2n + 1) p £ x£ , n ŒI 4 4

fi 2

6. 2 cos fi fi fi fi

x + cos x < 1 2 cos2x + 2 cos x – cos x – 1 < 0 2 cos x (cos x + 1) – (cos x + 1) < 0 (2 cos x – 1) (cos x + 1) < 0 1 -1 < cos x < 2

p (2n + 1) p < x < 2np + , n Œ I 6 2 7. 4 sin x – 1 £ 0 1 sin 2 x £ fi 4 1 sin x £ fi 2 p p fi np - £ x £ np + , n Œ I 6 6 fi

TR_03.indd 11

11. sin x + cos x > 1 pˆ 1 Ê sin Á x + ˜ > fi Ë 4¯ 2

1

np -

np -

10. |cos x| > |sin x| fi |tan x| < 1 p pˆ Ê fi x Œ Á np - , np - ˜ Ë 4 4¯

p p £ x £ np + , n Œ I 6 6

np -

3 2

p p £ x £ np + , n Œ I 6 6 9. |sin x| > |cos x| fi |tan x| > 1 ÏÊ p pˆ Ê p p ˆ¸ fi x Œ ÌÁ np - , np - ˜ » Á np + , np + ˜ ˝ Ë ¯ Ë 2 4 4 2¯˛ nŒI Ó fi

1. sin (3x – 1) > 0 fi 2n p < (3x –1) < (2n + 1) p 2n p + 1 (2n + 1)p + 1 1 fi

pˆ 1 Ê sin Á x + ˜ > Ë 6¯ 2

fi

2np +

p Ê pˆ 5p < Á x + ˜ < 2n p + 6 Ë 6¯ 6

fi 2n p < x < (2n + 1)p, n ΠI 14. sin x - 3 cos x < 1 fi

pˆ 1 Ê sin Á x - ˜ < Ë 3¯ 2

fi

2n p +

p Ê pˆ 5p < Á x - ˜ < 2n p + 6 Ë 3¯ 6

p < x < (2n + 1)p , n Œ I 2 15. sin2 x + sin x – 2 < 0 fi (sin x + 2) (sin x – 1) < 0 fi –2 < sin x < 1 fi

2n p +

2/10/2017 4:05:51 PM

3.12

Trigonometry Booster

fi

–1 < sin x < 1 p p fi 2np - < x < 2n p + , n Œ I 2 2 16. sin2 x + 3 sin x + 2 < 0 fi (sin x + 1) (sin x + 2) < 0 fi –2 < sin x < 1 fi x=j 17. cos2 x – cos x > 0 fi cos x (cos x – 1) > 0 fi cos x < 0, cos x > 1 fi cos x < 0 3p p fi 2n p + < x < 2n p + , n ŒI 2 2 18. sin x + cos x – cos 2 x > 0 fi (sin x – cos x) – (cos2 x – sin2 x) > 0 fi (sin x + cos x) (1 – cos x + sin x) > 0 fi (sin x + cos x) (sin x – cos x + 1) > 0 Case I: (sin x + cos x) > 0, (sin x – cos x + 1) > 0 pˆ pˆ 1 Ê Ê fi sin Á x + ˜ > 0, sin Á x - ˜ > Ë ¯ Ë ¯ 4 4 2 fi

pˆ p Ê pˆ p Ê 0 < Á x + ˜ < p, - < Á x - ˜ < Ë 4¯ 4 Ë 4¯ 2

fi

-

fi

0 0

fi

Êp ˆ tan Á + x ˜ < - 3 Ë4 ¯

fi

(sin x + cos x) 2(cos x - sin x) - 1) < 0

fi

p Êp p ˆ np - < Á + x ˜ < np Ë ¯ 2 4 3

fi

1 ˆ Ê (sin x + cos x) Á (cos x - sin x) ˜ |cos x| fi |tan x| > 1 p pˆ Ê p pˆ Ê x Œ Á np - , np - ˜ » Á np + , np + ˜ fi Ë 2 4¯ Ë 4 2¯ nŒI 3. cot x +

sin x ≥0 cos x - 2

Case I: 1 ˆ Ê (sin x + cos x) > 0, Á (cos x - sin x) ˜ 0, Á cos Á x + ˜ ˜ 0 fi sin 2x (–cos 4x) > 0 fi sin 2x (cos 4x) < 0 Case I: when sin 2x > 0, (cos 4x) < 0 3p p fi 0 < 2x < p , < 4x < 2 2

2/10/2017 4:05:54 PM

3.14

Trigonometry Booster

fi

01

>1

2 cos x > 1 1 cos x > 2 p p < x < 2 np + , n ΠI 3 3

2 np -

8. 2 ( 2 - 1) sin x - 2 cos 2x + 2 ( 2 - 1) < 0

fi

fi

cos x + 2 cos 2 x + 4 cos3 x - 3 cos x

fi

2( 2 - 1) sin x - 2(1 - 2sin 2 x) + 2 ( 2 - 1) < 0

fi

( 2 - 1) sin x - (1 - 2sin 2 x) +

fi

2 sin 2 x + ( 2 - 1) sin x +

fi

2 2 sin 2 x + 2 ( 2 - 1) sin x - 1 < 0

fi

2 2 sin 2 x + (2 - 2) sin x - 1 < 0

fi

2 2sin 2 x + 2sin x - 2 sin x - 1 < 0

fi

2sin x( 2 sin x + 1) - ( 2 sin x + 1) < 0

fi

(2 sin x - 1) ( 2 sin x + 1) < 0

fi

-

fi

2np -

1 2

< sin x <

( 2 - 1) 2

2 sin 2 x + (2 - 2) cos 2 x , x Π(0, 2p) fi

2 sin 2 x - 2 sin x cos x + (2 - 2) cos 2 x < 0

fi

2 tan 2 x - 2 tan x + (2 - 2) < 0

fi

tan 2 x - 2 tan x + ( 2 - 1) < 0

fi

tan x =

2 ± 2 - 4 ( 2 - 1) 0

TR_03.indd 15

, tan b =

3- 2

fi

Ê sin 2 x ˆ log 2 Á +2>0 Ë 2 ˜¯

fi fi

log2 sin 2x + 1 > 0 log2 sin 2x > –1

fi

sin 2 x >

fi

2n p +

2

fi

1 2

5p p ,n ŒI < 2x < 2n p + 6 6 5p p , n ŒI np + < x < np + 12 12

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TR_03.indd 16

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C H A P T E R

4

Inverse Trigonometric Functions

I NVERSE F UNCTION CONCEPT BOOSTER 4.1

y = f(x)

INTRODUCTION TO INVERSE FUNCTION A

f

1 2 3 4 5

y=

B

X¢

6 7 8 9 10 g

Let f : X Æ Y be a bijective function. If we can make another function g from Y to X, then we shall say that g is the inverse of f. 1 -1 i.e., g = f π f Thus, f–1( f (x)) = x Note (i) The inverse of a function exists only when the function f is bijective. (ii) If the inverse of a function is exists, then it is called an invertible function. (iii) The inverse of a bijective function is unique. (iv) Geometrically f–1(x) is the image of f (x) with respect to the line y = x. (v) Another way also we can say that f–1(x) is the symmetrical with respect to the line y = x. (vi) A function f (x) is said to be involution if for all x for which f (x) and f ( f (x)) are defined such that f ( f (x)) = x.

TR_04.indd 1

Y

x

y = f –1(x)

O

X

Y¢

(vii) If f is an invertible function, then (f–1)–1 = f. (viii) If f : A Æ B be a one one function, then f–1of = IA and fof–1 = IB, where IA and IB are the identity functions of the sets A and B respectively. (ix) Let f : A Æ B, g : B Æ C be two invertible functions, then gof is also invertible with (gof)–1 = (f–1og–1).

Rule to Find out the Inverse of a Function (i) First, we check the given function is bijective or not. (ii) If the function is bijective, then inverse exists, otherwise not. (iii) Find x in terms of y (iv) And then replace y by x, then we get inverse of f. i.e., f–1(x).

4.2

INVERSE TRIGONOMETRIC FUNCTIONS

We know that sine function is defined only for every real number and the range of sine function is [–1, 1]. Thus, the graph of f (x) = sin (x) is as follows

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4.2

Trigonometry Booster

Graph of f (x) = sin (x):

So the graph of f (x) = sin–1x is Y

X¢

–2p

Y

–p

p

2p

y=1 X y = –1

y = p/2 X¢

Y¢

From the graph, we can say that, it will be one one and onto only when we considered it in some particular intervals like Ê p p ˆ Ê p 3p ˆ Ê 3p 5p ˆ , ÁË - , ˜¯ , ÁË , ˜,Á ˜ , and so on. If we consider 2 2 2 2¯ Ë 2 2¯ the whole function, then it is not one one as well as onto. Also, when we think the inverse function, then domain and range are interchanged. So the graph of this function is as follows. Y

p

x=1

p p Thus, D f = [-1, 1] and R f = ÈÍ- , ˘˙ Î 2 2˚ Now, we shall discuss the graphs of other inverse trigonometric functions and their characteristics.

GRAPHS OF INVERSE TRIGONOMETRIC FUNCTIONS

Graph of f (x) = sin–1x. Y

X

p 3p 2 2p 5p 2 –3p

y = p/2 X¢

X

O

y = –p/2

Y¢

x=1

As a whole, inverse of this function does not exist. Its inverse exists only when, we restrict its range. p ˆ Ê p p ˆ Ê p 3p ˆ Ê 3p , - ˜ ,Á - , ˜ , Á , So the intervals are Á ˜, Ë 2 2¯ Ë 2 2¯ Ë 2 2 ¯ Ê 3p 5p ˆ , ÁË ˜ , and so on. 2 2¯ In the conventional mathematics, we consider it in Ê p pˆ . ÁË - , ˜¯ 2 2 Thus, sin inverse function is defined as È p p˘ sin -1:[-1, 1] Æ Í- , ˙ Î 2 2˚ È p p˘ Therefore, a function f :[-1, 1] Æ Í- , ˙ is defined as Î 2 2˚ f (x) = sin–1x.

TR_04.indd 2

Y¢

È p p˘ A function f :[-1, 1] Æ Í- , ˙ is defined as f (x) = Î 2 2˚ –1 sin x = arc sin x

–p 2

x = –1

x = –1

(i) sin–1x:

p 2

X¢

y = –p/2

4.3

3p 5p 2 2p 3p 2

X

O

x = –1

Y¢

x=1

Characteristics of ARC Sine Function 1. Df = [–1, 1] È p p˘ 2. R f = Í- , ˙ Î 2 2˚ 3. It is not a periodic function. 4. It is an odd function. since, sin–1(–x) = –sin–1x 5. It is a strictly increasing function. 6. It is a one one function. p 7. For 0 < x < , sin x < x < sin–1x. 2 (ii) cos–1x: A function f : [–1, 1] Æ [0, p] is defined as f (x) = cos–1x = arc cos x

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4.3

Inverse Trigonometric Functions

Graph of f (x) = cos–1x

Graph of f (x) = cot–1x Y

Y

O

y=p

y=p

y = p/2

y = p/2

X

X¢ x = –1

Y¢

5. 6. 7 (iii)

X

O

x=1

Y¢

Characteristics of ARC Cosine Function 1. 2. 3. 4.

X¢

Df = [–1, 1] [0, p] It is not a periodic function. It is neither even nor odd function since, cos–1(–x) = p – cos–1(x) It is a strictly decreasing function. It is a one one function. p For 0 < x < , 2 cos–1x < x < cos x tan–1x: p p A function f : R Æ ÊÁ - , ˆ˜ is Ë 2 2¯ defined as f (x) = tan–1x.

Graph of f (x) = tan–1x:

Characteristics of ARC Co-tangent Function 1. 2. 3. 4. 5. 6. 7. (v)

Df = R Rf = (0, p) It is not a periodic function. It is neither even nor odd function since, cot–1(–x) = p – cot–1x It is a strictly decreasing function. It is a one one function. p For 0 < x < , 2 cot x < x < cot–1x cosec–1x: A function È p p˘ f :( , 1] [1, ) Í , ˙ {0} Î 2 2˚ is defined as f (x) = cosec–1x. Graph of f (x) = cosec–1x.

Y

Y y = p/2 X¢

O

y = p/2

X y = –p/2

X¢

X

O

Y¢

Characteristics of ARC Tangent Function 1. Df = R Ê p pˆ 2. R f = Á - , ˜ Ë 2 2¯ 3. It is not a periodic function 4. It is an odd function. Since, tan–1(–x) = –tan–1x 5. It is a strictly increasing function. 6. It is a one one function. p 7. For 0 < x < , tan–1x < x < tan x. 2 (iv) cot–1x: A function f: R Æ (0, p) is defined as f (x) = cot–1x.

TR_04.indd 3

y = –p/2 x = –1 Y¢

x=1

Characteristics of ARC Co-secant Function 1. Df = (– , –1]

[1, )

È p p˘ 2. R f = Í- , ˙ - {0} Î 2 2˚ 3. It is an odd function, since cosec–1(–x) = –cosec–1(x) 4. It is a non periodic function. 5. It is a one one function. 6. It is a strictly decreasing function with respect to its domain.

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4.4

Trigonometry Booster

p , cosec–1 x < x < cosec x 2 (v) sec–1x: A function f :( , 1] [1, ) 7. For 0 < x <

Ïp ¸ [0, p ] Ì ˝ Ó2˛

is defined as f (x) = sec–1 x Graph of f (x) = sec–1 x Y

y=p p y= 2 X¢

4. cot–1x

Ê p˘ ÁË 0, ˙ 2˚

5. cosec–1x

Ê p˘ ÁË 0, ˙ 2˚

6. sec–1x

È pˆ Í0, 2 ˜¯ Î

Case II: When x < 0 Functions

Principal values

1. sin–1x

È p ˘ Í- 2 , 0˙ Î ˚

2. cosec–1x

È p ˆ Í- 2 , 0˜¯ Î

3. tan–1x

Ê p ˆ ÁË - , 0˜¯ 2

4. cos–1x

Èp ˘ Í2 , p˙ Î ˚

5. sec–1x

Êp ˘ ÁË , p ˙ 2 ˚

6. cot–1x

Èp ˆ Í 2 , p ˜¯ Î

X

O Y¢

Characteristics of ARC Secant Function 1. Df = (– , –1]

[1, ]

Ïp ¸ 2. R f = [0, p ] - Ì ˝ Ó2˛ 3. It is neither an even function nor an odd odd function, since sec–1(–x) = p – sec–1(x) 4. It is a non periodic function. 5. It is a one one function. 6. It is strictly decreasing function with respect to its domain. p 7. For 0 < x < , sec–1 x < x < sec x 2

4.4

p , " x Œ[-1, 1] 2 p (ii) tan -1 ( x) + cot -1 ( x) = , " x Œ R 2 p (iii) cosec -1 ( x) + sec -1 ( x) = , " x Œ R - (-1, 1) 2

CONVERSION OF INVERSE TRIGONOMETRIC FUNCTIONS

Case I: When x > 0 Functions 1. sin–1x 2. cos–1x 3. tan–1x

TR_04.indd 4

Step I: -1 -1 Ê 1 ˆ (i) sin ( x) = cosec Á ˜ , x Œ [–1, 1] – {0} Ë x¯

CONSTANT PROPERTY

(i) sin -1 ( x) + cos -1 ( x) =

4.5

Here, we shall discuss, how any inverse trigonometric function can be expressed in terms of any other inverse trigonometric functions.

Principal values È p˘ Í0, 2 ˙ Î ˚ p È ˘ Í0, 2 ˙ Î ˚ Ê pˆ ÁË 0, ˜¯ 2

-1 -1 Ê 1 ˆ (ii) cosec ( x) = sin Á ˜ , |x| ≥ 1 Ë x¯

Ê 1ˆ (iii) cos -1 ( x) = sec -1 Á ˜ , x Œ [–1, 1] – {0} Ë x¯ Ê 1ˆ (iv) sec -1 x = cos -1 Á ˜ , |x| ≥ 1 Ë x¯ Ï Ê 1ˆ cot -1 Á ˜ : x > 0 Ô Ë x¯ Ô -1 (v) tan ( x) = Ì Ô-p + cot -1 Ê 1 ˆ : x < 0 ÁË ˜¯ ÔÓ x Ï -1 Ê 1 ˆ :x>0 Ô tan ÁË x ˜¯ Ô -1 (vi) cot ( x) = Ì Ôp + tan -1 Ê 1 ˆ : x < 0 ÁË ˜¯ ÔÓ x

2/10/2017 4:12:19 PM

4.5

Inverse Trigonometric Functions

Step II:

(

Y

)

Ïcos -1 1 - x 2 : 0 £ x £1 Ô (i) sin x = Ì Ô -cos -1 ( 1 - x 2 ) : - 1 £ x < 0 Ó -1

(ii)

(iii)

(iv)

(v)

Ï -1 Ê 1 ˆ Ôsec Á ˜ : 0 £ x £1 Ë 1 - x2 ¯ Ô -1 sin x = Ì Ê 1 ˆ Ô -1 ˜ : -1 £ x < 0 Ô -sec Á Ë 1 - x2 ¯ Ó Ï sin -1 ( 1 - x 2 ) : 0 £ x £1 Ô cos -1 x = Ì ÔÓ p - sin -1 ( 1 - x 2 ) : - 1 £ x < 0 Ï Ê 1 ˆ : 0 < x £1 Ôcosec -1 Á ˜ Ë 1 - x2 ¯ Ô -1 cos x = Ì Ê 1 ˆ Ô -1 ˜ : -1 £ x < 0 Ôp - cosec Á Ë 1 - x2 ¯ Ó Ê x ˆ sin -1 x = tan -1 Á ˜ :-1 < x < 1 Ë 1 - x2 ¯

Ï Ê 1 - x2 ˆ Ô cot -1 Á ˜ :0 < x £ 1 ÁË x ˜¯ ÔÔ -1 (vi) sin x = Ì Ê 1 - x2 ˆ Ô -1 p cot + Á ˜ : -1 £ x < 0 Ô ÁË x ˜¯ ÔÓ

4.6

COMPOSITION OF INVERSE TRIGONOMETRIC FUNCTIONS AND TRIGONOMETRIC FUNCTIONS

È p p˘ (i) A function f : R Æ Í- , ˙ is Î 2 2˚ defined as f (x) = sin–1(sin x) Graph of f (x) = sin–1(sin x)

TR_04.indd 5

p

O

p 2

y= p

2p

X y =–

p 2

Y¢

1. Df = R È p p˘ 2. R f = Í- , ˙ Î 2 2˚ 3. It is an odd function. 4. It is a periodic function with period 2p p p Ï :- £ x £ Ô x 2 2 Ô 3 p Ô p-x : £ x£ p Ô 2 2 5. sin -1 (sin x) = Ì 3 5p p Ô x - 2p : £x£ Ô 2 2 Ô 3p p Ô-p - x :£x£2 2 Ó –1 (ii) cos (cos x) : A function f : R Æ [0, p] is defined as f (x) = cos–1(cos x) Graph of f (x) = cos–1(cos x): Y y=p O

COMPOSITION OF TRIGONOMETRIC FUNCTIONS AND ITS INVERSE

Let y = sin–1x fi x = sin y fi x = sin (sin–1x) fi sin (sin–1x) = x Therefore, sin (sin–1x) provide us a real value lies in [–1, 1] Hence, (i) sin (sin–1x) = x, |x| £ 1 (ii) cos (cos–1x) = x, |x| £ 1 (iii) tan (tan–1x) = x, x Œ R (iv) cot (cot–1x) = x, x Œ R (v) cosec (cosec–1x) = x, |x| ≥ 1 (vi) sec (sec–1x) = x, |x| ≥ 1

4.7

X¢

p y= 2

y=

p 2

X

X¢ Y¢

Df = R Rf = [0, p] It is neither an odd nor an even function. It is a periodic function with period 2p :0 £ x £ p Ï x Ô 2 p - x : p £ x £ 2p Ô 5. cos -1 (cos x) = Ì Ô x - 2p :2p £ x £ 3p ÔÓ - x :- p £ x £ 0 1. 2. 3. 4.

(iii) tan–1(tan x):

p Ê p pˆ Æ Á- , ˜ Ë 2 2¯ 2 is defined as f (x) = tan–1(tan x) Graph of f (x) = tan–1(tan x):

A function f : R - (2n + 1)

Y y= X¢

O

p 2

X p y=– 2

Y¢

2/10/2017 4:12:20 PM

4.6

Trigonometry Booster

1. D f = R - (2n + 1) Ê p pˆ 2. R f = Á - , ˜ Ë 2 2¯

1. Df = R – np, n Œ I È p p˘ 2. R f = Í- , ˙ - {0} Î 2 2˚ 3. It is an odd function. 4. It is a periodic function with period 2p 5. cosec–1(cosec x)

p , n ŒI 2

3. It is an odd function. 4. It is a periodic function with period p

p -p Ï : £x£ Ô x 2 2 Ô 3 p p Ô p-x: £ x£ Ô 2 2 =Ì 3 5p p Ô x - 2p : £x£ Ô 2 2 Ô -3p -p Ô- x - p : £x£ 2 2 Ó

-p p Ï : 1 =Ì a : xy < 0, x 2 + y 2 > 1 Ô Ô 2 2 Ó-p - a : x < 0, y > 0, x + y > 1 where a = sin -1 ( x 1 - y 2 + y 1 - x 2 )

2/10/2017 4:12:21 PM

4.7

Inverse Trigonometric Functions

(ii) sin–1x – sin–1y Ï x2 + y 2 £ 1 a : Ô 2 2 Ô p - a : x > 0, y < 0, x + y > 1 =Ì a : x y > 0, x 2 + y 2 > 1 Ô Ô-p - a : x < 0, y > 0, x 2 + y 2 >1 Ó Where a = sin -1 ( x 1 - y 2 - y 1 - x 2 ) (iii) cos–1x + cos–1y :x + y ≥ 0 Ïa =Ì 2 p a :x + y < 0 Ó where a = cos -1 ( xy - 1 - x 2 1 - y 2 ) (iv) cos–1x – cos–1y Ï a =Ì Ó -a

: :

x£ y x> y

where a = cos -1 ( xy + 1 - x 2 1 - y 2 ) (v) tan–1x + tan–1y a: xy < 1 Ï Ôp + a : x > 0, y > 0, xy > 1 Ô Ô -p + a : x < 0, y < 0, xy > 1 Ô = Ìp : x > 0, y > 0, xy = 1 Ô Ô2 Ôp : x < 0, y < 0, xy = 1 ÔÓ 2 Ê x+ yˆ where a = tan -1 Á Ë 1 - xy ˜¯ –1 –1 (vi) tan – tan y : x y > -1 Ï a Ôp + a : xy < -1, x > 0, y < 0 Ô Ô -p + a : xy < -1, x < 0, y > 0 Ô =Ìp : xy = -1, x > 0, y < 0 Ô Ô 2 Ô p : xy = -1, x < 0, y > 0 ÔÓ - 2 Ê x- yˆ where a = tan -1 Á Ë 1 + xy ˜¯

4.9

MULTIPLE ANGLES

-1 2 (i) sin (2x 1 - x )

1 1 Ï -1 £x£ : Ô 2 sin x 2 2 Ô 1 Ô -1 = Ì p - 2sin x : < x £1 2 Ô Ô 1 -1 Ô -p - 2sin x : - 1 £ x < 2 Ó

TR_04.indd 7

ÏÔ 2 cos -1 x :0 £ x £ 1 (ii) cos -1 (2x 2 - 1) = Ì -1 ÔÓ2p - 2 cos x : - 1 £ x < 0 Ïa : -1 < x < 2x ˆ Ô = Ì -p + a : x > 1 (iii) tan Á Ë 1 - x 2 ˜¯ Ô Óp + a : x < -1 -1 Ê

where a = 2 tan–1(x). Ï a : -1 £ x £ 1 2x ˆ Ô = Ì p -a :x >1 (iv) sin Á Ë 1 + x 2 ˜¯ Ô Ó-p - a : x < -1 -1 Ê

where a = 2 tan–1(x) 2 ÏÔ2 tan -1 ( x) :x ≥ 0 -1 Ê 1 - x ˆ cos = (v) Á ˜ Ì Ë 1 + x 2 ¯ ÔÓ -2 tan -1 ( x) : x £ 0

4.10

MORE MULTIPLE ANGLES

(i) sin–1(3x – 4x3) Ï -1 Ô 3 sin x Ô Ô = Ì p - 3sin -1 x Ô Ô -1 Ô-p - 3sin x Ó

:-

1 1 £x£ 2 2

1 : < x £1 2 : -1 £ x < -

1 2

(ii) cos–1(4x3 – 3x) 1 Ï -1 : £ x £1 Ô 3cos x 2 Ô 1 1 Ô -1 = Ì 2p - 3cos x : - £ x < 2 2 Ô 1 Ô -1 Ô-2p + 3cos x : - 1 £ x < - 2 Ó Ê 3x - x3 ˆ (iii) tan -1 Á ˜ Ë 1 - 3x 2 ¯ 1 1 Ï -1 : sin–1(3x – 1). Find the domain of f (x) = cos–1(2x + 4). p Find the range of f ( x) = 2cos -1 (3x + 5) + . 4 p -1 2 Find the range of f ( x) = 3 cos (- x ) - . 2 Solve for x: cos–1 x + cos–1 x2 = 0. Solve for x: [sin–1 x] + [cos–1 x] = 0, where x is a non negative real number and [,] denotes the greatest integer function. 2 ˆ -1 Ê x Find the domain of f ( x) = cos Á 2 ˜ . Ë x + 1¯ Solve for x: cos–1(x) > cos–1(x2). Find the domain of f ( x) = tan -1 ( 9 - x 2 ) . Find the range of the function p f ( x) = 2 tan -1 (1 - x 2 ) + 6 Find the range of f (x) = cot–1(2x – x2). Solve for x: [cot–1 x] + [cos–1 x] = 0, Find the number of solutions of sin{x} = cos{x}, " x Œ [0, 2p]

22. Find the range of f ( x) = 2 sin -1 (3x + 5) + 23. 24. 25. 26. 27. 28.

29. 30. 31. 32.

33. 34. 35.

Q. Find the domains of each of the following functions: Ê | x | -2 ˆ Ê 1- | x | ˆ + cos -1 Á 36. f ( x) = sin -1 Á ˜ Ë 3 ¯ Ë 4 ˜¯ –1 2 37. f (x) = sin (2x – 1) 38. f ( x) = 5p sin -1 x - 6(sin -1 x) 2 Ê 3 tan -1 x + p ˆ 39. f ( x) = log 2 Á ˜ Ë p - 4 tan -1 x ¯ 3 ˆ -1 Ê 40. f ( x) = cos Á Ë 2 + sin x ˜¯ 2 -1 Ê x + 1ˆ 41. f ( x) = sin Á ˜ Ë 2x ¯ 2 -1 Ê x + 1ˆ 42. f ( x) = cos Á 2 ˜ Ë x ¯ 43. f (x) = sin–1 (log2 (x2 + 3x + 4) Ê Ê x2 ˆ ˆ 44. f ( x) = sin -1 Á log 2 Á ˜ ˜ Ë 2 ¯¯ Ë –1 2 45. f (x) = sin [2 – 3x ] -1 1 1 46. f ( x) = + 3sin x + x x-2 47. f (x) = sin–1(log2 x2) x 1 48. f (x) = e x + sin -1 ÊÁ - 1ˆ˜ + Ë2 ¯ x

2/10/2017 4:12:23 PM

4.9

Inverse Trigonometric Functions

49. f ( x) = sin -1 (log x 2) -1 50. f ( x) = sin (log 2 x)

Q. Find the ranges of each of the following functions: 51. f (x) = sin–1(2x – 3) p 52. f ( x) = 2 sin -1 (2x - 1) 4 53. f (x) = 2 cos–1(–x2) – p 1 p 54. f ( x) = tan -1 (1 - x 2 ) 2 4 55. f (x) = cot–1 (2x – x2) 56. f (x) = sin–1 x + cos–1 x + tan–1 x 57. f (x) = sin–1 x + sec–1 x + tan–1 x p 58. f ( x) = 3 cot -1 x + 2 tan -1 x + 4 59. f (x) = cosec–1[1 + sin2 x] 60. f (x) = sin–1 (log2(x2 + 3x + 4) CONSTANT PROPERTY

61. Find the range of f (x) = sin–1 x + cos–1 x + tan–1 x 62. Solve for x: 4 sin–1 (x – 2) + cos–1 (x – 2) = p 63. Solve for x: p sin -1 ( x 2 - 2x + 1) + cos -1 ( x 2 - x) = 2 64. Find the number of real solutions of p tan -1 x( x + 1) + sin -1 x 2 + x + 1 = . 2 ˆ x 2 x3 -1 Ê + - ........˜ 65. If sin Á x 2 4 Ë ¯ 4 6 Ê ˆ p x x + cos -1 Á x 2 + - .......˜ = , for 2 4 Ë ¯ 2 0 < x < 2 , then find x. 66. Solve for x: sin–1 x > cos–1 x Q. Solve for x: 67. (sin–1 x)2 – 3 sin–1 x + 2 = 0 68. sin–1 x + sin–1 2y = p 69. cos–1 x + cos–1 x2 = 2p 70. cos–1 x + cos–1 x2 = 0 71. 4 sin–1(x – 1) + cos–1(x – 1) = p 1 ˆ p -1 Ê -1 2 72. cot Á 2 ˜ + tan ( x - 1) = Ë x - 1¯ 2 Ê x 2 - 1ˆ Ê 2x ˆ 2p 73. cot -1 Á + tan -1 Á 2 ˜ = ˜ Ë x - 1¯ 3 Ë 2x ¯ 3 p -1 -1 74. 4 sin x + cos x = 4 7p -1 -1 75. 5 tan x + 3 cot x = 4 76. 5 tan–1 x + 4 cot–1 x = 2p p 77. cot -1 x - cot -1 ( x + 1) = 2 78. [sin–1 x] + [cos–1 x] = 0

TR_04.indd 9

79. [tan–1 x] + [cot–1 x] = 0 80. [sin–1 cos–1 sin–1 tan–1 x] = 0 81. [sin–1 cos–1 sin–1 tan–1 x] = 1 82. (tan -1 x) 2 + (cot -1 x) 2 =

5p 2 8

CONVERSION OF INVERSE TRIGONOMETRIC FUNCTIONS

Ê1 Ê 3ˆ ˆ 83. Find the value of cos Á cos -1 Á ˜ ˜ . Ë 5¯ ¯ Ë2 Êp Ê 1ˆˆ 84. Find the value of sin Á + sin -1 Á ˜ ˜ . Ë 2¯ ¯ Ë4 85. If m is a root of x2 + 3x + 1 = 0, then find the value of Ê 1ˆ tan -1 (m) + tan -1 Á ˜ . Ë m¯ 86. Prove that cos (tan–1(sin (cot–1 x))) = Q. Solve for x: 87. 6 (sin–1 x)2 – p sin–1 x £ 0 2 tan -1 x + p £0 88. 4 tan -1 x - p 89. sin–1 x < sin–1 x2 90. cos–1 x > cos–1 x2 91. log2 (tan–1 x) > 1 92. (cot–1 x)2 – 5 cot–1 x + 6 > 0 93. sin–1 x < cos–1 x 94. sin–1 x > sin–1(1 – x) 95. sin–1 2x > cosec–1 x 96. tan–1 3x < cot–1 x 97. cos–1 2x³ sin–1 x 98. x2 – 2x < sin–1 (sin 2) Ê xˆ 99. sin -1 Á ˜ < cos -1 ( x + 1) Ë 2¯

x2 + 1 x2 + 2

100. tan –1 2x > 2 tan –1 x Ê Ê 1ˆˆ 101. tan (cos -1 x) £ sin Á cot -1 Á ˜ ˜ Ë 2¯ ¯ Ë COMPOSITION OF TRIGONOMETRIC FUNCTIONS AND ITS INVERSE

102. Let f (x) = sin–1 x + cos–1 x Then find the value of: Ê 1 ˆ (i) f Á 2 ˜ , m Œ R Ë m + 1¯ Ê m2 ˆ (ii) f Á 2 ˜ , m Œ R Ë m + 1¯ Ê m ˆ (iii) f Á 2 ˜ , m Œ R Ë m + 1¯ (iv) f (m2 – 2m + 6), m Œ R (v) f (m2 + 1), m Œ R

2/10/2017 4:12:24 PM

4.10

Trigonometry Booster

2p 103. If cos x + cos y = , then find the value of sin–1 3 –1 x + sin y. 104. If m is the root of x2 + 3x + 1 = 0, then find the value of Ê 1ˆ tan -1 (m) + tan -1 Á ˜ . Ë m¯ 105. Solve for x: -1

-1

Ê Ê 2x 2 + 5 ˆ ˆ -1 sin -1 Á sin Á 2 ˜ ˜ > sin (sin 3) Ë Ë x + 2 ¯¯ COMPOSITION OF INVERSE TRIGONOMETRIC FUNCTIONS AND TRIGONOMETRIC FUNCTIONS

106. Find the values of: (i) sin–1(sin 3) (ii) sin–1(sin 53) –1 (iii) sin (sin 7) (iv) sin–1(sin 10) –1 (v) sin (sin 20) 107. Find the values of: (i) cos–1(cos 2) (ii) cos–1(cos 3) –1 (iii) cos (cos 5) (iv) cos–1(cos 7) –1 (v) cos (cos 10) 108. Find the values of: (i) tan–1(tan 3) (ii) tan–1(tan 5) –1 (iii) tan (tan 7) (iv) tan–1(tan 10) –1 (v) tan (tan 15) 109. Find the value of cos–1(sin (–5)) 110. Find f ¢(x), where f (x) = sin–1(sin x) and –2p £ x £ p 111. Find f ¢(x), where f (x) = cos–1(cos x) and –p £ x £ 2p Ê Ê 2x 2 + 5 ˆ ˆ 112. Solve for x: sin -1 Á sin Á 2 ˜˜ < p - 3 Ë Ë x +1 ¯¯ 113. Find the integral values of x satisfying the inequality, x2 – 3x < sin–`1 (sin 2) 114. Find the value of sin–1 (si 50) + cos–1 (cos 50) + tan–1 (tan 50) Q. Find the values of: 115. sin–1(sin 1) + sin–1(sin 2) + sin–1(sin 3) 116. sin–1 (sin 10) + sin–1 (sin 20) + sin–1 (sin 30) + sin–1 (sin 40) –1 117. cos (cos 1) + cos–1(cos 2) + cos–1(cos 3) + cos–1(cos 4) –1 118. cos (cos 10) + cos–1 (cos 20) + cos–1 (cos 30) + cos–1 (cos 40) –1 119. sin (sin 10) + cos–1 (cos 10) 120. sin–1 (sin 50) + cos–1 (cos 50) 121. sin–1(sin 100) + cos–1(cos 100) 122. cos–1(sin (–5)) + sin–1(cos (–5)) 123. Find the number of ordered pairs of (x, y) satisfying the equations y = |sin x| and y = cos–1(cos x), where x Œ [–2p, 2p] 124. Let f (x) = cos–1(cos x) – sin–1(sin x) in [0, p]. Find the area bounded by f (x) and x-axis. 125. tan–1(tan 1) + tan–1(tan 2) + tan–1(tan 3) + tan–1(tan 4)

TR_04.indd 10

126. tan–1(tan 20) + tan–1(tan 40) + tan–1(tan 60) + tan–1(tan 80) –1 127. sin (sin 15) + cos–1(cos 15) + tan–1(tan 15) 128. sin–1(sin 50) + cos–1(cos 50) – tan–1(tan 50) 129. 3x2 + 8x < 2 sin–1 (sin 4) – cos–1(cos 4) Ê Ê 2x 2 + 4 ˆ ˆ -1 130. sin Á sin Á 2 ˜˜ < p - 3 Ë Ë x +1 ¯ ¯ SUM OF ANGLES

Ê 1ˆ Ê 1ˆ 131. Find the value of tan -1 Á ˜ + tan -1 Á ˜ . Ë 2¯ Ë 3¯ 132. Find the value of tan–1(1) + tan–1(2) + tan–1(3) Ê 5ˆ 133. Find the value of tan -1 (9) + tan -1 Á ˜ . Ë 4¯ 134. Find the value of Ê 4ˆ Ê 5ˆ Ê 63 ˆ sin -1 Á ˜ + sin -1 Á ˜ - sin -1 Á ˜ Ë 5¯ Ë 13 ¯ Ë 65 ¯ 1 1 p 135. Prove that 2 tan -1 ÊÁ ˆ˜ + tan -1 ÊÁ ˆ˜ = Ë 3¯ Ë 7¯ 4 136. If sin–1 x + sin–1 y + sin–1 z = p, prove that x 1 - x 2 + y 1 - y 2 + z 1 - z 2 = 2 xyz 137. If cos–1 x + cos–1 y + cos–1 z = p prove that x2 + y2 + z2 + 2xyz = 1 Ê xˆ Ê yˆ 138. If cos -1 Á ˜ + cos -1 Á ˜ = q , prove that Ë 2¯ Ë 3¯ 9x2 + 12 xy cos q + 4y2 = 36 sin2 q tan -1 (1) + tan -1 (2) + tan -1 (3) 139. Let m = cot -1 (1) + cot -1 (2) + cot -1 (3) Then find the value of (m –1)2013. 3p 4 p 1 1 141. Solve for x: sin ( x) + sin (2 x) = 3 142. Let f (x) = cos–1(x) 140. Solve for x: tan -1 (2x) + tan -1 (3x) =

143. 144. 145. 146. 147. 148.

Êx 3 - 3x 2 ˆ + cos -1 Á + ˜ , for 1 £ x £ 1 ÁË 2 ˜¯ 2 2 Then find f (2013) sin–1 x + sin–1(1 – x) = cos–1 x x2 – 4x > sin–1 (sin (p3/2]) + cos–1 (cos[p3/2]) cos (tan–1 x) = x sin (tan–1 x) = cos (cot–1(x + 1)) Ê xˆ sec -1 Á ˜ - sec -1 x = sec -1 2 Ë 2¯ Ê Ê Ê 3ˆ ˆ ˆ ˆ Ê cos Á tan -1 Á cot Á sin -1 Á x + ˜ ˜ ˜ ˜ + tan (sec–1 x) = 0 Ë Ë 2¯ ¯ ¯ ¯ Ë Ë

149. Find the smallest +ve integer x so that Ê Ê xˆ Ê 1 ˆˆ Êpˆ tan Á tan -1 Á ˜ + tan -1 Á = tan Á ˜ Ë 10 ¯ Ë x + 1˜¯ ˜¯ Ë 4¯ Ë

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4.11

Inverse Trigonometric Functions

150. Find the least integral value of k for which (k – 2) x2 + 8x + k + 4 > sin–1 (sin 12) + cos–1 (cos 12) holds for all x in R. Ê 1 - x2 ˆ Ê1+ xˆ and b = sin -1 Á 151 If a = 2 tan -1 Á ˜ ˜ for Ë1- x¯ Ë 1 + x2 ¯ 0 < x < 1, then prove that a + b = p. 152. Let f (x) = sin–1 (sin x), " x Œ [–p, 2p]. Then find f ¢(x). 153. Let f (x) = cos–1 (cos x), " x Œ [–2p, p]. Then find f ¢(x). È 3p 5p ˘ , 154. Let, f (x) = tan–1(tan x) , " x Œ Í˙ . Then find Î 2 2 ˚ f ¢(x).

Ê1 Ê 1ˆ ˆ 2 172. Prove that sin Á cos -1 Á ˜ ˜ = Ë 9¯ ¯ 3 Ë2 1 1 173. Prove that sin ÊÁ tan -1 ( 63)ˆ˜ = Ë4 ¯ 2 2 Ê 1 Ê -1 Ê 24 ˆ ˆ ˆ 3 174. Prove that cos Á Á tan Á ˜ ˜ ˜ = Ë ¯ Ë ¯ 4 7 Ë ¯ 10

1 p 155. Prove that sin -1 ÊÁ ˆ˜ + cot -1 (3) = . Ë 5¯ 4

Ê1 1 Ê 2ˆ ˆ 175. Prove that tan Á cos -1 Á ˜ ˜ = Ë 3¯ ¯ Ë2 5

Ê 3ˆ Ê 12 ˆ 156. Prove that 2 tan -1 Á ˜ + tan -1 Á ˜ = p . Ë 2¯ Ë 5¯

Ê 7 Ê 1ˆ p ˆ 176. Prove that tan Á 2 tan -1 Á ˜ - ˜ = Ë 5¯ 4 ¯ Ë 17

Q. Find the simplest form of:

Ê 3p 1 -1 Ê 4 ˆ ˆ 1 - 5 177. tan Á - sin Á - ˜ ˜ = Ë 5¯ ¯ Ë 4 4 2

Ê 1 + sin x + 1 - sin x ˆ -1 157. cot Á ˜ Ë 1 + sin x - 1 - sin x ¯ 158. sin -1 ( x 1 - x - x 1 - x 2 ) p p Ê sin x + cos x ˆ ,- < x < 159. sin -1 Á ˜ Ë ¯ 4 4 2 5p Ê sin x + cos x ˆ p 160. cos -1 Á ˜¯ , 4 < x < 4 Ë 2 Ê 1 + x2 + 1 - x2 ˆ 161. tan -1 Á ˜ ÁË 1 + x 2 - 1 - x 2 ˜¯ -1 Ê 3

4 ˆ 162. sin Á cos x + sin x˜ Ë5 ¯ 5 MULTIPLE ANGLES

Ê Ê 1ˆˆ 163. Find the value of sin Á 2 sin -1 Á ˜ ˜ Ë 4¯ ¯ Ë Ê Ê 1ˆ ˆ 164. Find the value of cos Á 2cos -1 Á ˜ ˜ Ë 3¯ ¯ Ë Ê Ê 1ˆ ˆ 165. Find the value of cos Á 2 tan -1 Á ˜ ˜ Ë 3¯ ¯ Ë Ê1 Ê 3ˆ ˆ 166. Find the value of sin Á cot -2 Á ˜ ˜ Ë 4¯ ¯ Ë2 Ê 3p Ê 3ˆ ˆ 167. Find the value of tan -1 Á - 2 tan -1 Á ˜ ˜ Ë 4¯ ¯ Ë 4 Ê 3 Ê 1ˆˆ 168. Prove that sin Á 2sin -1 Á ˜ ˜ = Ë 2¯ ¯ Ë 2 Ê Ê 1 ˆ ˆ 23 169. Prove that sin Á 3sin -1 Á ˜ ˜ = Ë 3 ¯ ¯ 27 Ë Ê1 Ê 1ˆ ˆ 3 170. Prove that cos Á cos -1 Á ˜ ˜ = Ë 8¯ ¯ 4 Ë2

TR_04.indd 11

Ê1 Ê 1 ˆˆ 3 5 171. Prove that cos Á cos -1 Á - ˜ ˜ = Ë 10 ¯ ¯ 10 Ë2

178. Find the integral values of x satisfying the inequation x2 – 3x < sin–1 (sin 2) 179. Find the value of x satisfying the inequation 3x2 + 8x < 2 sin–1 (sin 4) – cos–1 (cos 4) 180. For what value of x, Ïx 1 - 3x 2 ¸ ˝ f ( x) = cos -1 x + cos -1 Ì + 2 Ó2 ˛ is a constant function. MORE MULTIPLE ANGLES

Ê 2x ˆ 181. Let f ( x) = sin -1 Á + 2 tan -1 ( x) , x > 1 Ë 1 + x 2 ˜¯ Then find the value of f (2013) Ê 1 - x2 ˆ Ê1+ xˆ 182. Let f ( x) = 2 tan -1 Á + sin -1 Á ˜ ˜ Ë1- x¯ Ë 1 + x2 ¯ Ê 1 ˆ for 0 £ x < 1. Then find the value of f Á Ë 2014 ˜¯ Ê 6x ˆ Ê xˆ 183. Let f ( x) = sin -1 Á 2 + 2 tan -1 Á - ˜ Ë x + 9 ˜¯ Ë 3¯ is independent of x, then find the value of x 184. Find the interval of x for which the function Ê 1 - x2 ˆ -1 f ( x) = cos -1 Á ˜ + 2 tan ( x) is a constant Ë 1 + x2 ¯ function. 185. Find the interval of x for which the function f (x) = 3 cos–1(2x2 – 1) + 2 cos–1(4x3 – 3x) is independent of x 186. If tan–1 y : tan–1 x = 4 : 1, then express y as algebraic Ê 1 ∞ˆ function of x. Also, prove that tan Á 22 ˜ is a root of Ë 2 ¯ x4 – 6x2 + 1 = 0

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4.12

Trigonometry Booster

BOARD SPECIAL PROBLEMS

Q. Prove that: Ê 1 + x2 + 1 - x2 ˆ Ê p 1 ˆ + cos -1 x 2 ˜ 187. tan -1 Á ˜= ¯ ÁË 1 + x 2 - 1 - x 2 ˜¯ ÁË 4 2 aˆ a ˆ 2b Êp 1 Êp 1 188. tan Á + cos -1 ˜ + tan Á - cos -1 ˜ = Ë4 2 Ë4 2 b¯ b¯ a Ê p-qˆ Ê q-r ˆ Ê r- pˆ + tan -1 Á + tan -1 Á =p 189. tan -1 Á Ë 1 + pq ˜¯ Ë 1 + qr ˜¯ Ë 1 + pr ˜¯ where p > q > 0 and pr < –1 < qr Ê ab + 1ˆ Ê bc + 1ˆ Ê ca + 1ˆ 190. cot -1 Á + cot -1 Á + cot -1 Á =0 Ë a - b ˜¯ Ë b - c ˜¯ Ë c - a ˜¯ 2 ˆ Ê1 Ê 2x ˆ 1 -1 Ê 1 - y ˆ 191. tan Á sin -1 Á cos + ˜ Á ˜˜ Ë 1 + x2 ¯ 2 Ë 1 + y2 ¯ ¯ Ë2

Ê x+ yˆ , xy < 1 =Á Ë 1 - xy ˜¯ Ê 1- x ˆ Ê 1- y ˆ y-x Ê ˆ 192. tan -1 Á - tan -1 Á = sin -1 Á 2 2 ˜ Ë 1+ x ˜¯ Ë 1+ y ˜¯ Ë (1+ x )(1+ y ) ¯ 1 Ê ˆ 193. tan -1 Á tan 2A˜ + tan -1 (cot A) + tan -1 (cot 3 A) = 0 Ë2 ¯ -1 Ê

Ê b + a cos q ˆ a - b Êq ˆˆ tan Á ˜ ˜ = cos -1 Á 194. 2 tan Á Ë a + b cos q ˜¯ Ë a + b Ë 2¯¯ 195. tan (2 tan–1 a) = 2 tan (tan–1 a + tan–1 a3) Êx 3 - 3x 2 ˆ p 1 196. cos -1 x + cos -1 Á + ˜¯ = , < x < 1 Ë2 2 3 2 –1 –1 –1 197. If sin x + sin y + sin z = p, then prove that x 1 - x 2 + y 1 - y 2 + z 1 - z 2 = 2xyz 198. If cos–1 x + cos–1 y + cos–1 z = p, then prove that x2 + y2 + z2 + 2xyz = 1 Ê xˆ Ê yˆ 199. If cos -1 Á ˜ + cos -1 Á ˜ = q , then prove that Ë 2¯ Ë 3¯ 9x2 – 12 xy cos q + 4y2 = 36 sin2 q. 3p 200. If sin -1 x + sin -1 y + sin -1 z = , then prove that 2 2 2 2 x + y + z – 2xyz = 1. 3p 201. If sin -1 x + sin -1 y + sin -1 z = , then prove that 2 xy + yz + zx = 3. 3p 202. If sin–1 x + sin–1 y + sin–1 z = , then find the value of 2 9 x 2012 + y 2012 + z 2012 - 2013 2013 x +y + z 2013 203. If cos–1 x + cos–1 y + cos–1 z = 3p, then prove that, xy + yz + zx = 3.

TR_04.indd 12

204. If co–1 x + cos–1 y + cos–1 z = 3p, then find the value of Ê x 2013 + y 2013 + z 2013 + 6 ˆ Á ˜ Ë x 2014 + y 2014 + z 2014 ¯ p 205. If tan -1 x + tan -1 y + tan -1 z = , then prove that xy + 2 yz + zx = 1. p 206. If tan -1 x + tan -1 y = , then prove that x + y + xy = 1. 4 207. If tan–1 x + tan–1 y + tan–1 z = p, then prove that x + y + z = xyz. Ê 1 + x2 - 1 - x2 ˆ 208. If tan -1 Á ˜ = a , then prove that ÁË 1 + x 2 + 1 - x 2 ˜¯ x2 = sin 2a 209. Let m = tan2 (sec–1 2) + cot2 (cosec–1 3).. Then find the value of (m2 + m + 10). Ê 3 sin 2q ˆ p 1 = , then find the value of 210. If sin -1 Á 2 Ë 5 + 4 cos 2q ˜¯ 4 tan q. (tan -11 + tan -1 2 + tan -1 3) , then prove that 211. Let m = (cot -11 + cot -1 2 + cot -1 3) (m + 2)m + 1 = 64. Q. Solve for x:

3p 4 Ê ˆ x 1 x 1ˆ + Ê tan -1 Á + tan -1 Á = tan -1 ( - 7) ˜ Ë x ˜¯ Ë x - 1¯ p sin -1 (2x) + sin -1 ( x) = 3 p 1 Ê ˆ sin -1 Á ˜ + cos -1 x = Ë 5¯ 4 p -1 -1 sin (x) + sin (3x) = 3 Ê 1 ˆ Ê 1 ˆ Ê 2ˆ tan -1 Á + tan -1 Á = tan -1 Á 2 ˜ Ëx ¯ Ë 1 + 2x ˜¯ Ë 1 + 4x ˜¯

212. tan -1 (2x) + tan -1 (3x) = 213. 214. 215. 216. 217.

218. 2 tan–1(2x + 1) = cos–1 x 219. cos -1 x - sin -1 x = cos -1 ( x 3) 220. If tan–1 y : tan–1 x = 4 : 1, express y as an algebraic Êpˆ function of x. Hence, prove that tan Á ˜ is a root of Ë 8¯ x4 + 1 = 6x2

LEVEL II

(Mixed Problems)

1. The set of values of k for which x2 – kx + sin–1 (sin 4) > 0 for all real x is (a) {0} (b) (–2, 2) (c) R (d) None of these

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4.13

Inverse Trigonometric Functions

2. If x < 0 then value of -1

tan ( x) + tan Á ˜ = Ë x¯

p 2 (c) 0

p 2 (d) None of these (b) -

(a)

2p , then cos–1 x + cos–1 y is 3 p p 2p (a) (b) (c) (d) p 3 6 3 p Let f (x) = sin–1 x + cos–1 x. Then is equal to 2 Ê 1ˆ (a) f Á ˜ (b) f (k2 – 2k + 3), k Œ R Ë 2¯ Ê 1 ˆ (c) f Á ,kŒR (d) f (–2) Ë 1 + k 2 ˜¯ Which one of the following is correct? (b) tan 1 < tan–1 1 (a) tan 1 > tan–1 1 –1 (c) tan 1 = tan 1 (d) None If a sin–1 x – b cos–1 x = c, then the value of a sin–1 x + b cos–1 x is p ab + c(b - a ) (a) 0 (b) a+b p p ab - c(b - a ) (c) (d) a+b 2 The number of solutions of the equation sin–1 (1–x) – 2 p sin–1 x = is 2 (a) 0 (b) 1 (c) 2 (d) More than two The smallest and the largest values of Ê1 - xˆ , 0 £ x £ 1 are tan–1 Á Ë 1 + x ˜¯ p p p p p (a) 0, p (b) 0, (c) - , (d) , 4 4 4 4 2 Ê ˆ 3 The equation sin–1 x – cos–1 x = cos–1 Á ˜ has Ë 2 ¯ (a) No solution (b) Unique solution (c) Infinite number of solution (d) None If –p £ x £ 2p, then cos–1 (cos x) is (a) x (b) p – x (c) 2p + x (d) 2p – x

3. If sin–1 x + sin–1 y =

4.

5.

6.

7.

8.

9.

10.

Ê 1ˆ p 11. If sin -1 x + cot -1 Á ˜ = , then x is equal to Ë 2¯ 2 1 2 3 (c) (d) (a) 0 (b) 5 5 2 12. If cos [tan–1 {sin (cot–1 3 )}] = y, then the value of y is 4 2 (a) y = (b) y = 5 5 (c) y = -

TR_04.indd 13

2 5

(d) y =

1 , then the value of cos (cos–1x + 2 sin–1 x) is 5 24 24 1 1 (a) (d) (b) (c) 25 25 5 5 1 1 Ê ˆ Ê ˆ 14. tan -1 Á ˜ + tan -1 Á ˜ is equal to Ë 2¯ Ë 3¯ p p (b) (a) 4 2 p (c) (d) None of these 3 15. tan–1 a + tan–1 b, where a > 0, b > 0, ab > 1 is equal to 13. If x =

-1 Ê 1 ˆ

3 2

Ê a+bˆ (a) tan -1 Á Ë 1 - ab ˜¯

Ê a+bˆ (b) tan -1 Á -p Ë 1 - ab ˜¯

Ê a+bˆ (c) p + tan -1 Á Ë 1 - ab ˜¯

Ê a+bˆ (d) p - tan -1 Á Ë 1 - ab ˜¯

16. A solution to the equation p is 2 (a) x = 1 (b) x = –1 (c) x = 0 (d) x = p 17. All possible values of p and q for which 3p cos -1 ( p ) + cos -1 ( 1 - p ) + cos -1 ( 1 - q ) = 4 holds, is (a) p = 1, q = 1/2 (b) q >1, p = 1/2 (c) 0 < p < 1, q = 1/2 (d) None Êp 1 ˆ Êp 1 ˆ 18. tan Á + cos -1 x˜ + tan Á - cos -1 x˜ , Ë4 2 ¯ Ë4 2 ¯ tan -1 (1 + x) + tan -1 (1 - x) =

x π 0, is equal to (a) x

(b) 2x

(c)

2 x

(d)

x 2

19. The value of cot -1 (3) + cosec -1 ( 5) is p p p p (a) (b) (c) (d) 2 3 4 6 20. If

2n

2n

i =1

i =1

 sin -1xi = np , then  xi

is

(a) n n(n + 1) (c) 2

(b) 2n n(n - 1) (d) 2

(a) tan a (c) tan a

(b) cot a (d) cot a

Ê p uˆ 21. If u = cot -1 ( tan a ) - tan -1 ( tan a ), then tan Á - ˜ Ë 4 2¯ is equal to

Ê b ˆ Ê a ˆ 22. The value of tan -1 Á + tan -1 Á , if –C = 90, ˜ Ë b + c¯ Ë a + c ˜¯ in triangle ABC is p p p (b) (c) (d) p (a) 4 3 2

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4.14

Trigonometry Booster

Ê nˆ p 23. If cot -1 Á ˜ > , n Œ N, then the maximum value of Ëp¯ 6 ‘n’ is (a) 1 (b) 5 (c) 9 (d) None of these 24. sin–1 x > cos–1 x holds for 1 ˆ Ê (a) all values of x (b) x Œ Á 0, ˜ Ë 2¯ Ê 1 ˆ (c) Á ,1 (d) x = 0.75 Ë 2 ˜¯ Ê1 -1 Ê 1 ˆ ˆ 25. The value of cos Á cos Á ˜ ˜ is equal to Ë 8¯ ¯ Ë2 3 3 1 (b) (c) 4 4 16 26. The values of x satisfying Ê Ê 1 ˆˆ tan (sec -1 x) = sin Á cos -1 Á ˜ ˜ is Ë 5¯¯ Ë 3 5 3 (a) ± (c) ± (b) ± 5 3 5 (a)

(d) 4

(d) ±

3 5

5p 2 , then the value of x is 8 (a) 0 (b) –1 (c) –2 (d) –3 –1 28. The number of real solutions of cos x + cos–1 2x = –p are (a) 0 (b) 1 (c) 2 (d) infinitely many 29. Let a, b, c be positive real numbers and 27. If (tan -1 x) 2 + (cot -1 x) 2 =

Ê a (a + b + c) ˆ -1 Ê b( a + b + c ) ˆ q = tan -1 Á ˜¯ + tan ÁË ˜¯ Ë bc ac Ê c(a + b + c) ˆ + tan -1 Á ˜, ba Ë ¯ then the value of tan q is (a) 0 (b) 1 (c) –1 (d) None 30. The set of values of x satisfying the inequation tan2 (sin–1 x) > 1 is È 1 1 ˘ , (a) [–1, 1] (b) Í ˙ Î 2 2˚ Ê 1 1 ˆ È 1 1 ˘ , (d) [ -1, 1] - Á , (c) ( -1, 1) - Í ˜ ˙ Ë 2 2¯ Î 2 2˚ 31. The value of a for which ax2 + sin–1(x2 – 2x + 2) + cos–1(x2 – 2x + 2) = 0 has a real solution, is 2 -2 p -p (b) (c) (d) (a) p p 2 2 32. The value of È Ê ˆ˘ Ê 2 - 3ˆ -1 Ê 12 ˆ -1 sin -1 Ícot Ásin -1 Á cos sec 2 + + ˜ ˙ is ÁË ˜ Ë 4 ˜¯ ¯ ˙˚ ÍÎ Ë 4 ¯ p p p (a) 0 (b) (c) (d) 4 6 2

TR_04.indd 14

33. The number of positive integral solutions of Ê 1ˆ Ê 3 ˆ tan -1 x + cot -1 Á ˜ = sin -1 Á is Ë 10 ˜¯ Ë y¯ (a) 0 (b) 1 (c) 2 (d) 3 34. The value of È1 Ê Ê 63 ˆ ˆ Ô¸˘ ÔÏ cos Í cos Ìcos Á sin -1 Á ˜ ˜ ˝˙ is 2 Ë 8 ¯ ¯ ˛Ô˚˙ Ë ÓÔ ÎÍ 3 3 3 3 (b) (c) (d) (a) 8 4 2 16 35. If tan–1 x + tan–1 y + tan–1 z = p, then the value of 1 1 1 + + is yz zx xy 1 (a) 0 (b) 1 (c) (d) xyz xyz 1 -1 Ê ˆ 36. If x < 0, then tan Á ˜ is Ë x¯ (b) –cot–1(x) (a) cot–1(x) –1 (c) –p + cot (x) (d) None 37. The number of triplets satisfying sin–1 x + cos–1 y + sin–1 z = 2p, is (a) 0 (b) 2 (c) 1 (d) infinite 38. If x2 + y2 + z2 = r2, then Ê xy ˆ Ê zy ˆ Ê zx ˆ tan -1 Á ˜ + tan -1 Á ˜ + tan -1 Á ˜ is equal to .... Ë zr ¯ Ë xr ¯ Ë yr ¯ p (c) 0 (d) None (a) p (b) 2 39. If tan–1 x + tan–1 2x + tan–1 3x = p, then the value of x is (a) 0 (b) –1 (c) 1 (d) f 40. The number of solutions of the equation 1 + x2 + 2x sin (cos–1 y) = 0 is (a) 1 (b) 2 (c) 3 (d) 4 41. If a is the only real root of the equation x3 + bx2 + cx + -1 -1 Ê 1 ˆ 1 = 0, then the value of tan a + tan Á ˜ is equal to Ëa¯ p -p (b) (c) 0 (d) None (a) 2 2 42. If a, b, g are the roots of x3 + px2 + 2x + p = 0, the the general value of tan–1 a + tan–1 b tan–1 g is np (a) np (b) 2 (2n + 1)p (c) (d) depend on p 2 43. If [sin–1 (cos–1)(cos–1(sin–1(tan–1 x)))] = 1, where [,] = GIF, the value of x lies in (a) [tan sin cos 1, tan sin cos sin 1] (b) (tan sin cos 1, tan sin cos sin 1) (c) [–1, 1] (d) [sin cos tan 1, sin cos sin tan 1]

LEVEL IIA

(Problems For JEE Main)

1. Find the principal value of sin–1(sin 10) 2. Find the principal value of cos–1(cos 5)

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4.15

Inverse Trigonometric Functions

3. Find the value of tan–1(1) + tan–1(2) + tan–1(3) 4. Find x if sin–1 x > cos–1 x 5. Find x if sin–1 x < cos–1 x 6. Find x if 2sin -1 x = sin -1 (2x 1 - x 2 )

Ê 5ˆ Ê 12 ˆ p 28. If sin -1 Á ˜ + sin -1 Á ˜ = , then find x. Ë x¯ Ë x¯ 2 29. If x = sin–1 (b6 + 1) + cos–1 (b4 + 1) + tan–1 (a2 + 1) then find the value of pˆ pˆ Ê Ê sin Á x + ˜ + cos Á x + ˜ Ë ¯ Ë 4 4¯

7. Find x if 3 sin–1 x = p + sin–1(3x – 4x3) Ê 2x ˆ 8. Find x if 2 tan -1 x = p + tan -1 Á Ë 1 - x 2 ˜¯ Êp Ê 1ˆˆ 9. Find the value of cos Á + cos -1 Á - ˜ ˜ Ë 2¯ ¯ Ë6 10. Find the value of cos -1 (cos(2 cot -1 ( 2 - 1))) 1 Ê ˆ 11. Find the value of  tan -1 Á . 2˜ + + r r 1 Ë ¯ r =0 2 r -1 ˆ -1 Ê 12. Find the value of  tan Á ˜. Ë 1 + 22r -1 ¯ r =1 n

13. Find the value of

Ê r - r - 1ˆ ˜ r (r + 1) ¯

 sin -1 ÁË r =1

14. Find the value of Ê a x - yˆ Ê a -a ˆ tan -1 Á 1 + tan -1 Á 2 1 ˜ Ë a1 y + x ¯˜ Ë 1 + a1a2 ¯ Ê an - an –1 ˆ Ê a - a2 ˆ + tan -1 Á 3 + .... + tan -1 Á ˜ ˜ Ë 1 + a3a2 ¯ Ë 1 + an an –1 ¯

15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26

Ê 1ˆ + tan -1 Á ˜ , where x, y, a1, a2, … an Œ R+ Ë an ¯ p2 Find x if (tan -1 x) 2 + (cot -1 x) 2 = 8 Find the maximum value of f (x), if f (x) = (sec–1 x)2 + (cosec–1 x)2 Find the minimum value of f (x), if f (x) = (sin–1 x)3 + (cos–1 x)3 Find x if [cot–1 x] + [cos–1 x] = 0 Find x if [sin–1 x] + [cos–1 x] = 0. Find x if [tan–1 x] + [cot–1 x] = 2 Find x if [sin–1(cos–1(sin–1(tan–1 x)))] = 1 Find the range of f (x) = sin–1 x + tan–1 x + cot–1 x Find the range of f (x) = sin–1 x + cos–1 x + tan–1 x. Find the range of f (x) = sin–1 x + sec–1 x + tan–1 x p If tan -1 (2x) + tan -1 (3x) = , then find x 4 Ê 1ˆ -1 -1 Ê 4 ˆ If cos x = cot Á ˜ + tan -1 Á ˜ , Ë 3¯ Ë 7¯ then find x.

Ê nˆ p 27. If cot -1 Á ˜ > , where n Œ N, then find the maximum Ëp¯ 6 value of n.

TR_04.indd 15

30. Find the number of integral values of k for which the equation sin–1 x + tan–1 x = 2k + 1 has a solution. p -1 -1 31. If sin x + sin y = , then find the value of 2 Ê 1 + x4 + y 4 ˆ Á 2 ˜ Ë x - x2 y 2 + y 2 ¯ 32. If cos–1 x + cos–1(2x) + cos–1(3x) and x satisfies the equation ax3 + bx2 + cx = 1 then find the value of a2 + b2 + c2 + 10. 33. If f (x) = sin–1 x + tan–1 x + x2 + 4x + 5 such that Rf = [a, b], find the value of a + b + 5. 34. If cot -1 ( cos a ) + tan -1 ( cos a ) = x , then sin x is (b) cot2(a/2)

(a) 1

Êaˆ (d) cot Á ˜ Ë 2¯ [JEE Main, 2002] Ê x ˆ Ê ˆ 35. The domain of sin –1 Á log 3 Á ˜ ˜ is Ë 3¯ ¯ Ë (a) [1, 9] (b) [–1, 9] (c) [–9, 1] (d) [–9, –1] [JEE Main, 2002] 36. The trigonometric equation sin–1 x = 2 sin–1 a has a solution for 1 (a) all real values (b) |a| < 2 1 1 1 < |a| < (c) |a| £ (d) 2 2 2 [JEE Main, 2003] 37. The domain of the function sin -1 ( x - 3) f (x) = is 9 - x2 (c) tan a

(a) [1, 2]

(b) [2, 3)

(c) [2, 3] (d) [1, 2) [JEE Main, 2004] 38. Let f: (–1, 1) Æ B be a function defined as Ê 2x ˆ , then f is both one one and onto, f (x) = tan -1 Á Ë 1 - x 2 ˜¯ when B lies in p p (b) ÊÁ 0, ˆ˜ (a) ÈÍ0, ˆ˜ ¯ Ë 2 2¯ Î p p p p (c) ÊÁ – , ˆ˜ (d) ÈÍ – , ˘˙ Ë 2 2¯ Î 2 2˚ [JEE Main, 2005]

2/10/2017 4:12:30 PM

4.16

Trigonometry Booster -1 Ê

yˆ 39. If cos x - cos Á ˜ = a , then 4x2 – 4xy cos a + y2 is Ë 2¯ (a) 4 (b) 2 sin a (c) –4 sin2 a (d) 4 sin2 a [JEE Main, 2005] p –1 Ê x ˆ -1 Ê 5 ˆ 40. If sin Á ˜ + cosec Á ˜ = , then x is Ë 5¯ Ë 4¯ 2 (a) 4 (b) 5 (c) 1 (d) 3 [JEE Main, 2007] 41. Find the value of Ê Ê 5ˆ Ê 2ˆ ˆ cot Á cosec -1 Á ˜ + tan -1 Á ˜ ˜ is Ë 3¯ Ë 3¯ ¯ Ë 3 5 6 4 (b) (c) (d) (a) 17 17 17 17 [JEE Main, 2008] –1

LEVEL III

(Problems for JEE Advanced)

1. Find the domain of Ê | x| - 2 ˆ Ê 1 - | x| ˆ f (x) = sin -1 Á + cos -1 Á Ë 3 ˜¯ Ë 4 ˜¯ 2. Find the domain of f (x) = 5p sin -1 x - 6(sin -1 x) 2 3. Find the domain of f (x) = sin–1 (log2 (x2 + 3x + 4)). 4. Solve for x: cos–1 x + cos–1 x2 = 2p 5. Solve for x: p Ê 1 ˆ cot -1 Á 2 ˜ + tan -1 ( x 2 - 1) = 2 Ë x - 1¯ 6. Solve for x: Ê x 2 - 1ˆ Ê 2x ˆ 2p cot -1 Á + tan -1 Á 2 ˜ = Ë 2x ˜¯ 3 Ë x - 1¯ 7. Solve for x: Ê Ê 2x 2 + 4 ˆ ˆ sin -1 Á sin Á 2 ˜˜ < p - 3 Ë Ë x +1 ¯¯ 8. Solve for x: x2 – 4x > sin–1 (sin [p3/2]) + cos–1 (cos [p3/2]) 9. Solve for x: Ê Ê Ê 3ˆ ˆ ˆ ˆ Ê cos Á tan -1 Á cot Á sin -1 Á x + ˜ ˜ ˜ ˜ + tan (sec–1 x) = 0 Ë Ë 2¯ ¯ ¯ ¯ Ë Ë 10. Solve for x: Ê Ê 1 ˆˆ Ê xˆ Êpˆ tan Á tan -1 Á ˜ + tan -1 Á ˜ = tan ÁË 4 ˜¯ ˜ Ë ¯ 10 x 1 + Ë ¯ Ë ¯ 2 Ê -1 Ê 1 + x ˆ -1 1 - x ˆ and b = sin Á 11. If a = 2 tan Á ˜ for ˜ Ë1 - x¯ Ë 1 + x2 ¯ 0 < x < 1, then prove that a + b = p

TR_04.indd 16

12. Find the range of f (x) = 2 sin–1 (2x – 3) 13. Find the range of p f (x) = 2 sin -1 (2x - 1) 4 14. Find the range of f (x) = 2 cos–1 (–x)2 – p 15. Find the range of p 1 f (x) = tan -1 (1 - x 2 ) 2 4 –1 16. Find the range of f (x) = cot (2x – x2). 17. Find the range of f (x) = sin–1 x + cos–1 x + tan–1 x 18. Find the range of f (x) = sin–1 x + sec–1 x + tan–1 x 19. Find the range of p f (x) = 3 cot -1 x + 2 tan -1 x + 4 –1 1 20. Prove that sin (cot (tan (cos x))) = x, " x Œ (0, 1] 21. Prove that sin (cosec–1(cot (tan–1 x))) = x, " x Œ (0, 1] 22. Find the value of sin–1(sin 5) + cos–1(cos 10) + tan–1(tan (–6)) + cot–1 (cot (–10)) 23. If U = cot -1 ( cos 2q ) - tan -1 ( then prove that sin U = tan2 q 24. Prove that aˆ Êp 1 Êp tan Á + cos -1 ˜ + tan Á Ë4 2 Ë4 b¯ 25. Prove that

cos 2q ) ,

1 a ˆ 2b cos -1 ˜ = 2 b¯ a

Ê cos x + cos y ˆ Ê Ê xˆ Ê yˆˆ = 2 tan -1 Á tan Á ˜ tan Á ˜ ˜ cos -1 Á ˜ Ë 2¯ Ë 2¯¯ Ë Ë 1 + cos x cos y ¯ 26. Prove that Ê a - b Ê xˆˆ Ê b + a cos x ˆ 2 tan -1 Á tan Á ˜ ˜ = cos -1 Á Ë ¯ 2 ¯ Ë a + b cos x ˜¯ Ë a+b 27. If tan–1 x, tan–1 y, tan–1 z are in AP then prove that (x + z) y2 + 2y(1 – xz), where y Œ (0, 1), xz < 1, x > 0 and z > 0. 28. Prove that 46p ˆ Ê 33p ˆ Ê sin -1 Á sin + cos -1 Á cos ˜ ˜ Ë Ë 7 ¯ 7 ¯ 13p ˆ Ê Ê 19p ˆ ˆ -1 Ê + tan -1 Á - tan ˜¯ + cot ÁË cot ÁË ˜ Ë 8 8 ¯ ˜¯ 13p 7 29. Solve for x and y: =

2p p , cos -1 x - cos -1 y = . 3 3 2 2ˆ 1+ x - 1- x ˜ , then prove that 1 + x 2 + 1 - x 2 ˜¯

sin -1 x + sin -1 y = Ê -1 30. If y = tan Á ÁË

x2 = sin (2y).

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4.17

Inverse Trigonometric Functions

31. Prove that Ê1 Êa ˆˆ a3 b3 Ê1 Ê b ˆˆ sec 2 Á tan -1 Á ˜ ˜ cosec 2 Á tan -1 Á ˜ ˜ + 2 Ë b ¯¯ Ë2 Ëa ¯¯ Ë2 2 = (a + b)(a2 + b2) 32. Find the minimum value of n, if Ê n 2 - 10n + 21.6 ˆ p cot -1 Á ˜¯ > , n Œ N Ë p 6 33. Prove that ÏÔ ÏÔ ¸ÔÔ¸ Ê 2- 3ˆ -1 Ê 12 ˆ sin -1 Ìcot Ìsin -1 Á + sec -1 ( 2)˝˝ = 0 ˜ + cos ÁË ˜ Ë 4 ¯ 4 ¯ ˛ÔÔ˛ ÓÔ ÓÔ 34. Solve for x: [sin–1(cos–1(sin–1(tan–1 x)))] = 1 where [,] = GIF 35. Find the interval for which Ê 2x ˆ 2 tan -1 x + sin -1 Á is independent of x. Ë 1 + x 2 ˜¯ 36. If x = cosec (tan–1(cos (cot–1(sec (sin–1 a))))) and y = sec (cot–1(sin (tan–1(cosec (cos–1 a))))), where a Œ [0, 1], then find the relation between x and y. 37. Find the sum of the infinite series. Ê 1ˆ Ê 1ˆ Ê 1ˆ tan -1 Á ˜ + tan -1 Á ˜ + tan -1 Á ˜ + º Ë 3¯ Ë 7¯ Ë 13 ¯ 38. Find the sum of Ê 2 - 1ˆ Ê 3 - 2ˆ Ê 1 ˆ + sin -1 Á + sin -1 Á sin -1 Á ˜ ˜ ˜ Ë 2¯ Ë Ë 6 ¯ 12 ¯ Ê n - n - 1ˆ ...... sin -1 Á ˜ Ë n ¥ n + 1¯

to

39. Find the sum of infinite series: cot–1(2.12) + cot–1(2.22) + cot–1(2.32) + … Ê xˆ Ê yˆ 40. If cos -1 Á ˜ + cos -1 Á ˜ = q , then prove that Ë 2¯ Ë 3¯ 9x2 – 2 xy cos q + 4y2 = 36 sin2 q [Roorkee, 1984] Ê1 ˆ Ê ˆ 5 [Roorkee, 1986] 41. Evaluate: tan Á cos -1 Á ˜ ˜ Ë 3 ¯¯ Ë2 Note: No questions asked between 1987 and 1991. 42. Solve for x: sin[2 cos–1{cot (2 tan–1 x)}] [Roorkee, 1992] 43. Find all positive integral solutions of

tan -1 x + cos -1

y ˆ Ê 3 ˆ Ê = sin -1 Á Á 2˜ Ë 10 ˜¯ Ë 1+ y ¯ [Roorkee, 1993]

TR_04.indd 17

44. If cos–1 x + cos–1 y + cos–1 z = p, then find the value of x2 + y2 + z2 + 2xyz. [Roorkee, 1994] 45. Convert the trigonometric function sin[2 cos–1{cot (2 tan–1 x)}] into an algebraic function f (x). Then from the algebraic function f (x), find all values of x for which f (x) is zero. Also, express the values of x in the form of a ± b , where a and b are rational numbers. [Roorkee, 1995] Note No questions asked in 1996. Ê 3 sin 2q ˆ 1 46. If q = tan -1 (2 tan 2q ) - sin -1 Á , 2 Ë 5 + 4 cos 2q ˜¯ then find the general value of q [Roorkee, 1997] Note No questions asked in 1998. 47. Using the principal values, express the following expression as a single angle Ê 142 ˆ Ê 1ˆ Ê 1ˆ 3 tan -1 Á ˜ + 2 tan -1 Á ˜ + sin -1 Á Ë 2¯ Ë 5¯ Ë 65 5 ˜¯ [Roorkee, 1999]

48. Solve for x:

Ê ax ˆ Ê bx ˆ sin -1 Á ˜ + sin -1 Á ˜ = sin -1 x Ë c¯ Ë c¯ where a2 + b2 = c2, c π 0

[Roorkee, 2000]

49. Solve for x: p 2 [Roorkee, 2001] 50. Let x1, x2, x3, x4 be four non zero numbers satisfying the equation cos -1 ( x 6) + cos -1 (3 3 x 2 ) =

Ê aˆ Ê bˆ Ê cˆ Êdˆ p tan –1 Á ˜ + tan –1 Á ˜ + tan –1 Á ˜ + tan –1 Á ˜ = Ë x¯ Ë x¯ Ë x¯ Ë x¯ 2 then prove that 4

(i)

Ê 1ˆ

(ii)

’ ( xi ) = abcd

(iv) P( x1 + x2 + x3 ) = abcd

i =1 4

(iii)

4

 xi = 0 i =1

 ÁË x ˜¯ = 0 i =1

i

51. Let cos–1 (x) + cos–1 (2x) + cos–1 (3x) = p If x satisfies the cubic equation ax3 + bx2 + cx – 1 = 0, then find the value of (a + b + c + 2). Ê1 Ê 4ˆ ˆ 52. If x = sin(2 tan -1 2), y = sin Á tan -1 Á ˜ ˜ then prove Ë 3¯ ¯ Ë 2 that y2 = 1 – x

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4.18

Trigonometry Booster

LEVEL IV

16. Find the greatest and least value of the function f (x) = (sin–1 x)3 + (cos–1 x)3. p 17. Solve for x: sin -1 x + sin -1 2x = 2 18. Solve for x: Ê 1 ˆ Ê 1 ˆ Ê 2ˆ tan -1 Á + tan -1 Á = tan -1 Á 2 ˜ ˜ ˜ Ë 1 + 2x ¯ Ë 1 + 4x ¯ Ëx ¯

(Tougher Problems for JEE Advanced)

1. Prove that sin–1(cos (sin–1 x)) + cos–1(sin (cos–1 x)) 2. Prove that tan -1{cosec (tan -1 x) - tan (cot -1 x)} =

1 tan -1 x 2

where x π 0 3. Prove that tan (tan–1 x + tan–1 y + tan–1 z) = cot (cot–1 x + cot–1 y + cot–1 z). 4. Prove that sin (cot–1(tan (cos–1 x))) = " x Œ (0, 1] 5. Prove that sin (cosec–1(cot (tan–1 x))) = x " x Œ (0, 1] 6. Find the value of sin–1(sin 5) + cos–1 (cos 10) + tan–1(tan (–6)) + cot–1(cot (1–10)) 7. Find the simplest value of Êx 3 - 3x 2 ˆ Ê1 ˆ cos -1 x + cos -1 Á + ˜¯ , " x ŒÁ , 1˜ Ë2 Ë2 ¯ 2 8. Find the value of Ê 5-2 6ˆ Ê 1 ˆ - tan -1 Á tan -1 Á ˜ ˜ Ë 2¯ Ë 1+ 6 ¯ 9. Let m = sin–1(a6 + 1) + cos–1(a4 + 1) – tan–1 (a2 + 1) Then find the image of the line x + y = m about the yaxis. (3p )3 -1 3 -1 3 -1 3 10. If (sin x) + (sin y ) + (sin z ) = 8 then find the value of (3x + 4y – 5z + 2) n 1ˆ Ê 11. Let S =  cot -1 Á 2r +1 + r ˜ Ë ¯ 2 r =1 Then find lim ( S ) n

12. Find the value of Ê Ê n Ê 4 ˆˆˆ lim Á tan Á  tan -1 Á 2 ˜˜ n Ë 4r + 3 ˜¯ ¯ ¯ Ë r =1 Ë 13. Find the number of solution of the equation Ê 2x ˆ 2 sin -1 Á = p x3 Ë 1 + x 2 ˜¯ Ê xˆ Ê yˆ 14. If cos -1 Á ˜ + cos -1 Á ˜ = a , then prove that, Ë a¯ Ë b¯ x2

-

2xy y2 cos a + 2 = sin 2a ab b

a2 15. If sin x + sin–1 y + sin–1 z = p, then prove that, –1

x 1 - x 2 + y 1 - y 2 + z 1 - z 2 = 2 xyz

TR_04.indd 18

19 Solve for x: tan–1(x – 1) + tan–1(x) + tan–1(x + 1) = tan–1(3x) p -1 Ê 1 ˆ -1 20. Solve for x: sin Á ˜ + cos x = Ë 5¯ 4 21. Solve for x: Ê x 2 - 1ˆ Ê 2x ˆ 2p cos -1 Á 2 ˜ + tan -1 Á 2 ˜ = Ë x - 1¯ 3 Ë x + 1¯ 22. Solve for x: 2 Ê 1 - a2 ˆ -1 Ê 1 - b ˆ 2 tan -1 x = cos -1 Á cos ˜ Á ˜, Ë 1 + a2 ¯ Ë 1 + b2 ¯

a > 0, b >0 23. Solve for x: cot–1 x + cot–1(n2 – x + 1) = cot–1(n – 1) 24. Solve for x: Ê x - 1ˆ Ê 2x - 1ˆ Ê 23 ˆ tan -1 Á + tan -1 Á = tan -1 Á ˜ Ë x + 1˜¯ Ë 2x + 1˜¯ Ë 36 ¯ 25. Solve for x: Ê xˆ Ê xˆ sec -1 Á ˜ - sec -1 Á ˜ = sec -1b - sec -1a Ë a¯ Ë b¯ 26. Find the sum of Ê

8n

ˆ

 tan -1 ÁË n4 - 2n2 + 5 ˜¯ n =1

27. Find the number of real solutions of the equation p sin -1 (e x ) + cos -1 ( x 2 ) = 2 28. Find the number of real roots of sin( x) = cos -1 (cos x ) in (0, 2p ) Ê 1ˆ Ê 1ˆ Ê 1ˆ Ê 1ˆ p 29. If tan -1 Á ˜ + tan -1 Á ˜ + tan -1 Á ˜ + tan -1 Á ˜ = Ë 3¯ Ë 4¯ Ë 5¯ Ë n¯ 4 where n Œ N, then find n 30. If a is the real root of x3 + bx2 + cx + 1 = 0 where b < c, then find the value of Ê 1ˆ tan -1 (a ) + tan -1 Á ˜ Ëa¯ 3 2 31. If the equation x + bx + cx + 1 = 0 has only one root a, then find the value of 2 tan–1 (cosec a) + tan–1 (2 sin a sec2 a) Q. Solve the following inequalities: 32. sin–1 x > cos–1 x 33. cos–1 x > sin–1 x

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4.19

Inverse Trigonometric Functions

34. 35. 36. 37.

(cos–1 x)2 – 5(cot–1 x) + 6 > 0 tan2(sin–1 x) > 1 4(tan–1 x)2 – 8(tan–1 x) + 3 < 0 4 cot–1 x – (cot–1 x)2 – 3 ≥ 0

Ê Ê 2x 2 + 4 ˆ ˆ -1 B 43. Prove that Ï Ï ¸ÔÔ¸ Ê 2- 3ˆ Ô Ô -1 Ê 12 ˆ -1 sin -1 Ìcot Ìsin -1 Á sec ( 2) + ˜ + cos Á ˝˝ ˜ ÁË 4 ˜¯ Ë 4 ¯ ÔÓ ÔÓ ÔÔ ˛˛ =0 44. Find the domain of the function f (x) = sin–1(cos–1 x + tan–1 x + cot–1 x)

53. Solve Ê 3 sin 2q ˆ 1 q = tan -1 (2 tan 2q ) - sin -1 Á 2 Ë 5 + 4 cos 2q ˜¯ 54. Simplify Ê x cos q ˆ Ê cos q ˆ - cot -1 Á tan -1 Á ˜ Ë 1 - x sin q ¯ Ë x - sin q ˜¯ 55. Solve Ê x 2 - 1ˆ Ê 2x ˆ Ê 2x ˆ 2p cos -1 Á 2 ˜ + sin -1 Á 2 ˜ + tan -1 Á 2 ˜ = Ë ¯ Ë x - 1¯ 3 x +1 Ë x + 1¯ 56. Prove that

Ê yˆ 2p -1 Ê x ˆ + sin -1 Á 1 - ˜ + tan -1 y = 45. If sin Á ˜ 4¯ 3 Ë 2 ¯ Ë

Ê xz ˆ Ê yz ˆ Ê xy ˆ p tan -1 Á ˜ + tan -1 Á ˜ + tan -1 Á ˜ = Ë xr ¯ Ë zr ¯ 2 . Ë yr ¯ 2 2 2 2 where x + y + z = r . 10 3 ˆ Ê mˆ -1 Ê = cot -1 Á ˜ 57. If  tan Á 2 ˜ Ë n¯ Ë 9r + 3r - 1¯ r =1

then find the maximum value of (x2 + y2 + 1) 46. Find the number of integral ordered pairs (x, y) satisfying the equation

58. If the sum

Ê 1ˆ Ê 1ˆ Ê 1ˆ tan -1 Á ˜ + tan -1 Á ˜ = tan -1 Á ˜ Ë x¯ Ë 10 ¯ Ë y¯ È Ê 10 ˆ˘ a 47. Let Ícot Á  cot -1 (k 2 + k + 1)˜ ˙ = ¯ ˚˙ b ÎÍ Ë k =1 where a and b are co-prime, then find the value of (a + b + 10). 48. If p > q > 0, pr < –1 < qr, then prove that Ê p-qˆ Ê q-r ˆ Ê r - pˆ tan -1 Á + tan -1 Á + tan -1 Á =p Ë 1 + pq ˜¯ Ë 1 + qr ˜¯ Ë 1 + rp ˜¯ 49. Consider the equation (sin–1 x)3 + (cos–1 x)3 = ap3 find the values of ‘a’ so that the given equation has a solution. Ê x2 ˆ 50. If the range of the function f ( x) = cot -1 Á 2 ˜ is Ë x + 1¯ Êb ˆ (a, b), find the value of Á + 2˜ Ëa ¯

TR_04.indd 19

Ê c(a + b + c) ˆ + tan -1 Á ˜ = p , where a, b, c > 0 ab Ë ¯

-1

f ( x) = 8sin x + 8cos x 41. Find the set of values of k for which x2 – kx + sin–1 (sin 4) > 0, for all real x 42. If

Ê Ê p ˆˆ 51. If tan–1 y = 4 tan–1 x, Á x < tan Á ˜ ˜ , find y as an Ë 8 ¯¯ Ë Êpˆ algebraic function of x and hence prove that tan Á ˜ Ë 8¯ is a root of the equation x4 – 6x2 + 1 = 0 52. Prove that Ê a (a + b + c) ˆ Ê b(a + b + c) ˆ tan -1 Á + tan -1 Á ˜ ˜ bc ac Ë ¯ Ë ¯

where m and n are co-prime, find the value of (2m + n + 4) 10 10

Ê aˆ

  tan -1 ÁË b ˜¯ = mp , then find the value of

b =1 a =1

(m + 4) 1 ( x + 1) (sin -1 x + cos -1 x + tan -1 x) + 2 p x + 2x + 10 such that the maximum value of f (x) is m, then find the value of (104 m – 90). 60. Let m be the number of solutions of sin (2x) + cos (2x) + cos x + 1 = 0 in p and 0< x< 2 È Ê Ê Ê 7p ˆ ˆ Ê 7p ˆ ˆ ˘ n = sin Í tan -1 Á tan Á ˜ ˜ + cos -1 Á cos Á ˜ ˜ ˙ Ë 6 ¯¯ Ë 3 ¯¯ ˚ Ë Ë Î 59. Let f (x) =

then find the value of (m2 + n2 + m + n + 4) n

61. Let f (n) =

Ê

Ê 1ˆ

ˆ

 ÁË cot -1 ÁË k ˜¯ - tan -1 (k )˜¯

k =-n 10

such that

 ( f (n) + f (n - 1)) = ap

n=2

then find the value of (a = 1)

2/10/2017 4:12:33 PM

4.20

Trigonometry Booster

Integer Type Questions

Comprehensive Link Passages

1. If the solution set of

In these questions, a passage (paragraph) has been given followed by questions based on each of the passage. You have to answer the questions based on the passage given.

Ê Ê 2x + 4 ˆ ˆ sin -1 Á sin Á 2 ˜ ˜ < p - 3 is Ë x + 1 ¯¯ Ë 2

(a, b), where a, b Œ I, then find (b – a + 5) 2. If a sin–1 x – b cos–1 x = c, such that the value of a sin–1 mp ab + c(a - b) x + b cos–1 x is , m Œ N , then find the a+b value of (m2 + m + 2) 3 If m is a root of x2 + 3x +1 = 0, such that the value kp -1 -1 Ê 1 ˆ , k Œ I , then find the of tan (m) + tan Á ˜ is Ë m¯ 2 value of (k + 4) 4. Find the number of real solutions of sin -1 ( x 2 - 2x + 1) + cos -1 ( x 2 - x) =

p 2

5. Let cos–1(x) + cos–1(2x) + cos–1(3x) = p If x satisfies the cubic equation ax3 + bx2 + cx + d = 0, then find the value of (b + c) – (a + d) 6. Consider a, b, g are the roots of x3 – x2 – 3x + 4 = 0 such that an–1 a + tan–1 b + tan–1 g = q p If the positive value of tan (q) is , where p and q are q natural numbers, then find the value of (p + q) 7. If M is the number of real solution of cos–1 x + cos–1 (2x) + p = 0 and N is the number of values of x satisfying Ê 5ˆ Ê 12 ˆ p the equation sin -1 Á ˜ + sin -1 Á ˜ = , then find the Ë x¯ Ë x¯ 2 value of M + N + 4 8. Find the value of È Ê1 ˆ Ê1 ˆ˘ 4 cos Ícos -1 Á ( 6 - 2)˜ - cos -1 Á ( 6 + 2)˜ ˙ Ë4 ¯ Ë4 ¯˚ Î 9. Find the value of Ê 5 ˆ 5 cot Á  cot -1 (k 2 + k + 1)˜ Ë k =1 ¯ p 10. Let 3 sin -1 (log 2 x) + cos -1 (log 2 y ) = 2 11p and sin -1 (log 2 x) + 2cos -1 (log 2 y ) = 6 Ê 1 ˆ 1 then find the value of Á 2 + 2 + 2˜ y Ëx ¯ 11. If a and b are the roots of x2 + 5x – 44 = 0, then find the value of cot (cot–1 a + cot–1 b) 12. If x and y are positive integers satisfying Ê 1ˆ Ê 1ˆ Ê 1ˆ tan -1 Á ˜ + tan -1 Á ˜ = tan -1 Á ˜ , then find the numË x¯ Ë 7¯ Ë y¯ ber of ordered pairs of (x, y)

TR_04.indd 20

Passage 1 Function

Domain

sin–1x

[–1, 1]

tan–1x

R

Co-domain È p 3p ˘ ÍÎ 2 , 2 ˙˚ Ê p 3p ˆ ÁË , ˜¯ 2 2

[–1, 1] cos–1x cot–1x R 1. sin–1 (–x) is (a) –sin–1 x (b) p + sin–1 x (c) 2p – sin–1 x

[p, 2p] [p, 2p]

-1 2 (d) 2p - cos 1 - x , x > 0

2. If f (x) = 3 sin–1 x – 2 cos–1 x, then f (x) is (a) even function (b) odd function (c) neither even nor odd (d) even as well as odd function. 3. The minimum value of (sin–1 x)3 – (cos–1 x)3 is (a) -

63p 3 8

(b)

63p 3 8

125p 3 125p 3 (d) 32 32 4. The value of sin–1 x + cos–1 x is p 3p 5p 7p (a) (b) (c) (d) 2 2 2 2 5 p 3 p È ˘ 5. If the co-domain of sin–1 x is Í , - ˙ such that Î 2 2 ˚ 5p -1 -1 , then the co-domain of cos–1 x is sin x + cos x = 2 (a) [4p, 5p] (b) [3p, 4p] (c) [6p, 7p] (d) [5p, 6p] (c)

Passage II We know that corresponding to every bijection function f: A Æ B, there exist a bijection. g: B Æ A defined by g(y) = x if and only if f (x) = y The function g: B Æ A is called the inverse of function f: A Æ B and is denoted by f–1. Thus, we have f (x) = y fi f–1(y) = x We know that trigonometric functions are periodic functions and hence, in general all trigonometric functions are not bijectives. Consequently, their inverse do not exist. However, if we restrict their domains and co-domains, they we can make the bijectives and also we can find their inverse.

2/10/2017 4:12:33 PM

4.21

Inverse Trigonometric Functions

Now, answer the following questions. 1. sin–1(sin q) = q, for all q belonging to È p p˘ (a) [0, p] (b) Í - , ˙ Î 2 2˚ È p ˘ (c) Í - , 0˙ (d) None of these Î 2 ˚ –1 2. cos (cos q) = q, for all q belonging to Ê p pˆ (a) [0, p] (b) Á - , ˜ Ë 2 2¯ p p È ˘ (c) Í - , ˙ (d) None of these Î 2 2˚ 3. tan–1(tan q) = q, for all q belonging to (a) [0, p] Ê p pˆ (c) ÁË - , ˜¯ 2 2

È p p˘ (b) Í - , ˙ - {0} Î 2 2˚ (d) None of these

4. cosec–1(cosec q) = q, for all q belonging to È p p˘ Ê p pˆ (a) Í - , ˙ (b) Á - , ˜ - {0} Ë 2 2¯ Î 2 2˚ (c) [0, p] (d) (0, p) –1 5. sec (sec q) = q, for all q belonging to p p (b) (0, p) – (a) [0, p] – 2 2 (c) (0, p) (d) None of these 6. sin–1(sin x) = x, for all x belonging to (a) R (b) j È p 3p ˘ (c) [–1,1] (d) Í , ˙ Î2 2 ˚ –1 –1 7. The value of sin (sin 2) + cos (cos 2) is p (a) 0 (b) 2 p (c) (d) None of these 2

{}

{}

Passage III Let f (x) = sin{cot–1(x + 1)} – cos (tan–1 x) and a = cos (tan–1(sin (cot–1 x))) and b = cos (2 cos–1 x + sin–1 x) 1. The value of x for which f (x) = 0 is (a) –1/2 (b) 0 (c) 1/2 (d) 1 2 2. If f (x) = 0, then a is equal to (a) 1/2 (b) 2/3 (c) 5/9 (d) 9/5 26 3. If a 2 = , then b2 is equal to 51 (a) 1/25 (b) 24/25 (c) 25/26 (d) 50/51 Passage IV Every bijective (one-one onto function) f: A Æ B there exists a bijection g: B Æ A is defined by g(y) = x if and only if f (x) = y.

TR_04.indd 21

The function g : B Æ A is called the inverse of function f : A Æ B and is denoted by f–1. If no branch of an inverse trigonometric function is mentioned, then it means the principal value branch of that function. 1. The value of cos{tan–1(tan 2)} is (a) 1/ 5 (b) –1/ 5 (c) cos 2 (d) –cos 2 2. If x takes negative permissible value then sin–1 x is (a) cos -1 ( 1 - x 2 )

(b) - cos -1 ( 1 - x 2 )

-1 2 (c) cos -1 ( x 2 - 1) (d) p - cos ( 1 - x ) . 1 3. If x + = 2 , then the value of sin–1 x is x p 3p p (a) (b) (c) p (d) 4 2 2

Passage V ap 2 …(i) 4 p2 and cos -1 x ◊ (sin -1 y ) 2 = …(ii) 16 where –1 £ x, y £ 1. Then 1. The set of values of ‘a’ for which the equation (i) holds good is 4ˆ È 4ˆ Ê (a) Á 0, 2 + ˜ (b) Í0, 1 + ˜ Ë p¯ p¯ Î

Let cos -1 x + (sin -1 y ) 2 =

4 (d) È0, -1 + ˜ˆ ÍÎ p¯ 2. The set of values of ‘a’ for which equations (i) and (ii) posses solutions (a) (– , 2] [2, ) (b) (–2, 2) 4˘ È (c) 2, 1 + (d) R ÍÎ p ˙˚ 3. The values of x and y, the system of equations (i) and (ii) posses solutions for integral values of ‘a’ Ï Ï ¸ Ê 2ˆ ¸ Ê 2ˆ (b) Ìcos Á p ˜ , - 1˝ (a) Ìcos Á p ˜ , 1˝ Ë ¯ Ë ¯ 4 4 Ó ˛ Ó ˛ (c) R

Êp2ˆ ÔÏ Ô¸ (c) Ìcos Á ˜ , ±1˝ Ë 4¯ ÔÓ Ô˛

(d) {(x, y): x ΠR, y ΠR}

Matrix Match (For JEE-Advanced Examination Only) Given below are matching type questions, with two columns (each having some items) each. Each item of column I has to be matched with the items of column II, by encircling the correct match(es). Note: An item of Column I can be matched with more than one items of Column II. All the items of Column II have to be matched.

2/10/2017 4:12:34 PM

4.22

Trigonometry Booster

1. Match the following columns: (A) (B) (C) (D)

Column I The principal value of sin–1 (sin 20) is The principal value of sin–1 (sin 10) is The principal value of cos–1 (cos 10) is The principal value of cos–1 (cos 20) is

5. Match the following columns: Column II (P) (20 – 6p)

Column I (A) The value of tan–1 + tan–1 2 + tan–1 3 is

(Q)

(3p – 10)

(B) The value of

(R)

(4p – 10)

(S)

(5p – 20)

2. Match the following columns: Column I Column II (A) The range of (P) (0, p) f (x) = 3 sin–1 x + 2 cos–1 x is (Q) È p 3p ˘ (B) The range of , f (x) = sin–1 x + cos–1 x + tan–1 ÎÍ 2 2 ˚˙ x is (C) The range of (R) Ê p 3p ˆ ÁË - , ˜ -1 f (x) = sin x + p is 2 2¯ (D) The range of f (x) = 2 tan -1 x + sin -1 x

(S) [0, p]

Ê 1ˆ Ê 1ˆ tan -1 1 + tan -1 Á ˜ + tan -1 Á ˜ Ë 2¯ Ë 3¯ is (C) The value of Ê 5ˆ tan -1 (9) + tan -1 Á ˜ is Ë 4¯ (D) The value of

(S)

-

p 2

6. Match the following columns: (A) (B)

(D)

is

(R) p

Ê 2x ˆ 2 tan -1 x - tan -1 Á , x >1 Ë 1 - x 2 ˜¯ is

(C)

Ê 1ˆ + sec -1 Á ˜ Ë x¯

Column II (P) 3p 4 p (Q) 2

Column I Column II 3 –1 3 (sin x) + (cos x) is maxi- (P) 1 x= mum at 2 –1

(sin–1 x)2 + (cos–1 x)2 is mini- (Q) mum at (sin–1 x) – (cos–1 x) is mini- (R) mum at (tan–1 x)2 + (cot–1 x)2 is mini- (S) mum at

x=1 x = –1 x=0

3. Match the following columns: Column I Column II –1 –1 (A) sin (sin x) = sin (sin x), if (P) –1 £ x £ 1 (B) cos (cos – 1 x) = cos x), if (C) tan (tan – 1 x) = tan x), if (D) cot (cot – 1 x) = cot x), if

–1

(cos (Q) 0 £ x £ 1

–1

(tan (R) 0 < x < p

–1

(cot (R)

-

p p £x£ 2 2

4. Match the following columns: Column I The value of (A) Ê 1ˆ Ê 1ˆ sin -1 Á ˜ + cos -1 Á - ˜ Ë 2¯ Ë 2¯ (B)

TR_04.indd 22

Column II (P)

Ê 1ˆ Ê 1ˆ sin -1 Á - ˜ + cos -1 Á ˜ Ë 2¯ Ë 2¯

(Q)

(C)

tan -1 ( 3) + cot -1 ( - 3)

(R)

(D)

Ê 1 ˆ Ê 1 ˆ (S) sin -1 Á + cos -1 Á ˜ Ë 2013 ¯ Ë 2013 ˜¯

7p 6 5p 6 p 6 p 2

Assertion and Reason Codes: (A) Both A and R are individually true and R is the correct explanation of A (B) Both A and R are individually true and R is not the correct explanation of A. (C) A is true but R is false. (D) A is false but R is true. 3p 1. Assertion (A): If sin -1 x + sin -1 y + sin -1 z = , then 2 the value of (x2013 + y2013 + z2013) -

9 (x

2014

+y

2014

+ z 2014 )

Reason (R): Maximum value of sin–1 x is (a) A

(b) B

(c) C

is zero.

p 2 (d) D

-1 -1 Ê 2x ˆ 2. Assertion (A): The value of 2 tan x - tan Á Ë 1 - x 2 ˜¯ is p Reason (R): x > 1 (a) A (b) B (c) C (d) D

2/10/2017 4:12:35 PM

4.23

Inverse Trigonometric Functions

3. Assertion (A): The value of

Questions Asked In Previous Years’ JEE-Advanced Examinations

1ˆ p tan (p ) + tan Á ˜ is Ë p¯ 2 -1

4.

5.

6.

7.

8.

9.

-1 Ê

Reason (R): P is the root of x2 + 2013x + 2014 = 0. (a) A (b) B (c) C (d) D Assertion (A): 2p If sin -1 x + sin -1 y = , then the value of cos–1 x + 3 p cos–1 y is 3 p Reason (R): sin -1 x + cos -1 x = , when x Œ [–1, 1] 2 (a) A (b) B (c) C (d) D Assertion (A): The value of cos–1 (cos 10) is (2p – 5) Reason (R): The range of cos–1 x is [0, p] (a) A (b) B (c) C (d) D Assertion (A): If cos–1 x + cos–1 y + cos–1 z = p, then x2 + y2 + z2 + 2xyz =1 Reason(R): For –1 £ x, y, z £ 1 (a) A (b) B (c) C (d) D Assertion (A): p 1 If tan -1 (2x) + tan -1 (3x) = , then x is 4 6 Reason (R): For 0 < 2x, 3x < 1 (a) A (b) B (c) C (d) D 2 Ê1 - x ˆ Assertion (A): cos Á = 2 tan -1 x 2˜ Ë1 + x ¯ Reason (R): for x ≥ 0 (a) A (b) B (c) C (d) D Assertion (A): sin–1 (3x – 4x3) = p – 3 sin–1 x 1 Reason (R): for < x £ 1 2 (a) A (b) B (c) C (d) D

10. Assertion (A): cos–1 (4x3 – 3x) = 2p – 3 cos–1 (x) 1 1 Reason (R): For - £ x < 2 2 (a) A (b) B (c) C (d) D 1 Ê ˆ 11. Assertion (A): cot -1 (x) = tan -1 Á ˜ , x > 0 Ë x¯ -1

-1 Ê 1 ˆ

Reason (R): cot (x) = p + tan Á ˜ , x < 0 Ë x¯ (a) A (b) B (c) C (d) D 12. Assertion (A): If a. b are the roots of x2 – 3x + 2 = 0, then sin–1 a exists but not sin–1 b, where a > b Reason (R): Domain of sin–1 x is [–1, 1] (a) A (b) B (c) C (d) D

TR_04.indd 23

1. Let a, b, c be positive real numbers such that q = tan -1 + tan -1

a(a + b + c) bc b( a + b + c ) c(a + b + c) + tan -1 ca ab

Then tan q is equal to? 2. The numerical value of

[IIT-JEE, 1981]

Ï Ê 1ˆ p ¸ tan –1 Ì2 tan -1 Á ˜ - ˝ is ? Ë 5¯ 4 ˛ Ó

[IIT-JEE, 1981] 3. Find the value of cos (2 cos–1 x + sin–1 x) at x = 1/5, p p -1 where 0 £ cos–1 x £ p and – £ sin x £ 2 2 [IIT-JEE, 1981] È Ê 4ˆ Ê 2ˆ ˘ 4. The value of tan Ícos -1 Á ˜ + tan -1 Á ˜ ˙ is Ë ¯ Ë 3¯ ˚ 5 Î (a) 6/17 (b) 17/6 (c) –17/6 (d) –6/17 [IIT-JEE, 1983] 5. No questions asked between 1984 and 1985. Ê Ê 2p ˆ ˆ 6. The principal value of sin -1 Á sin Á ˜ ˜ is Ë 3 ¯¯ Ë p 2p 2p 5p (a) (b) (c) (d) 3 3 3 3 [IIT-JEE, 1986] 7. No questions asked between 1987 and 1988. 8. The greater of the two angles A = 2 tan -1 (2 2 - 1)

9. 10.

11. 12.

Ê 1ˆ Ê 3ˆ and B = 3 sin -1 Á ˜ + sin -1 Á ˜ is....... Ë 3¯ Ë 5¯ [IIT-JEE, 1989] No questions asked between 1990 and 1998. The number of real solutions of p tan -1 x( x + 1) + sin -1 x 2 + x + 1 = is 2 (a) 0 (b) 1 (c) 2 (d) [IIT-JEE, 1999] No questions asked in 2000. ˆ x 2 x3 -1 Ê + - .......˜ If sin ÁË x ¯ 2 4 Ê ˆ p x 4 x6 + cos -1 Á x 2 + - ....˜ = , Ë ¯ 2 2 4 for 0 < | x| < 2 , then x is (a) 1/2 (b) 1 (c) –1/2

13. Prove that cos (tan–1(sin (cot–1 x))) =

(d) –1 [IIT-JEE, 2001] x2 + 1 x2 + 2 [IIT-JEE, 2002]

2/10/2017 4:12:35 PM

4.24

Trigonometry Booster

p is 6 Ê 1 1˘ È 1 3ˆ (a) Á - , ˙ (b) Í - , ˜ Ë 2 2˚ Î 4 4¯ È 1 1˘ È 1 1˘ (c) Í - , ˙ (d) Í - , ˙ Î 4 4˚ Î 4 2˚ [IIT-JEE, 2003] 15. If sin (cot–1(x + 1)) = cos (tan–1 x), then the value of x is (a) –1/2 (b) 1/2 (c) 0 (d) 9/4 [IIT-JEE, 2004] 16. No questions asked in between 2005 to 2006. 17. Match the following columns: Let (x, y) be such that p sin -1 (ax) + cos -1 ( y ) + cos -1 (bxy ) = 2 -1 14. The domain of f (x) = sin (2x) +

(A) (B) (C) (D)

Column I If a = 1 and b = 0, then (x, y) If a = 1 and b = 1, then (x, y) If a = 1 and b = 2, then (x, y) If a = 2 and b = 2, then (x, y)

18. If 0 < x < 1, then

Column II (P) lies on the circle x2 + y2 = 1 (Q) lies on (x2 – 1)(y2 – 1) = 0 (R) lies on the line y = x (S) lies on (4x2 – 1)(y2 – 1) = 0 [IIT-JEE, 2007]

1+ x 2 ¥ [{x cos (cot -1 x) + sin (cot -1 x)}2 - 1]1/2 equals x (a) (b) x 1 + x2

(c) x 1 + x 2

1 + x2 [IIT-JEE, 2008] 19. No questions asked between 2009 to 2010. (d)

Ê Ê sin q ˆ ˆ 20. Let f (q ) = sin Á tan -1 Á ˜˜ Ë cos 2q ¯ ¯ Ë d p p ( f (q )) where - < q < . Then the value of d (tan q ) 4 4 is....... [IIT-JEE, 2011] 21 No questions asked in 2012. n Ê 23 Ê ˆˆ –1 22 The value of cot Á  cot Á1 +  2k ˜ ˜ is ÁË n =1 Ë k =1 ¯ ˜¯ (a) 23/25

(b) 25/23

(c) 23/24

(d) 24/23 [IIT-JEE, 2013]

23. The value of Ê 1 Ê cos (tan -1 y ) + y sin (tan -1 y ) ˆ 2 ˆ 4 y + Á 2Á ˜ ˜ -1 -1 Ë y Ë cot (sin y ) + tan (sin y ) ¯ ¯ is.........

1/2

[IIT-JEE, 2013] -1

-1

24 If cot (sin 1 - x ) = sin (tan ( x 6)), x π 0 then the value of x is....... [IIT-JEE, 2013] 25. The number of positive solutions satisfying the equation Ê 1 ˆ Ê 1 ˆ Ê 2ˆ tan -1 Á + tan -1 Á = tan -1 Á 2 ˜ ˜ ˜ Ë Ë 2x + 1¯ Ë 4x + 1¯ x ¯ 2

is.... 26. No questions asked in 2015.

[IIT-JEE, 2014]

A NSWERS

LEVEL II 1. 6. 11. 16. 21. 26. 31. 36. 41.

(d) (b) (c) (b) (a) (b) (b) (c) (b)

2. 7. 12. 17. 22. 27. 32. 37. 42.

(b) (b) (c) (b) (a) (b) (a) (c) (a)

3. 8. 13. 18. 23. 28. 33. 38. 43.

(b) (d) (a, c) (c) (b) (a) (a) (b) (a)

4. 9. 14. 19. 24. 29. 34. 39.

(b) (a) (a) (c) (c, d) (a) (c) (c)

5. 10. 15. 20. 25. 30. 35. 40.

(d) (c) (d) (b) (a) (c) (b) (a)

17 18. 19. 20. 21.

LEVEL III 6. (8p – 21) p 7. 3

TR_04.indd 24

7p 3 , when x = –1 8 1 p3 and minimum Value = , when x = 2 32 1 3 x= 2 7 x=3 1 -1 x = 0, , 2 2 3 x= 10 x = 2 - 3, 3 a-b x= 1 + ab x = n2 – n + 1, n 4 x= 3 x = ab

16. Maximum Value

22. 23. 24. 25.

2/10/2017 4:12:36 PM

4.25

Inverse Trigonometric Functions

1 26. x = , y = 1 2 27. x = 1, y = 2 ; x = 2, y = 7 29. –p Ê 1 ˘ 32. x Œ Á ,1 Ë 2 ˙˚ 1 ˘ Ê 33. x Œ Á -1, Ë 2 ˙˚ 34. x Œ (– , cot 3) (cot 2, ) 1 ˆ Ê 1 ˆ Ê 35. x Œ Á ,1 -1, ˜ Ë 2 ˜¯ ÁË 2¯ Ê 1ˆ Ê 3ˆ 36. tan Á ˜ < x < tan Á ˜ Ë 2¯ Ë 2¯ 37. cot (3) £ x £ cot (1) 38. x Œ (–1, 1) 39. n = 5 40. n =8 41. k = f 44. tan (sin (cos (sin 1))) £ x < tan (sin (1)). 45. [1, ) È 1 7˘ 49. a Œ Í , ˙ Î 32 8 ˚ 50. 53. 54. 55. 56. 57. 58. 59. 60. 61.

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11, 12.

LEVEL I 1. Given, f (x) = 3 x + 5 fi f ¢(x) = 3 > 0 fi f is strictly increasing function. fi f is one one function Also, Rf = R = Co-domain fi f is onto function. Thus, f is a bijective function. Hence, f –1 is exists. Let y = 3x + 5 y-5 fi x= 3

7 4 3 x=1 3 9 5, where M = 0, N = 1 2 7 8 9 6

COMPREHENSIVE LINK PASSAGES

Passage I:

1. 5. Passage II: 1. 5. Passage III: 1. Passage IV: 1. Passage V: 1.

(c) (a) (b) (a) (a) (d) (b)

2. (b)

3. (a)

4. (c)

2. 6. 2. 2. 2.

3. 7. 3. 3. 3.

4. (a)

(a) (c) (c) (b) (c)

(c) (a) (b) (b) (c)

MATRIX MATCH

x = y = a2 + 1 q Œ np + tan–1 (–2), n Œ Z q 1 , x= 3 p 2 36 29 4 6 100

1.(A) Æ (P); (B) Æ (Q); (C) Æ (R); (D) Æ (P) 2.(A) Æ (S); (B) Æ (P); (C) Æ (Q); (D) Æ (R) 3.(A) Æ (P); (B) Æ (Q); (C) Æ (S); (D) Æ (R) 4.(A) Æ (Q); (B) Æ (R); (C) Æ (P); (D) Æ (S) 5.(A) Æ (R); (B) Æ (Q); (C) Æ (P); (D) Æ (R) 6.(A) Æ (R); (B) Æ (P); (C) Æ (Q); (D) Æ (Q) ASSERTION AND REASON

1. (a) 6. (a) 11. (b)

H INTS

TR_04.indd 25

INTEGER TYPE QUESTIONS

AND

2. (a) 7. (a) 12. (a)

3. (a) 8. (a)

4. (a) 9. (a)

5. (a) 10. (a)

S OLUTIONS x-5 . 3 2. Given, f (x) = x2 + 2 fi f ¢(x) = 2x > 0 for every x > 0 fi f is strictly increasing function. fi f is one one function. Also, Rf = (2, ) = Co-domain fi f is ont function. Thus, f is a bijective function. Therefore, the inverse of the given function exists. Let y = x2 + 2 fi x2 = y – 2 fi x= y-2 Thus, f –1 ( x) =

2/10/2017 4:12:36 PM

4.26

Trigonometry Booster

So its inverse is exists. Let y = 2x (x + 1)

–1 Hence, f ( x) = x - 2

3. Given, f (x) =

x2 x2 + 1

X¢

Y

O

X

Y¢

x=

x=

1 ± 1 + 4 log 2 ( y ) 2

fi

x=

1 + 1 + 4 log 2 ( y ) 2

fi fi fi fi fi

x=

Ê1 + yˆ 1 log10 Á 2 Ë 1 - y ˜¯

1 Ê1 + xˆ log10 Á . 2 Ë 1 - x ˜¯ 6. Given, f (x) = x + sin x fi f ¢(x) = 1 + cos x ≥ 0 for all x in R. fi f is strictly increasing function fi f is a one one function. Also, the range of a function is R. fi f is a onto function Thus, f is a bijective function. Hence, f–1 exists. Therefore, f–1 (x) = x – sin x 7. Given, f (x) = x2 – 4x + 9 Thus, f –1 ( x) =

x . 1- x

4. Given, f (x) = 2x(x – 1). Y

O

10 x - 10 - x 102x - 1 = . 10 x + 10 - x 102x + 1 y × 102x + y = 102x – 1 102x(y – 1) = – y – 1 y +1 y +1 102x = = y -1 1- y Ê y + 1ˆ 2x = log10 Á Ë 1 - y ˜¯

Let y =

fi Rf = (0, 1) = Co-domain fi f is onto function. Thus, f is a bijective function. fi f –1(x) is exists.

X¢

fi

1 + 1 + 4 log 2 (x) 2 5. Since f is a bijective function, so its inverse exists.

y (1 - y )

Hence, f -1 ( x) =

y = 2x - x x2 – x = log2(y) x2 – x – log2(y) = 0

Thus, f -1 (x) =

1 f (x) = 1 - 2 fi x +1 2x > 0, " x Œ R + f ¢ (x) = 2 fi 2 ( x + 1) fi f is strictly increasing function. fi f is a one one function. x2 Also, let y = 2 x +1 2 fi y ◊ x + y = x2 fi x2(y – 1) = –y y y fi x2 = = ( y – 1) (1 - y ) fi

2

fi fi fi

X

Y¢

fi f ¢(x) = 2 ¥ (2x – 1) ¥ log, 2 > 0 for all x in [1, ) fi f is strictly increasing function. fi f is a one one function. Also, Rt = [1, ) fi Rf = [1, )= Co-domain fi f is onto function. Thus, f is a bijective function.

Y

X¢

O

X

x(x – 1)

TR_04.indd 26

Y¢

fi f ¢(x) = 2x – 4 ≥ 0 for all x in Df fi f is strictly increasing function. fi f is a one one function. Also, Rf = [5, ) = Co-domain fi f is onto function.

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4.27

Inverse Trigonometric Functions

Therefore, f is a bijective function. Hence, its inverse is exists. Let y = x2 – 4x + 3 fi x2 – 4x + (5 – g) = 9 fi

x=

4 ± 16 – 4(5 – y ) 2

fi

x=

4 ± 4y + 16 – 20 2

fi

x=

4 ± 4( y – 1) = 2 ± ( y – 1) 2

fi

x = 2 + ( y - 1) , since x ≥ 2

fi

f –1 ( x) = 2 + ( x - 1)

8. Consider the function È 1 ˆ ÍÎ 4 , ˜¯ , where 1 f (x) = x 2 4 Clearly, f is a one one and onto function. So its inverse is exists. 1 Let its inverse be f -1 : È , ˆ˜ [0, ) . ÍÎ 4 ¯ f :[0, )

1 . 4 Consequently, we can say that, the two sides of the given equation are inverse to each other. Thus, the intersection point is the solution of the given equation. f (x) = x 1 x2 - = x fi 4 1 fi x2 - x = 4 2 1ˆ 1 Ê fi ÁË x – ˜¯ = 2 2

fi

f -1 (x) = x +

1ˆ 1 Ê ÁË x – ˜¯ = ± 2 2 1 1 fi x= ± 2 2 Hence, the solutions are fi

1 1 1 ¸ Ï1 , Ì + ˝ 2 2 2 2˛ Ó 9. Clearly, f is bijective So, its inverse exists Let y = 3x + 5 y-5 fi x= 3 x-5 Thus, f –1 (x) = 3

TR_04.indd 27

10. Clearly, f is bijective So, its inverse exists x Let y = x -1 fi xy – y = x fi x(y – 1) = y y fi x= ( y – 1) x Thus, f –1 (x) = ( x – 1) 11. Clearly, f is bijective So its inverse exists Let y = x2 + 1 fi

x = y -1 Thus, f –1 (x) = x - 1 12. Since f is bijective, so its inverse exists 2 x - 2- x Let y = x 2 + 2- x fi

y 2 x - 2- x = 1 2 x + 2- x

fi

y + 1 2 x - 2- x + 2 x + 2- x = y - 1 2 x - 2- x - 2 x - 2- x

fi fi

y +1 2.2 x = - -x y -1 2.2 y +1 22x = 1– y

fi

Ê y + 1ˆ 2x = log 2 Á Ë 1 – y ˜¯

fi

x=

1 Ê y + 1ˆ log 2 Á 2 Ë 1 – y ˜¯

Thus, f –1 (x) =

1 Ê x + 1ˆ log 2 Á 2 Ë 1 – x ˜¯

13. Clearly, f is bijective. So, its inverse exists x Let y = 2 x +1 2 fi yx – x + y = 0 fi

x=

1 ± 1 – 4y 2 2y

2 Thus, f –1 (x) = 1 + 1 – 4x 2x

14. We have –1 £ 3x + 5 £ 1 fi –6 £ 3x £ –1 4 -2 £ x £ fi 3

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4.28

Trigonometry Booster

4˘ È D f = x Œ Í - 2, - ˙ 3˚ Î 15. We have x -1 £ £1 x +1 x £1 Case I: When x +1 x fi -1£ 0 x +1 –1 fi £0 x +1 1 fi ≥0 x +1 fi x > –1 fi

x Case II: When ≥ -1 x +1 x fi +1≥ 0 x +1 2x + 1 fi ≥0 x +1 È 1 ˆ fi x ( , 1) Í , ˜ Î 2 ¯ 1 Hence, D f ÈÍ , ˆ˜ . Î 2 ¯ 16. We have x2 + 1 -1 £ £1 2x x2 + 1 fi £1 2x | x 2 + 1| fi £1 |2x| | x 2 + 1| £1 2 | x| fi x2 + 1 £ 2 |x| fi |x|2 – 2 |x| + 1 £ 0 fi (|x| – 1)2 £ 0 fi (|x| – 1)2 = 0 fi (|x| – 1) = 0 fi |x| = 1 fi x = ±1 Hence, Df = {–1, 1} fi

| x| - 1 £1 2 fi –2 £ |x| – 1 £ 2 fi –1 £ |x| £ 3 fi |x| £ 3 ({ |x| ³ –1 is rejected) fi –3 £ x £ 3 Hence, Df [–3, 3] 18. We have –1 £ (log2 x) £ 1 fi 2–1 £ x £ 21 17. We have -1 £

TR_04.indd 28

1 £x£2 2 È1 ˘ Hence, D f = Í , 2˙ Î2 ˚ 19. We have –1 £ log3 x2 £ 1 fi

fi fi fi fi fi fi

4–1 £ x2 £ 41 1 £ x2 £ 4 4 1 £ | x| £ 2 2 1 | x| £ 2 and | x| ≥ 2 1 1 –2 £ x £ 2 and x ≥ and x £ 2 2 1˘ È1 ˘ È x Í 2, ,2 Î 2 ˚˙ ÎÍ 2 ˚˙

1˘ È1 ˘ È ÍÎ 2, 2 ˙˚ ÍÎ 2 , 2˙˚ 20. Given, sin–1 x + sin–1 y = p It is possible only when each term of the given equation provides the maximum value. p p Thus, sin -1 x = and sin -1 y = 2 2 Êpˆ Êpˆ fi x = sin Á ˜ = 1 and y = sin Á ˜ = 1 Ë 2¯ Ë 2¯ Hence, the solutions are x = 1 and y = 1. 3p 21. Given sin -1 x + sin -1 y + sin -1 z = 2 It is possible only when each term will provide us the maximum value. p p Thus, sin -1 x = , sin -1 y = 2 2 p and sin -1 z = 2 fi x = 1, y = 1 and z = 1 Hence, the value of 9 x 2013 + y 2013 + z 2013 - 2014 2014 x +y + z 2014 9 =1+1+1– 1+1+1 =3–3 =0 p p 22. We have – £ sin -1 (3x + 5) £ 2 2 fi –p £ 2 sin–1 (3x + 5) £ p p p p fi –p + £ 2 sin -1 (3x + 5) + £ p + 4 4 4 3p 5p fi £ f ( x) £ 4 4 È 3p 5p ˘ Hence, R f = Í , Î 4 4 ˙˚ Hence, D f

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4.29

Inverse Trigonometric Functions

23. We have sin–1 x > sin–1 (3x – 1) fi x > (3x – 1) fi 2x – 1 < 0 1 fi x< 2 1ˆ È fi x Œ Í -1, ˜ 2¯ Î 24. We have –1 £ 2x + 4 £ 1 fi –5 £ 2x £ –3 5 3 - £x£fi 2 2 È 5 3˘ Hence, D f = Í - , - ˙ Î 2 2˚ 25. We have 0 £ cos–1 (3x + 5) £ p fi 0 £ 2 cos–1 (3x + 5) £ 2p p p p fi £ 2 cos -1 (3x + 5) + £ 2p + 4 4 4 È p 9p ˘ Hence, R f = Í , Î 4 4 ˙˚ p 26. We have £ cos -1 ( - x 2 ) £ p 2 3p fi £ 3cos -1 ( - x 2 ) £ 3p 2 p p 3p p fi - £ 3 cos -1 ( - x 2 ) - £ 3p – 2 2 2 2 5p fi p £ f ( x) £ 2 È 5p ˘ Hence, R f = Íp , . Î 2 ˙˚ 27. Given, cos–1 x + cos–1 x2 = 0 It is possible only when each term will provide us the minimum value. So, cos–1 x + cos–1 x2 = 0 fi x = 1 and x2 = 1 fi x = 1 and x = ±1 Hence, the solution is x = 1. 28. Given, [sin–1 x] + [cos–1 x] and x ≥ 0 fi [sin–1 x] = 0 and [cos–1 x] = 0 fi x Œ [0, sin 1) & x Œ (cos 1, 1] fi x Œ (cos 1, sin 1) x2 £1 29. We have -1 £ 2 x +1 fi

x2 £1 x +1

fi

|x2 | £1 | x 2 + 1|

fi fi

TR_04.indd 29

2

x2 £1 x2 + 1 x2 + 1 ≥ x2

30.

31.

32.

33.

fi 1>0 Hence, x Œ R We have cos–1 (x) > cos–1 (x2) fi x < x2 fi x2 – x > 0 fi x(x– 1) > 0 fi x Œ [–1, 0) Since tan–1 x is defined for all real values of x, so 9 – x2 £ 0 fi x2 – 9 £ 0 fi (x + 3)(x – 3) £ 0 fi –3 £ x £ 3 Hence, Df = [–3, 3] p p We have – £ tan -1 (1 - x 2 ) £ 2 4 p fi – p £ 2 tan -1 (1 - x 2 ) £ 2 p p p p fi – p + £ 2 tan -1 (1 - x 2 ) + £ + 6 6 2 6 5p 2p fi £ f ( x) £ – 6 3 p p˘ 5 2 È Hence, R f = Í , Î 6 3 ˙˚ We have f (x) = cot–1 (2x – x2) fi f (x) = cot–1 (1 – (x2 – 2x + 1)) fi f (x) = cot–1 (1 – (x – 1)2) Since (1 – (x –)2) £ 1 and 0 £ cot–1 x £ p and cot–1 x is strictly decreasing function so, cot–1 (1) £ cot–1 (1 – (x – 1)2) £ cot–1 (0) p fi £ f (x) £ p 4 p Hence, R f = È , p ˘ ÍÎ 4 ˙˚ where [,] = GIF

34. We have [cot–1 x] + [cos–1 x] = 0 fi [cot–1 x] = 0 & [cos–1 x] = 0 fi 0 £ cot–1 x < 1 & 0 £ cos–1 x < 1 fi x Œ (cot 1, ) & x Œ (cos 1, 1] fi x Œ (cot 1, 1] 35. We have sin {x] = cos {x}, " x Œ [0, 2p] fi tan {x} = 1 p fi {x} = tan -1 (1) = 4 Hence, the number of solutions = 6 (Since {x} is a periodic function with period 1, it has one solution between 0 to 1. So, there is six solutions between 0 to 6.28). 36. We have, Ê | x| - 2 ˆ -1 £ Á £1 Ë 3 ˜¯

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4.30

Trigonometry Booster

fi fi fi

–3 £ |x| – 2 £ 3 –1 £ |x| £ 5 –5 £ x £ 5 Ê 1 – | x| ˆ £1 Also, -1 £ Á Ë 4 ˜¯ fi –4 £ 1 – |x| £ 4 fi –4 £ |x| – 1 £ 4 fi –3 £ |x| £ 5 fi –5 £ x £ 5 From (i) and (ii), we get –5 £ x £ 5 Thus, Df = [–5, 5] 37. Given, f (x) = sin–1 (2x2 – 1) So, –1 £ (2x2 – 1) £ 1 fi 0 £ 2x2 £ 2 fi 0 £ x2 £ 1 fi 0 £ |x| £ 1 fi –1 £ x £ 1 Thus, Df = [–1, 1] -1 -1 2 38. Given, f (x) = 5p sin x - 6(sin x) We have 5p sin–1 x – 6(sin–1 x) ≥ 0 fi (5p – 6(sin–1 x)) sin–1 x ≥ 0 fi (6(sin–1 x) – 5p) sin–1 x £ 0 5p fi 0 £ sin -1 x £ 6 Ê 5p ˆ fi 0 £ x £ sin Á ˜ Ë 6¯ 1 fi 0£ x£ 2 Also, –1 £ x £ 1

È 1˘ Thus, D f = Í0, ˙ Î 2˚

Ê 3 tan -1 x + p ˆ 39. Given, f ( x) = log 2 Á ˜ Ë p - 4 tan -1 x ¯ -1 3 tan x + p >0 So, p - 4 tan -1 x fi fi fi fi

3 tan -1 x + p 0 Furthermore, –1 £ logx 2 £ 1 which is also true for x π 1 and x > 0 Hence, Df = (0, 1) (1, ) 50. Given, f (x) = sin -1 (log 2 x) We have sin–1 (log2 x) ≥ 0 fi (log2 x) ≥ 0 fi x≥1 Also, –1 £ log2 x £ 1 1 fi £x£2 2 Hence, Df = [1, 2] 51. Given, f (x) = sin–1 (2x – 3) È p p˘ R f = Í- , ˙ Î 2 2˚ 52. Given, f (x) = 2 sin -1 (2x - 1) -

p p £ sin -1 (2x - 1) £ 2 2

p 4

–p £ 2 sin–1 (2x – 1) £ p p p p - p - £ 2 sin -1 (2x - 1) - £ p – 4 4 4 5p Ê p ˆ 3p -1 £ Á 2 sin (2x - 1) - ˜ £ 4 Ë 4¯ 4 -

5p 3p £ f (x) £ 4 4

È 5p 3p ˘ So, R f = Í , Î 4 4 ˙˚ 53. Given, f (x) = 2 cos–1 (–x2) – p = 2(p – cos–1 (x2)) – p = p – 2 cos–1 (x2) p 1 54. Given, f (x) = tan -1 (1 - x 2 ) 2 4 Now, – < (1 – x2) £ 1 fi tan–1(– ) < tan–1(1 – x2) £ tan–1(1) p p fi - < tan -1 (1 - x 2 ) £ 2 4

2/10/2017 4:12:40 PM

4.32

Trigonometry Booster

p 1 p < tan -1 (1 - x 2 ) £ 4 2 8 p p 1 p p p fi - - < tan -1 (1 - x 2 ) - £ 4 4 2 4 8 4 p p fi - < f ( x) £ 2 8 Ê p p˘ So, R f = Á - , - ˙ Ë 2 8˚ –1 55. Given, f (x) = cot (2x – x2) = cot–1 (1 – (x – 1)2) Clearly, – < (1 – (x – 1)2) £ 1 fi cot–1 (1) £ cot–1 ((1 – (x – 1)2)) £ cot–1 (– ) p fi £ cot -1 ((1 - ( x - 1) 2 )) £ p 4 p fi £ f ( x) £ p 4 Èp ˘ So, R f = Í , p ˙ Î4 ˚ 56. Given, f (x) = sin–1 x + cos–1 x + tan–1 x Df = [–1, 1] So, Rf = [f (–1), f (1)] Èp p p p ˘ =Í - , + ˙ Î2 4 2 4˚ È p 3p ˘ =Í , Î 4 4 ˙˚ fi

57. Given, f (x) = sin–1 x + sec–1 x + tan–1 x Thus, Df = {–1, 1} So, Rf = {f (–1), f (1)}

{ {

p p p p +0- , + 2 4 2 4 3p 3p , = 4 4 = -

}

}

58. Given, f (x) = 3cot -1 x + 2tan -1 x +

p 4

= 2(tan -1 x + cot -1 x) + cot -1 x + p p + cot -1 x + 2 4 5p -1 = cot x + 4 Thus, 0 £ cot–1 x £ p 5p 5p 5p fi £ cot -1 x + £p + 0+ 4 4 4 5p 9p fi £ f ( x) £ 4 4 5p 9p ˘ È So, R f = Í , Î 4 4 ˙˚ 59. Given, f (x) = cosec-1 [1 + sin2 x]. Clearly, 1 £ (1 + sin2 x) £ 2 Rf = [cosec–1 (2), cosec–1 (1)] =2¥

TR_04.indd 32

È ˘ Ê 1ˆ = Ísin -1 Á ˜ , sin -1 (1) ˙ Ë ¯ 2 Î ˚ Èp p ˘ =Í , ˙ Î6 2˚

-

p 4

60. Given, f (x) = sin–1 (log2 (x2 + 3x + 4)) Clearly, Df = [–2, –1] Thus, Rf = [f (–2), f (–1)] p Èp p ˘ =Í , ˙= Î2 2˚ 2 –1 61. We have, f (x) = sin x + cos–1 x + tan–1 x is defined only when –1 £ x £ 1 Now, f (1) = sin–1(1) + cos–1(1) + tan–1(1) p p 3p = +0+ = 2 4 4 –1 and f (–1) = sin (–1) + cos–1(–1) + tan–1(–1) p p 3p p =- +p - =p = 2 4 4 4 È p 3p ˘ Thus, R f = Í , Î 4 4 ˙˚ 62. We have 4 sin–1(x –1) + cos–1(x – 2) = p p fi 3 sin -1 ( x - 2) + = p 2 p 1 fi 3 sin ( x - 2) = 2 p fi sin -1 ( x - 2) = 6 Êpˆ 1 fi ( x - 2) = sin Á ˜ = Ë 6¯ 2 1 5 x=2+ = fi 2 2 5 Hence, the solution is x = 2 p -1 63. As we know that, if sin ( f ( x)) + cos -1 ( g ( x)) = , 2 then

{}

f (x) = g(x) fi (x2 – 2x + 1) = (x2 – x) fi 2x – x = 1 fi x=1 Hence, the solution is x = 1 64. We have tan -1 x( x + 1) + sin -1 x 2 + x + 1 =

p 2

1 p Ê ˆ + sin -1 x 2 + x + 1 = Á 2 ˜ 2 Ë x + x + 1¯

fi

cos –1

fi

1 Ê ˆ = x2 + x + 1 Á 2 ˜ x + x + 1 Ë ¯

fi fi fi

x2 + x + 1 = 1 x2 + x = 0 x(x + 1) = 0

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4.33

Inverse Trigonometric Functions

fi x = 0 and –1 Hence, the number of solutions is 2 p 65. As we know that, if sin -1 ( f ( x)) + cos -1 ( g ( x)) = , 2 then f (x) = g(x) fi fi

Ê ˆ Ê ˆ x 2 x3 x 4 x6 + - ...˜ = Á x 2 + - ...˜ ÁË x ¯ Ë ¯ 2 4 2 4 2 2 4 Ê ˆ Ê ˆ x x x x x Á1 - + - ...˜ = x 2 Á1 + - ...˜ Ë ¯ Ë ¯ 2 4 2 4

Ê 1 ˆ Ê 1 ˆ = x2 Á Á x˜ x2 ˜ ÁË 1 + ˜¯ 1+ ˜ Á 2 Ë 2¯ 2 ˆ Ê Ê 2x ˆ 2x fi ÁË x + 2 ˜¯ = Á x 2 + 2 ˜ Ë ¯ ÏÊ 1 ˆ Ê x ˆ ¸ x ÌÁ fi ˜ -Á 2 ˜˝ = 0 ÓË x + 2 ¯ Ë x + 2 ¯ ˛ Ê 1 ˆ Ê x ˆ x = 0 and Á = fi Ë x + 2 ˜¯ ÁË x 2 + 2 ˜¯ fi x = and x = 1 66. We have sin–1 x > cos–1 x fi 2 sin–1 x > sin–1 x + cos–1 x p fi 2 sin -1 x > 2 p fi sin -1 x > 4 Êpˆ fi x > sin Á ˜ Ë 4¯ 1 x> fi 2 fi

x

Ê 1 ˘ x ŒÁ ,1 Ë 2 ˙˚ 67. (sin–1 x)2 – 3 sin–1 x + 2 = 0 fi (sin–1 x – 1)(sin–1 x – 2) = 0 fi (sin–1 x – 1) = 0, (sin–1 x – 2) = 0 fi sin–1 x = 1, 2 fi sin–1 x = 1 fi x sin (1) 68. Given equation is sin–1x + sin–12y = p. It is possible only when p p fi sin -1 x = , sin -1 (2y ) = 2 2 fi x = 1, 2y = 1 1 fi x = 1, y = 2 69. Given equation is cos–1 x + cos–1 x2 = 2p. It is possible only when fi cos21 x = p, cos–1 (x2) = p fi x = –1, x2 = –1 fi x=j fi

TR_04.indd 33

70. Given equation is cos–1 x + cos–1 x2 = 0 It is possible only when fi cos–1 x = 0, cos–1 (x2) = 0 fi x = 1 and x2 = 1 fi x=1 71. Given equation is 4 sin–1 (x –1) + cos–1 (x – 1) = p p fi 3 sin -1 ( x - 1) + = p 2 p 1 fi 3 sin ( x - 1) = 2 p 1 fi sin ( x - 1) = 6 Êpˆ 1 fi ( x - 1) = sin Á ˜ = Ë 6¯ 2 3 fi x= 2 3 Hence, the solution is x = 2 72. Given equation is p Ê 1 ˆ cot -1 Á 2 ˜ + tan -1 ( x 2 - 1) = 2 Ë x - 1¯ It is possible only when 1 = x2 - 1 fi x2 - 1 fi (x2 –1)2 = 1 fi (x2 – 1) = ± 1 fi x2 = 1 ± 1 = 2, 0 x = {- 2, 0, 2} fi 73. Given equation is Ê x 2 - 1ˆ Ê 2x ˆ 2p cot -1 Á + tan -1 Á 2 ˜ = Ë 2x ˜¯ 3 Ë x - 1¯ fi

Ê 2x ˆ Ê 2x ˆ 2p tan -1 Á 2 ˜ + tan -1 Á 2 ˜ = 3 Ë x - 1¯ Ë x - 1¯

fi

Ê 2x ˆ 2p 2 tan -1 Á 2 ˜ = 3 Ë x - 1¯

fi

Ê 2x ˆ p tan -1 Á 2 ˜ = Ë x - 1¯ 3

fi fi fi fi

p Ê 2x ˆ tan -1 Á =2˜ 3 Ë1 – x ¯ p 2 tan –1 x = 3 p tan –1 x = 6 1 Ê pˆ x = tan Á - ˜ = Ë 6¯ 3

Hence, the solution is x = -

1 . 3

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4.34

Trigonometry Booster

74. Given equation is 4 sin -1 x + cos -1 x = fi fi fi fi

3p 4

p 3p = 2 4 p 3 sin -1 x = 4 p -1 sin x = 12 3 -1 Êpˆ x = sin Á ˜ = Ë 12 ¯ 2 2 3 sin -1 x +

79.

80.

75. Given equation is 5 tan -1 x + 3cot -1 x = fi fi fi fi

7p 4

3p 7p = 2 4 p 2 tan -1 x = 4 p -1 tan x = 8 Êpˆ x = tan Á ˜ = ( 2 - 1) Ë 8¯ 2 tan -1 x +

76. Given equation is 5 tan–1 x + 4 cot–1 x = 2p fi tan–1 x + 2p = 2p fi tan–1 x = 0 fi x = tan (0) = 0 Hence, the solution is x = 0. 77. Given equation is p cot -1 x - cot -1 ( x + 1) = 2 p -1 Ê 1 ˆ -1 Ê 1 ˆ tan Á ˜ - tan Á = fi Ë x¯ Ë x + 1˜¯ 2 fi

1 Ê 1 ˆ Á x x +1 ˜ p tan -1 Á ˜= Á1 + 1 ¥ 1 ˜ 2 x x + 1¯ Ë

fi

1 Ê ˆ p tan -1 Á 2 = Ë x + x + 1˜¯ 2

fi

1 Ê ˆ ÁË x 2 + x + 1˜¯

fi

x2 + x + 1 =

Êpˆ tan Á ˜ Ë 2¯ 1

=0

fi x2 + x = 1 = 0 So, no real values of x satisfies the above equation. Hence, the solution is x = j 78. Given equation is [sin–1 x] + [cos–1 x] = 0 It is possible only when [sin–1 x] = 0, [cos–1 x] = 0

TR_04.indd 34

81. 82.

fi 0 £ sin–1 x £ 1 and 0 £ cos–1 x £ 1 fi 0 £ x £ sin (1) and cos (1) £ x £ 1 fi x Œ [cos (1) , sin (1)] Given equation is [tan–1 x] + [cot–1 x] = 0 It is possible only when [tan–1 x] = 0 and [cot–1 x] = 0 fi 0 £ tan–1 x £ 1 and 0 £ cot–1 x £ 1 fi cot (1) £ x £ tan (1) Hence, x Œ [cot (1), tan (1)] Given equation is [sin–1 cos–1 sin–1 tan–1 x] = 0 fi 0 £ sin–1 (cos–1 (sin–1 (tan–1 x))) < 1 fi 0 £ (cos–1 (sin–1 (tan–1 x))) < sin (1) fi cos (sin (1)) < (sin–1 (tan–1 x)) £ 1 fi sin (cos (sin (1))) < (tan–1 x) £ sin (1) fi tan (sin (cos (sin (1)))) < x £ tan (sin (1)) Do yourself. Given equation is 5p 2 (tan -1 x) 2 + (cot -1 x) 2 = 8 5p 2 fi (tan -1 x + cot -1 x) 2 - 2 tan -1 x ◊ cot -1 x = 8 2 p2 Êp ˆ 5p -1 fi , a = tan x - 2a Á - a˜ = Ë2 ¯ 4 8 2 Êp ˆ 3p 2a Á - a ˜ + =0 fi Ë2 ¯ 8 3p 2 ap – 2a 2 + =0 fi 8 fi 8ap – 16a2 + 3p2 = 0 fi 16a2 – 8ap – 3p2 = 0 fi 16a2 – 12ap + 4ap – 3p2 = 0 fi 4a(4a – 3p) + p(4a – 3p) = 0 fi (4a + p) (4a – 3p) = 0 3p p fi a= ,4 4 3p p –1 fi tan x = ,4 4 3 p Ê ˆ Ê pˆ x = tan Á ˜ , tan Á - ˜ fi Ë 4¯ Ë 4¯ x = –1 1 Ê 3ˆ cos -1 Á ˜ = q Ë 5¯ 2 Ê 3ˆ cos -1 Á ˜ = 2q fi Ë 5¯ 3 cos (2q ) = fi 5 3 fi 2 cos 2q – 1 = 5

83. Let

2/10/2017 4:12:42 PM

4.35

Inverse Trigonometric Functions

fi

2 cos 2q = 1 +

3 8 = 5 5

4 5 2 cos q = fi 5 84. We have Êp Ê 1ˆ ˆ sin Á + sin -1 Á ˜ ˜ Ë 2¯ ¯ Ë4 p Ê ˆ Ê 1ˆ = sin Á + q ˜ , q = sin -1 Á ˜ Ë4 ¯ Ë 2¯ p 1 Ê ˆ = sin Á + q ˜ , sin q = Ë4 ¯ 2 Êpˆ Êpˆ = sin Á ˜ cos(q ) + cos Á ˜ sin(q ) Ë 4¯ Ë 4¯ 1 1 = cos(q ) + sin(q ) 2 2 1 3 1 1 = ¥ + ¥ 2 2 2 2 fi

cos 2q =

=

3 +1

2 2 85. Let m1 and m2 be the roots of x2 + 3x + 1 = 0 Thus, m1 + m2 = –3 < 0 and m1 ◊ m2 = 1 It is possible only when both are negative. Ê 1ˆ Thus, tan -1 (m) + tan -1 Á ˜ Ë m¯ –1 = tan (m) – p + cot–1 (m) = tan–1 (m) + cot–1 (m) – p p = -p 2 p =2 86. We have cos (tan–1(sin (cot–1 x))) = cos (tan–1(sin q)), cot q = x Ê Ê 1 ˆˆ = cos Á tan -1 Á 2 ˜˜ Ë 1 + x ¯¯ Ë = cos j , tan j = =

Ê 1 ˆ Á 2˜ Ë 1+ x ¯

x2 + 1 x2 + 2

87. Given, 6(sin–1 x)2 – p sin–1 x £ 0 fi sin–1 x (6 sin–1 x – p) £ 0 p fi 0 £ sin -1 x £ 6 1 fi 0£ x£ 2

TR_04.indd 35

88. Given, in-equation is 2 tan -1 x + p £0 4 tan -1 x - p p p - £ tan -1 x £ fi 2 4 fi – 0 fi x > 1 and x < 0 fi x Œ [–1, 0) 90. Given in-equation is fi cos–1 x > cos–1 x2 fi x2 > x fi x2 – x > 0 fi x(x – 1) > 0 fi –1 £ x < 0 91. Given in-eqution is log2 (tan–1 x) > 1 fi tan–1 x > 2 fi x > tan (2) Hence, the solution is (tan 2, ) 92. Given in-equation is (cot–1 x)2 – 5 cot–1 x + 6 > 0 fi (cot– x – 2) (cot–1 x – 3) > 0 fi (cot–1 x – 2) < 0, (cot–1 x – 3) > 0 fi x > cot (2), x < cot (3) fi x Œ (cot 2, cot 3) 93. Given in-equation is sin–1 x < cos–1 x p fi 2 sin -1 x < 2 p -1 fi sin x < 4 1 x< fi 2 1 ˆ È x Œ Í -1, fi ˜ 2¯ Î 94. Given in-equation is sin–1 x > sin–1 (1 – x) x > (1 – x) 2x > 1 1 x> 2 Ê1 ˘ Hence, the solution is x Œ Á , 1˙ Ë2 ˚ 95. Given in-equation is sin–1 2x > cosec–1 x Ê 1ˆ sin -1 (2x) > sin -1 Á ˜ fi Ë x¯ 1 fi 2x > x

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4.36

Trigonometry Booster

fi fi

1 >0 x 2x 2 - 1 >0 x

( 2 x + 1)( 2 x - 1) >0 x Ê 1 ˆ Ê 1 ˘ x Á , 0˜ Á ,1 fi Ë 2 ¯ Ë 2 ˙˚ 96. Given in-equation is tan–1 3x < cot–1 x Ê 1ˆ tan -1 (3x) < tan -1 Á ˜ fi Ë x¯ Ê 1ˆ (3x) – Á ˜ < 0 fi Ë x¯ fi

( 3 x + 1)( 3 x - 1) x2 5x2 – 1 < 0

( 5 x + 1)( 5 x - 1) < 0 1 1 2 tan –1 x fi fi fi fi

Ê 1 – x 2 - 1ˆ xÁ ˜ >0 Ë 1 - x2 ¯

fi

Ê x2 ˆ xÁ 2 ˜ > 0 Ë x - 1¯

fi

Ê ˆ x3 ÁË ( x – 1)( x + 1) ˜¯ > 0

fi

x Œ (–1, 0)

- (p - 1) < ( x - 1) < (p - 1)

fi 1 - (p - 1) < x < 1 + (p - 1) 99. Given in-equation is

Ê Ê 1ˆˆ tan (cos -1 x) £ sin Á cot -1 Á ˜ ˜ Ë 2¯ ¯ Ë

TR_04.indd 36

fi fi fi fi fi

Ê xˆ sin -1 Á ˜ < cos -1 ( x + 1) Ë 2¯

fi

fi

Ê xˆ sin -1 Á ˜ < sin -1 ( 1 - ( x + 1) 2 ) Ë 2¯

fi

fi

Ê xˆ 2 ÁË ˜¯ < ( 1 - ( x + 1) ) 2

fi

Ê xˆ 2 ÁË ˜¯ < 1 - ( x + 1) 2

fi

x2 < - 2x - x 2 4

2

(1, )

101. Given in-equation is

fi

fi

Ê 2x ˆ tan -1 (2x) > tan -1 Á Ë 1 - x 2 ˜¯ Ê 2x ˆ (2x) > Á Ë 1 - x 2 ˜¯ 1 ˆ Ê (2x) Á1 >0 Ë 1 - x 2 ˜¯

1 - x2 1 £ x 5 Ê 1 - x2 ˆ 1 ÁË 2 ˜¯ £ 5 x Ê 1 - x2 1 ˆ ÁË 2 - ˜¯ £ 0 5 x 5 - 5x 2 - x 2 £0 5x 2 6x 2 - 5 £0 x2 6x 2 - 5 ≥0 x2 Ê 5 5ˆ x ŒR - Á - , Ë 6 6 ˜¯

p 102. As we know that sin -1 x + cos -1 x = , for every x in 2 [–1, 1] 1 £1 (i) Since 0 < 2 m +1 Ê 1 ˆ p so, f Á 2 ˜ = Ë m + 1¯ 2

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4.37

Inverse Trigonometric Functions

(ii) Since, 0 £

m2 sin (sin 3) Ë Ë x + 2 ¯¯ fi

Ê Ê 2x 2 + 5 ˆ ˆ sin -1 Á sin Á p - 2 ˜˜ x + 2 ¯¯ Ë Ë < sin–1 (sin (p – 3))

fi fi

TR_04.indd 37

Ê 2x 2 + 5 ˆ p -Á 2 ˜ >p -3 Ë x +2¯ Ê 2x 2 + 5 ˆ Á 2 ˜ p - 3 Ë x + 1 ¯¯¯ Ë Ë Ê Ê 2x 2 + 5 ˆ ˆ fi Áp - Á 2 ˜˜ < p - 3 Ë x +1 ¯¯ Ë Ê 2x 2 + 5 ˆ fi -Á 2 ˜ < -3 Ë x +1 ¯ Ê 2x 2 + 5 ˆ fi Á 2 ˜ >3 Ë x +1 ¯ Ê 2x 2 + 5 ˆ - 3˜ > 0 fi Á 2 Ë x +1 ¯ -1 Ê

fi

Ê 2x 2 + 5 – 3x 2 - 3 ˆ Á ˜ >0 x2 + 1 Ë ¯

fi

x2 < 2

- 20 x2 + 1 Ë ¯

fi fi

x 0, 2x ◊ 3x = 6x2 < 1 1 x< fi 6 p 3p Then = tan -1 (2x) + tan -1 (3x) < 4 2 So, it is not possible. Case III: When x > 0, 2x ◊ 3x >1 1 x> fi 6 3p Then = tan -1 (2x) + tan -1 (3x) 4 Ê 5x ˆ 3p p + tan -1 Á = fi 4 Ë 1 - 6x 2 ˜¯ p Ê 5x ˆ 3p tan -1 Á = -p = 2˜ 4 4 Ë 1 - 6x ¯ Ê 5x ˆ fi ÁË 1 - 6x 2 ˜¯ = -1 fi 6x2 – 5x – 1 = 0 fi x = 1, –1/6 Thus, x = 1 is a solution. 141. We have p sin -1 ( x) + sin -1 (2x) = 3 Ê 3ˆ sin -1 x + sin -1 (2x) = sin -1 Á ˜ fi Ë 2 ¯ Ê ˆ 3 sin -1 x – sin -1 Á ˜ = - sin -1 (2x) fi Ë 2 ¯ fi

fi fi fi fi fi fi

Êx ˆ 3 sin -1 Á 1 - x 2 ˜ = sin -1 ( - 2x) Ë2 ¯ 2 Êx ˆ 3 1 - x 2 ˜ = - 2x ÁË ¯ 2 2 5x = 3 1 - x 2 25 x3 = 3(1 – x2) 28x2 = 3 3 x=± 2 7

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4.41

Inverse Trigonometric Functions

3 , negative value of x does not satisfy 2 7 the given equation. 142. We have Êx 3 - 3x 2 ˆ f (x) = cos -1 (x) + cos -1 Á + ˜ Ë2 ¯ 2 Ê 1ˆ = cos -1 ( x) + cos -1 Á ˜ - cos -1 ( x) Ë 2¯ fi

x=

=

p 3

p 3 143. sin–1 x + sin–1 (1 – x) = cos–1 x p fi sin -1 (1 - x) = - 2sin -1 x 2 Êp ˆ fi (1 - x) = sin Á - 2 sin -1 x˜ Ë2 ¯ –1 fi (1 – x) = cos (2 sin x) fi (1 – x) = 1 – 2x2 fi x(2x – 1) = 0 fi x = 0, (2x – 1) = 0 1 fi x = 0, 2 2 –1 144. x – 4x > sin (sin[p3/2]) + cos–1(cos [p3/2]) fi x2 – 4x > sin–1 (sin 5.5) + cos–1(cos 5.5) fi x2 – 4x > (5.5 – 2p) + (2p – 5.5) fi x2 – 4x > 0 fi x (x – 4) > 0 fi x < 0 and x > 4 fi x Œ (– , 0) (4, ) 145. cos (tan–1 x) = x 1 fi =x 2 x +1 Now, f (2013) =

fi fi fi fi

x2(x2 + 1) = 1 x4 + x2 – 1 = 0 -1 ± 5 x2 = 2 5 -1 x2 = 2

5 -1 2 146. sin (tan–1 x) = cos (cot–1 (x + 1)) x x +1 = fi 2 x +1 1 + ( x + 1) 2 x2 ( x + 1) 2 = fi x 2 + 1 1 + ( x + 1) 2 fi

fi

TR_04.indd 41

x=±

( x + 1) 2 x2 -1= -1 x +1 1 + ( x + 1) 2 2

–1 -1 = x + 1 1 + ( x + 1) 2 fi x2 + 1 = 1 + (x + 1)2 fi x2 = (x + 1)2 fi 2x = –1 1 fi x=2 x Ê ˆ 147. sec -1 Á ˜ - sec -1 x = sec -1 2 Ë 2¯ Ê xˆ Ê 1ˆ Ê 1ˆ cos -1 Á ˜ = cos -1 Á ˜ + cos -1 Á ˜ fi Ë 2¯ Ë x¯ Ë 2¯ fi

2

fi

Ê1 1 1 3ˆ Ê xˆ cos -1 Á ˜ = cos -1 Á ◊ - 1 - 2 ◊ Ë 2¯ Ë2 x 2 ˜¯ x

fi

1 3ˆ Ê xˆ Ê 1 1 ÁË ˜¯ = ÁË ◊ - 1 - 2 ◊ 2 2 x 2 ˜¯ x

fi

x-

1 1 = - 3 1- 2 x x 2

fi fi fi fi fi fi

1ˆ 1ˆ Ê Ê ÁË x - ˜¯ = 3 ÁË1 - 2 ˜¯ x x 1 3 2 x + 2 -2=3- 2 x x 4 2 x + 2 -5=0 x x4 – 5x2 + 4 = 0 (x2 – 1)(x2 – 4) = 0 x = ±2, ±1

Ê Ê Ê 3ˆ ˆ ˆ ˆ Ê 148. cos Á tan -1 Á cot Á sin -1 Á x + ˜ ˜ ˜ ˜ + tan(sec -1 x) = 0 Ë 2¯ ¯ ¯ ¯ Ë Ë Ë Ê 2x + 3 ˆ 2 fi ÁË ˜ = x -1 2 ¯ fi fi

(2x + 3)2 = 4(x2 –1) 12x + 9 = –4 13 x=fi 12 149. Given equation is Ê Ê xˆ Ê 1 ˆˆ Êpˆ tan Á tan -1 Á ˜ + tan -1 Á = tan Á ˜ ˜ ˜ Ë ¯ Ë 4¯ 10 x + 1 Ë ¯ Ë ¯ 1 ˆ Ê x + Á 12 x + 1 ˜ p tan -1 Á fi ˜= Á1 - x ◊ 1 ˜ 4 12 x + 1 ¯ Ë fi fi fi fi fi

x 2 + x + 12 =1 11x + 12 x2 + x + 12 = 11x + 12 x2 – 10x = 0 x(x – 10) = 0 x = 0, 10

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4.42 150. (x – 2)x2 + 8x + k + 4 > sin–1(sin 12) + cos–1(cos 12) fi (k –2)x2 + 8x + k + 4 > (12 – 4p) + (4p – 12) fi (k – 2)x2 + 8x + (k + 4) > 0 For all x in R, D ≥ 0 fi 64 – 4(k – 2)(k + 4) ≥ 0 fi 16 – (k – 2)(k + 4) ≥ 0 fi (k – 2)(k + 4) – 16 £ 0 fi k2 + 2k – 24 £ 0 fi (k + 6)(k – 4) £ 0 fi –6 £ k £ 4 Thus, the least integral value of k is –6 151. Do yourself. 152. Given, f (x) = sin–1 (sin x), " x Œ [–p, 2p]. = (–p –x) – x + (p – x) + (x – 2p) = –2p – 2x Thus, f ¢(x) = –2 153. Given, f (x) = cos–1 (cos x), " x Œ [–2p, p] = (x + 2p) – x + x = (x + 2p) Thus, f ¢(x) = 1 154. Given, 3p 5p ˘ f (x) = tan–1 (tan x), " x Œ È ÍÎ 2 , 2 ˙˚ = (x + p) + x + (x – p) + (x – 2p) = (4x – 2p) Thus, f ¢(x) = 4 155. We have Ê 1 ˆ sin -1 Á ˜ + cot -1 (3) Ë 5¯ Ê 1ˆ Ê 1ˆ = tan -1 Á ˜ + tan -1 Á ˜ Ë 2¯ Ë 3¯ Ê 1 1 ˆ + -1 Á 2 3 ˜ = tan Á 1 1˜ Á1 - ◊ ˜ Ë 2 3¯ Ê 5ˆ p Á ˜ = tan -1 Á 6 ˜ = tan -1 (1) = 5 4 Á ˜ Ë 6¯ Ê 3ˆ Ê 12 ˆ 156. We have 2 tan -1 Á ˜ + tan -1 Á ˜ Ë 2¯ Ë 5¯ 3ˆ Ê 2◊ Á 2 ˜ + tan -1 Ê 12 ˆ = tan -1 Á ÁË ˜¯ 9˜ 5 Á1 - ˜ Ë 4¯ Ê 12 ˆ Ê 12 ˆ = tan -1 Á - ˜ + tan -1 Á ˜ Ë 5¯ Ë 5¯ Ê 12 ˆ Ê 12 ˆ = p – tan -1 Á ˜ + tan -1 Á ˜ Ë 5¯ Ë 5¯ =p

TR_04.indd 42

Trigonometry Booster

Ê 1 + sin x + 1 - sin 157. cot -1 Á Ë 1 + sin x - 1 - sin

xˆ ˜ x¯

Ê

Ê xˆ Ê xˆ Ê xˆ Ê xˆ ˆ cos Á ˜ + sin Á ˜ + cos Á ˜ - sin Á ˜ ˜ Ë 2¯ Ë 2¯ Ë 2¯ Ë 2¯ = cot Á ˜ Á cos ÁÊ x ˜ˆ + sin ÁÊ x ˜ˆ - cos ÁÊ x ˜ˆ + sin ÁÊ x ˜ˆ ˜ Ë 2¯ Ë 2¯ Ë 2¯ Ë 2¯ ¯ Ë Ê Ê xˆ ˆ cos Á ˜ ˜ Ë 2¯ –1 Á = cot Á ˜ Ê Á sin Á x ˆ˜ ˜ Ë 2¯ ¯ Ë Ê Ê xˆˆ x = cot –1 Á cot Á ˜ ˜ = Ë 2¯ ¯ 2 Ë –1 Á

158. sin -1 ( x 1 - x - x 1 - x 2 ) = sin -1 ( x 1 - ( x ) 2 - x 1 - x 2 ) = sin -1 ( x) - sin -1 ( x ) p p -1 Ê sin x + cos x ˆ 159. sin Á ˜¯ , - 4 < x < 4 Ë 2 1 Ê 1 ˆ = sin -1 Á sin x + cos x˜ Ë 2 ¯ 2 Ê p ˆˆ Ê = sin -1 Á sin Á x + ˜ ˜ Ë Ë 4¯¯ pˆ Ê =Áx + ˜ Ë 4¯ 5p Ê sin x + cos x ˆ p 160. cos -1 Á ˜¯ , 4 < x < 4 Ë 2 1 Ê 1 = cos -1 Á cos x + sin Ë 2 2 p ˆˆ Ê Ê = cos -1 Á cos Á x - ˜ ˜ Ë Ë 4¯¯ pˆ Ê =Áx - ˜ Ë 4¯

ˆ x˜ ¯

Ê 1 + x2 + 1 - x2 ˆ 161. tan -1 Á ˜ ÁË 1 + x 2 - 1 - x 2 ˜¯ Put x2 = cos 2p Ê 1 + cos 2q + 1 - cos 2q ˆ = tan -1 Á ˜ Ë 1 + cos 2q - 1 - cos 2q ¯ Ê cos q + sin q ˆ = tan -1 Á Ë cos q - sin q ¯˜ Ê 1 + tan q ˆ = tan -1 Á Ë 1 - tan q ˜¯ Ê Êp ˆˆ = tan -1 Á tan Á + q ˜ ˜ ¯¯ Ë Ë4 Êp ˆ = Á + q˜ Ë4 ¯

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4.43

Inverse Trigonometric Functions

Êp 1 ˆ = Á + cos -1 ( x 2 )˜ Ë4 2 ¯ 4 ˆ -1 Ê 3 162. sin Á cos x + sin x˜ Ë5 ¯ 5 –1 = sin (sin a cos x + cos a sin x) = sin–1(sin (x + a)) = (x + a) -1 Ê 3 ˆ = x + tan Á ˜ Ë 4¯ -1 Ê 1 ˆ 163. Let sin Á ˜ = q Ë 4¯ 1 sin q = fi 4 Now, sin (2q) = 2 sin q ◊ cos q 1 1 = 2 ¥ ¥ 14 16 =

15 8

Ê 1ˆ 164. Let cos -1 Á ˜ = q Ë 3¯ 1 cos q = fi 3 Now, cos (2q) = 2 cos2 q – 1 2 = -1 9 7 =9 Ê 1ˆ 165. Let tan -1 Á ˜ = q Ë 3¯ 1 tan q = fi 3 Now, 1 – tan 2q cos (2q) = 1 + tan 2q =

Ê 1ˆ 1– Á ˜ Ë 3¯

2

Ê 1ˆ 1+ Á ˜ Ë 3¯

2

9 -1 9 +1 8 4 = = 10 5 =

-1 Ê 3 ˆ 166. Let cot Á ˜ = q Ë 4¯ Êqˆ Now, sin Á ˜ Ë 2¯

TR_04.indd 43

=

1 – cos q 2 1–

= 1– = 1– = =

cos q sin q ◊ cosec q 2 cot q cosec q 2 cot q 1 + cot 2q 2

1 3/4 12 1 + 9/16

1 3 15 2 2 = 5 =

Ê 3ˆ 167. Let tan -1 Á ˜ = q Ë 4¯ 3 fi tan q = 4 We have Ê 3p ˆ tan Á - 2q ˜ = Ë 4 ¯

Ê 3p ˆ tan Á ˜ - tan (2q ) Ë 4¯ Ê 3p ˆ 1 + tan Á ˜ ◊ tan (2q ) Ë 4¯

-1 - tan (2q ) 1 - tan (2q ) 2 tan q -1 1 - tan 2q = 2 tan q 11 - tan 2q =

tan 2q - 2 tan q - 1 1 – tan 2q - 2 tan q 9 6 – –1 = 16 4 9 6 1– – 16 4 41 = 17 =

Ê 1ˆ 168. Let sin -1 Á ˜ = q Ë 2¯ 1 Then sin q = 2 Ê Ê 1ˆˆ Now, sin Á 2 sin -1 Á ˜ ˜ Ë 2¯ ¯ Ë

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4.44

Trigonometry Booster

= sin (2q) = 2 sin q cos q = 2¥

fi

1 3 3 ¥ = 2 2 2

Ê 1ˆ 169. Let sin -1 Á ˜ = q Ë 3¯ 1 Then sin q = 3 Ê Ê 1ˆ ˆ Now, sin Á 3 sin -1 Á ˜ ˜ Ë 3¯ ¯ Ë = sin (3q) = 3 sin q – 4 sin3 q 1 Ê 1ˆ = 3◊ - 4◊ Á ˜ Ë 3¯ 3 4 23 =1= 27 27

3

TR_04.indd 44

3 2 5

=

fi fi fi

fi

1 Ê 1ˆ cos -1 Á ˜ = q Ë 8¯ 2 1 Then cos -1 ÊÁ ˆ˜ = 2q Ë 8¯ 1 fi cos (2q ) = 8 1 2 fi 2 cos q - 1 = 8 9 fi 2 cos 2q = 8 9 fi cos 2q = 16 3 fi cos q = 4 Ê1 Ê 1ˆ ˆ 3 fi cos Á cos -1 Á ˜ ˜ = Ë 8¯ ¯ 4 Ë2 1 Ê 1ˆ 171. Let cos -1 Á - ˜ = q Ë 10 ¯ 2 Ê 1ˆ Then cos -1 Á - ˜ = 2q Ë 10 ¯ 1 cos (2q ) = fi 10 1 2 fi 2 cos q - 1 = 10 1 9 fi = 2 cos 2q = 1 10 10 9 fi cos 2q = 20 3 cos q = fi 2 5 cos q =

fi

fi

170. Let

fi

172. Let

3 5 3 5 = 10 2 5¥ 5

fi fi

Ê1 Ê 1 ˆˆ 3 5 cos Á cos -1 Á - ˜ ˜ = Ë 10 ¯ ¯ 10 Ë2 1 Ê 1ˆ cos -1 Á ˜ = q Ë 9¯ 2 Ê 1ˆ cos -1 Á ˜ = 2q Ë 9¯ 1 cos (2q ) = 9 1 9 1 10 2 cos 2 (q ) = 1 + = 9 9 5 2 cos (q ) = 9 5 4 sin 2 (q ) = 1 - = 9 9 2 sin (q ) = 3 Ê1 Ê 1ˆ ˆ 2 sin Á cos -1 Á ˜ ˜ = Ë 9¯ ¯ 3 Ë2 2 cos 2 (q ) - 1 =

1 tan -1 ( 63) = q 4 fi tan -1 ( 63) = 4q

173. Let

fi

tan (4q ) = 63

fi

tan (4q ) = 63

sin (4q ) = 63 cos (4q ) sin (4q ) cos (4q ) 1 = = fi 1 8 63 1 Now, cos (4q ) = 8 1 2 fi 2 cos (2q ) - 1 = 8 1 9 fi 2 cos 2 (2q ) = 1 + = 8 8 9 2 fi cos (2q ) = 16 3 fi cos (2q ) = 4 3 2 fi 2 cos (q ) - 1 = 4 7 2 fi 2 cos (q ) = 4 7 fi cos 2 (q ) = 8 7 fi cos (q ) = 2 2 fi

…(i)

…(ii)

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4.45

Inverse Trigonometric Functions

63 8

Also, sin (4q ) =

63 8

fi

2 sin (2q ) cos (2q ) =

fi

2 sin (2q ) ¥

fi

sin (2q ) =

fi

2 sin (q ) cos (q ) =

fi

2 sin (q ) ¥

fi

7 63 = sin (q ) ¥ 12 2

fi

sin (q ) =

fi

2 4 1 sin (q ) = 2 2 1 Ê1 ˆ sin Á tan -1 ( 63)˜ = Ë4 ¯ 2 2 1 Ê 24 ˆ tan -1 Á ˜ = q Ë 7¯ 4

fi fi 174. Let

3 63 = , from (i) 4 8

63 12

7 63 = 12 2 2

3 2 12

sin (q ) =

-1 Ê

24 ˆ fi tan Á ˜ = 4q Ë 7¯ 24 fi tan (4q ) = 7 sin (4q ) cos (4q ) 1 fi = = 24 7 25 7 Now, cos (4q ) = 25 7 fi 2 cos 2 (2q ) - 1 = 25 7 32 fi = 2 cos 2 (2q ) = 1 + 25 25 32 fi cos 2 (2q ) = 50 32 cos (2q ) = fi 50 fi fi fi

TR_04.indd 45

32 2 cos (q ) - 1 = 50 8 18 = 2 cos 2 (q ) = 1 + 10 10 9 cos 2 (q ) = 10 2

cos (q ) =

fi

Ê1Ê 3 Ê 24 ˆ ˆ ˆ cos Á Á tan -1 Á ˜ ˜ ˜ = Ë 7 ¯¯¯ Ë4Ë 10

175. Let fi

63 , from (ii) 12

3 10

fi

fi fi fi fi fi fi fi

1 Ê 2ˆ cos -1 Á ˜ = q Ë 3¯ 2 Ê 2ˆ cos -1 Á ˜ = 2q Ë 3¯ 2 cos (2q ) = 3 2 1 - tan q 2 = 1 + tan 2q 3 2 + 2 tan2 q = 3 – 3 tan2 q 5 tan2 q = 1 1 tan 2q = 5 1 tan q = 5 Ê1 1 Ê 2ˆ ˆ tan Á cos -1 Á ˜ ˜ = Ë 3¯ ¯ Ë2 5

Ê 7 Ê 1ˆ p ˆ 176. We have tan Á 2 tan -1 Á ˜ - ˜ = Ë 5¯ 4 ¯ Ë 17 Let

Ê 1ˆ 2 tan -1 Á ˜ = q Ë 5¯

fi

ˆ ˜ tan Á =q 1 ˜ ÁË 1 – ˜¯ 25

fi

Ê 2 ˆ Á ˜ 10 5 tan q = Á 5 ˜ = = 1 24 12 ÁË 1 – ˜¯ 25

Ê

-1 Á

2 5

pˆ Ê Now, tan Á q – ˜ Ë 4¯ tan q – 1 = 1 + tan q 5 -1 7 12 = =5 17 +1 12 177. Let

1 -1 Ê 4 ˆ sin Á - ˜ = q Ë 5¯ 4

fi

Ê 4ˆ sin -1 Á - ˜ = 4q Ë 5¯

fi

sin (4q ) = -

4 5

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4.46

Trigonometry Booster

fi

2 tan (2q ) 4 =2 5 1 + tan (2q )

fi

tan (2q ) 2 =2 5 1 + tan (2q )

fi

fi

2 tan2(2q) + 5 tan (2q) + 2 = 0 1 tan (2q ) = - , - 2 fi 2 1 when tan (2q ) = 2 2 tan q 1 =fi 2 2 1 – tan q fi

tan2 q – 4 tan q – 1 = 0

fi

tan q = 2 - 5

Ê 3p 1 -1 Ê 4 ˆ ˆ Now, tan Á - sin Á - ˜ ˜ Ë 5¯ ¯ Ë 4 4 Ê 3p ˆ - q˜ = tan Á Ë 4 ¯

4p + 1 3 + 2 2

fi –0.2 < x < 3.3 Thus, the integral values of x are 0, 1, 2, 3. 179. Given in-equation is 3x2 + 8x < 2 sin–1 (sin 4) – cos–1 (cos 4) fi 3x2 + 8x < 2(p – 4) – (2p – 4) fi 3x2 + 8x + 4 < 0 fi 3x2 + 6x + 2x + 4 < 0 fi 3x(x + 2) + 2(x + 2) < 0 fi (3x + 2) (x + 2) < 0 2 fi -2 < x < 3 180. We have ÔÏ x 1 - 3x 2 Ô¸ ˝ f ( x) = cos -1 x + cos -1 Ì + 2 ÓÔ 2 ˛Ô 1ˆ ˜ 4¯

Ê 1ˆ = cos -1 (x) + cos -1 Á ˜ - cos -1 (x) Ë 2¯

Êp ˆ = –tan Á + q ˜ Ë4 ¯

Ê 1ˆ = cos -1 Á ˜ Ë 2¯

Ê 1 + tan q ˆ = –Á Ë 1 - tan q ˜¯

p 3 = constant function. Hence, the result. 181. We have f (x) –1 Ê 2x ˆ + 2 tan -1 (x) = sin Á Ë 1 + x 2 ˜¯ =

Ê1+ 2 – 5ˆ = –Á ˜ Ë1- 2 + 5¯ Ê 3 – 5ˆ =Á ˜ Ë1- 5 ¯ Ê (3 – 5)(1 + 5) ˆ =Á ˜¯ -4 Ë

= p – 2 tan–1 (x) + 2 tan–1 (x) =p Hence, the value of f (2013) = p 182. We have 2 Ê1 + xˆ -1 Ê 1 - x ˆ f (x) = 2 tan –1 Á sin + Á ˜ Ë 1 - x ˜¯ Ë 1 + x2 ¯

1 = - (3 + 2 5 - 5) 4 1 = - (– 2 + 2 5) 4 Ê1 – 5ˆ =Á Ë 2 ˜¯

= 2(tan -1 (1) + tan -1 (x)) +

178. Given equation is x2 – 3x < sin–1 (sin 2) fi x2 – 3x < (p – 2)

TR_04.indd 46

4p + 1 sin–1 > sin–1x + cos–1x p 2 sin –1 x > fi 2 p –1 fi sin x > 4 1 Êpˆ x > sin Á ˜ = fi Ë 4¯ 2 Also, the domain of sin–1 x is [–1, 1]

220. We have, tan–1 y = 4 tan–1 x Ê 4x - 4x ˆ tan -1 y = tan -1 Á ˜ Ë 1 - 6x 2 + x 4 ¯ 4x(1 - x 2 ) fi y= 1 - 6x 2 + x 4 Which is a function of x. p Let tan -1 x = 8 Êpˆ fi x = tan Á ˜ Ë 8¯ p 1 fi tan y = 4 tan -1 x = 2 4x(1 - x 2 ) fi 1 - 6x 2 + x 4 3

fi

fi

1 – 6x2 + x4 = 0

fi

Êpˆ x = tan Á ˜ is a root of 1 + x4 = 6x2 Ë 8¯

LEVEL IIA 1. We have sin–1(sin 10) = sin–1(sin (3p – 10)) = (3p – 10) 2. We have cos–1(cos 5) = cos–1(cos (2p – 5) = (2p – 5) 3. We have tan–1(1) + tan–1(2) + tan–1(3) Ê 2+3 ˆ = tan -1 (1) + p + tan -1 Á Ë 1 - 2.3 ˜¯

TR_04a.indd 52

p 4 p = 4 p = 4 p = 4 =p =

Ê 1 ˘ Thus, the solution set is x Œ Á ,1 Ë 2 ˙˚ 5. Given, sin–1 x < cos–1 x fi sin–1 x + sin–1 x < sin–1 x + cos–1 x p fi 2 sin -1 x < 2 p -1 fi sin x < 4 1 Êpˆ x < sin Á ˜ = fi Ë 4¯ 2 Also, the domain of sin–1 x is [–1, 1] 1 ˆ È Thus, the solution set is x Œ Í -1, ˜ 2¯ Î 6. Given, 2 sin -1 x = sin -1 (2x 1 - x 2 ) p p Now, the range of sin -1 (2x 1 - x 2 ) is ÈÍ - , ˘˙ Î 2 2˚ p p Thus, - £ 2 sin -1 x £ 2 2 p p -1 fi - £ sin x £ 4 4 Ê pˆ Êpˆ fi sin Á - ˜ £ x £ sin Á ˜ Ë 4¯ Ë 4¯ p Ê ˆ Êpˆ – sin Á ˜ £ x £ sin Á ˜ fi Ë 4¯ Ë 4¯ 1 1 – £x£ fi 2 2 È 1 1 ˘ Hence, the solution set is Í , 2 2 ˙˚ Î 7. Given, 3 sin–1x = p + sin–1(3x – 4x3) Now, the range of p + sin–1(3x – 4x3)

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4.53

Inverse Trigonometric Functions

È p 3p ˘ is Í , Î 2 2 ˚˙ p 3p Thus, £ 3 sin -1 x £ 2 2 p p fi £ sin -1 x £ 6 2 p Ê ˆ Êpˆ fi sin Á ˜ £ x £ sin Á ˜ Ë 6¯ Ë 2¯ fi

10. We have cos -1 (cos(2cot -1 ( 2 - 1))) Ê Ê Ê 2 -1 ˆˆˆ = cos -1 Á cos Á 2cos -1 Á ˜˜˜ ÁË ÁË Ë 4 - 2 2 ¯ ˜¯ ˜¯ Ê Ê Ê 2( 2 - 1) 2 ˆ ˆ ˆ = cos -1 Á cos Á cos -1 Á - 1˜ ˜ ˜ Ë (4 - 2 2) ¯ ¯ ¯ Ë Ë Ê Ê Ê 2(3 – 2 2) - 4 + 2 2 ˆ ˆ ˆ = cos -1 Á cos Á cos -1 Á ˜˜˜ (4 - 2 2) Ë ¯¯¯ Ë Ë

1 £ x £1 2

È1 ˘ Hence, the solution set is Í , 1˙ Î2 ˚ Ê 2x ˆ 8. Given, 2 tan -1 x = p + tan -1 Á Ë 1 - x 2 ˜¯

Ê Ê Ê 2 – 2 2 ˆˆˆ = cos -1 Á cos Á cos -1 Á ˜˜˜ Ë (4 - 2 2) ¯ ¯ ¯ Ë Ë Ê Ê Ê 2(1 – 2) ˆ ˆ ˆ = cos -1 Á cos Á cos -1 Á ˜˜˜ Ë 2 2( 2 - 1) ¯ ¯ ¯ Ë Ë

Ê 2x ˆ Now, the range of p + tan -1 Á Ë 1 - x 2 ˜¯ p 3p ˆ is ÊÁ , Ë 2 2 ˜¯ p 3p Thus, £ 2 tan -1 x £ 2 2 p p 3 fi £ tan -1 x £ 4 4 Êpˆ Êpˆ fi tan Á ˜ £ x < tan Á ˜ Ë 4¯ Ë 2¯ Êpˆ Ê 3p ˆ and tan Á ˜ < x £ tan Á ˜ Ë 2¯ Ë 4¯ fi 1 £ x < and – < x £ –1 Therefore, the solution set is x Œ (– , –1] » [1, ) 9. We have Êp Ê 1ˆ ˆ cos Á + cos -1 Á - ˜ ˜ Ë 2¯ ¯ Ë6

Ê Ê Ê 1 ˆˆˆ = cos -1 Á cos Á p - cos -1 Á Ë 2 ˜¯ ˜¯ ˜¯ Ë Ë Ê p ˆˆ Ê = cos -1 Á cos Á p - ˜ ˜ Ë Ë 4¯¯ Ê Ê 3p ˆ ˆ = cos -1 Á cos Á ˜ ˜ Ë 4 ¯¯ Ë Ê 3p ˆ =Á ˜ Ë 4¯ 11. We have Ê

r =0

=

pˆ Êp = cos Á + p - ˜ Ë6 3¯

=

pˆ Ê = cos Á p - ˜ Ë 6¯ Êpˆ = - cos Á ˜ Ë 6¯ =-

1 2

Ê

1

ˆ

 tan -1 ÁË 1 + r (r + 1) ˜¯

r =0

=

Ê 5p ˆ = cos Á ˜ Ë 6¯

ˆ

1

 tan -1 ÁË 1 + r + r 2 ˜¯

Êp Ê 1ˆ ˆ = cos Á + p - cos -1 Á ˜ ˜ Ë 2¯ ¯ Ë6

Ê p 2p ˆ = cos Á + ˜ Ë6 3 ¯

TR_04a.indd 53

Ê Ê Ê 1 ˆˆˆ = cos -1 Á cos Á cos -1 Á ˜ Ë Ë 2 ¯ ¯˜ ¯˜ Ë

Ê (r + 1) - r ˆ

 tan -1 ÁË 1 + r (r + 1) ˜¯

r =0

 (tan -1 (r + 1) - tan -1 (r ))

r =0 n

 (tan -1 (r

1)

tan -1 (r )), n

r =0

= (tan–1(2) – tan–1(1)) + (tan–1(3) – tan–1(2)) + (tan–1(4) – tan–1(3) + (tan–1(5) – tan–1(4)) + … + ) tan–1 (n + 1) – tan–1(n)) = (tan–1(n + 1) – tan–1(1)), n Æ Ê (n + 1) - 1 ˆ tan -1 Á ,n Ë 1 + (n + 1) ◊ 1˜¯ Ê n ˆ tan -1 Á ,n Ë n + 2 ˜¯

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4.54

Trigonometry Booster

= tan–1(1), when n Æ p = 4 12. We have 2 r -1 ˆ -1 Ê tan  Á ˜ Ë 1 + 22r -1 ¯ r =1

Ê yˆ = tan -1 (a1 ) - tan -1 Á ˜ + tan -1 (a2 ) - tan -1 ( a1 ) Ë x¯ tan–1(a3) – tan–1(a2) + … + tan–1(an) – tan–1(an–1) + cot–1(an) Ê yˆ = tan –1 (an ) + cot -1 (an ) - tan -1 Á ˜ Ë x¯

n Ê ˆ 2 r -1 = Â tan -1 Á r r -1 ˜ Ë1+ 2 ◊ 2 ¯ r =1

=

Ê yˆ = cot -1 Á ˜ Ë x¯

n Ê 2 r - 2 r -1 ˆ = Â tan -1 Á ˜ Ë 1 + 2r ◊ 2r -1 ¯ r =1 n

= Â (tan (2 ) - tan (2 -1

r

-1

r -1

Ê xˆ = tan -1 Á ˜ Ë y¯

))

r =1

= (tan–1(2) – tan–1(1)) + (tan–1(22) – tan–1(2)) + (tan–1(23) – tan–1(22)) + … + (tan–1(2n) – tan–1(2n–1)) = tan–1(2n) – tan–1(1) p = tan -1 (2n ) 4 13. We have n Ê r - r - 1ˆ  sin -1 ÁË r (r + 1) ˜¯ r =1 Ê r - r -1 ˆ =  tan -1 Á ˜ Ë 1 + r ◊ r - 1¯ r =1 n

15. We have (tan -1 x) 2 + (cot -1x) 2 =

= Â (tan -1 ( r ) - tan -1 ( r - 1)) r =1

= (tan–1(1) – tan–1(0) + (tan–1(2) – tan–1(1)) + (tan–1(3) – tan–1(2)) + (tan–1(4) – tan–1(3)) + … + (tan–1(n) – tan–1 (n – 1)) = tan–1(n) 14. We have Ê a x - yˆ Ê a -a ˆ tan -1 Á 1 + tan -1 Á 2 1 ˜ Ë a1 y + x ¯˜ Ë 1 + a1a2 ¯ Ê an - an –1 ˆ Ê a - a2 ˆ + tan -1 Á 3 + .... + tan -1 Á ˜ ˜ Ë 1 + a3 a2 ¯ Ë 1 + an an –1 ¯ Ê 1ˆ + tan -1 Á ˜ Ë an ¯ y ˆ Ê a1 Á x ˜ + tan -1 Ê a2 - a1 ˆ = tan -1 Á ÁË 1 + a a ˜¯ y˜ 1 2 Á 1 + a1 ◊ ˜ Ë x¯ Ê an - an –1 ˆ Ê a - a2 ˆ + tan -1 Á 3 + .... + tan -1 Á ˜ ˜ Ë 1 + a3 a2 ¯ Ë 1 + an an –1 ¯ Ê 1ˆ + tan -1 Á ˜ Ë an ¯

p2 8

fi

(tan -1 x + cot -1 x) 2 - 2 tan -1 x ◊ cot -1 x =

fi

p Êpˆ -1 -1 ÁË ˜¯ - 2 tan x ◊ cot x = 2 8

2

2

p2 8

p2 p2 - 2 tan -1 x ◊ cot -1 x = 4 8 p2 -1 -1 2 tan x ◊ cot x = 8 2 p tan -1 x ◊ cot -1 x = 16 2 Êp ˆ p a Á - a˜ = , where a = tan -1 x Ë2 ¯ 16 Êp ˆ 16a Á - a˜ = p 2 Ë2 ¯

fi fi fi fi

n

TR_04a.indd 54

p Ê yˆ - tan -1 Á ˜ Ë x¯ 2

fi fi fi fi

16a2 – 8ap + p2 = 0 (4a – p)2 = 0 (4a – p) = 0 p a= fi 4 p fi tan -1 x = 4 fi x=1 Hence, the solution is x = 1 16. Given, f (x) = (sec–1 x)2 + (cosec–1 x)2 = (sec–1 x + cosec–1 x)2 – 2 sec–1 x ◊ cosec–1 x 2

Êpˆ Êp ˆ = Á ˜ - 2 sec -1 x Á - sec -1 x˜ Ë2 ¯ Ë 2¯ p2 - p ◊ sec -1 x + 2(sec –1 x) 2 4 Ê p p2 ˆ = 2 Á (sec –1 x) 2 - ◊ sec -1 x + 2 8 ˜¯ Ë =

2/10/2017 4:11:39 PM

Inverse Trigonometric Functions 2 Ê p Êpˆ p2 p2 ˆ –1 2 -1 = 2 Á (sec x) - 2 ◊ sec x ◊ + Á ˜ + ˜ 4 Ë 4¯ 8 16 ¯ Ë 2

pˆ p2 Ê = 2 Á sec –1 x – ˜ + Ë 4¯ 8 p2 at x = 1. 4

Maximum value of f (x) is 17. Given,

f (x) = (sin–1 x)3 + (cos–1 x)3 = (sin–1 x + cos–1 x) ((sin–1 x)2 + (cos–1 x)2 – sin–1 x cos–1 x) = (sin–1 x + cos–1 x) ((sin–1 x + cos–1 x)2 – 3 sin–1 x cos–1 x) 2 p ÊÊ p ˆ Êp ˆˆ = Á Á ˜ - 3 sin -1 x Á - sin -1 x˜ ˜ Ë2 ¯¯ 2 ËË 2¯ p ÊÊ p ˆ Êp ˆˆ = Á Á ˜ - 3a Á - a˜ ˜ Ë2 ¯¯ 2 ËË 2¯ 2

=

ˆ p Ê p 2 3ap + 3a 2 ˜ ÁË ¯ 2 4 2

=

p Ê Ê 2 ap ˆ p 2 ˆ 3Á a ˜ + ˜¯ 2 ÁË Ë 2¯ 4

=

2 2 3p 2 ˆ pÊ Ê 2 p Êpˆ ˆ p Á 3Á a - 2 ◊ ◊ a + Á ˜ ˜ + ˜ Ë 6¯ ¯ 2Ë Ë 4 4 16 ¯

=

2 pÊ Ê pˆ p 2 3p 2 ˆ 3 a – Á ÁË ˜ ˜¯ + 2Ë 4 4 16 ¯

=

pÊ Ê pˆ p2ˆ Á 3 ÁË a – ˜¯ + ˜ 2Ë 4 16 ¯ 2

2 Ê 3p Ê pˆ p3ˆ =Á a– ˜ + ˜ Á Ë 2 Ë 4¯ 32 ¯ 2

=

3p Ê pˆ p3 p3 a – + ≥ Á ˜ 2 Ë 4¯ 32 32

p3 Hence, the minimum value of f (x) is 32 18. Given, [cot–1 x] + [cos–1 x] = 0 It is possible only when fi cot–1 x = 0 and cos–1 x = 0 fi 0 £ cot–1 x < 1 and 0 £ cos–1 x < 1 fi x Œ (cot 1, ) and x Œ (cos 1, 1] Thus, the solution is x (cot 1, 1] 19. Given, [sin–1 x] + [cos–1 x] = 0 It is possible only when fi [sin–1 x] = 0 and [cos–1 x] = 0 fi 0 £ sin–1 x < 1 and 0 £ cos–1 x < 1 fi x Œ [0, sin 1) and x Œ (cos 1, 1] Thus, the solution is x Œ (cos 1, sin 1)

TR_04a.indd 55

4.55 20. Given, [tan–1 x] + [cot–1 x] = 2 The range of [tan–1 x] is {–2, –1, 0, 1} and [cot–1 x] is {0, 1, 2, 3 } Case I: When [cot–1 x] = 1 and [tan–1 x] = 1 fi 1 £ cot–1 x < 2 and 1 £ tan–1 x < 2 fi x Œ (cot 2, cot 1] and x Œ [tan 1, tan 2) fi x Œ j ({ cot 1 < tan 1) Case II: When [cot–1 x] = 3 and [tan–1 x] = –1 fi 3 £ cot–1 x < 4 and –1 £ tan–1 x < 0 fi x Œ (cot 4, cot 3] and x Œ [–tan 1, 0) fi x Œ j, ({ cot 3 < –tan 1) Case III: When [cot–1 x] = 2 and [tan–1 x] = 0 fi 2 £ cot–1 x < 3 and 0 £ tan–1 x < 1 fi x Œ (3, cot 2] and x Œ [0 tan 1) fi x Œ j, ({ cot 2 < tan 1) Thus, there is no such value of x, where the equation is valid. 21. Given, [sin–1 (cos–1(sin–1(tan–1 x)))] = 1 fi 0 £ sin–1 (cos–1(sin–1(tan–1 x))) < 1 fi 0 £ (cos–1(sin–1(tan–1 x))) < sin 1 fi cos (sin 1) < (sin–1(tan–1 x)) £ 1 fi sin (cos (sin 1)) < (tan–1 x) £ sin 1 fi tan (sin (cos (sin 1))) < x £ tan (sin 1) fi x £ (tan (sin (cos (sin 1))), tan (sin 1)] 22. Given, f (x) = sin–1 x + tan–1 x + cot–1 x It is defined for –1 £ x £ 1 Thus, f (–1) = sin–1 (–1) + tan–1 (–1) + cot–1 (–1) p p p = - - +p 2 4 4 = –p + p = 0 and f (1) = sin–1(1) + tan–1(1) + cot–1(1) p p p = + + 2 4 4 =p Thus, range of f (x) is = [f (–1), f (1)] = [0, p] 23. Given, f (x) = sin–1 x + cos–1 x + tan–1 x p = + tan -1 x 2 As we know that, range of tan–1 x Ê p pˆ is Á - , ˜ Ë 2 2¯ Thus, the range of f (x) is (0, p) 24. Given, f (x) = sin–1 x + sec–1 x + tan–1 x The domain of f (x) is {–1, 1} Now, f (1) = sin–1(1) + sec–1(1) + tan–1(1) p p 3p = +0+ = 2 4 4

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4.56

Trigonometry Booster

and f (–1) = sin–1(–1) + sec–1(–1) + tan–1(–1) p p =- +p -02 4 p = 4 p 3p Thus, the range of f (x) is , 4 4 p 25. Given, tan -1 (2x) + tan -1 (3x) = 4 p –1 Ê 2x + 3x ˆ fi tan Á = Ë 1 - 2x ◊ 3x ˜¯ 4

{ }

fi

Ê 5x ˆ ÁË 1 - 6 x 2 ˜¯ = 1

fi fi fi

6x2 + 5x – 1 = 0 6x2 + 6x – x – 1 = 0 (6x – 1) (x + 1) = 0 1 fi x = -1, 6 1 Hence, the solution set is -1, 6 4 Ê ˆ Ê 1ˆ 26. Given, cos -1 x = cot -1 Á ˜ + tan -1 Á ˜ Ë 3¯ Ë 7¯

{ }

fi

Ê 3ˆ Ê 1ˆ cos -1 x = tan -1 Á ˜ + tan -1 Á ˜ Ë 4¯ Ë 7¯

Ê 3 1 ˆ + 4 7 ˜ cos x tan = fi Á 3 1˜ ÁË 1 - ◊ ˜¯ 4 7 Ê 25 ˆ fi cos -1 x = tan -1 Á ˜ = tan -1 (1) Ë 25 ¯ p -1 fi cos x = 4 1 fi x= 2 1 Hence, the solution is x = 2 p -1 Ê n ˆ 27. Given, cot Á ˜ > Ëp¯ 6 -1

-1 Á

n Êpˆ < cot Á ˜ Ë 6¯ p n fi < 3 p p n< fi 3 p fi n< = 3.14 ¥ 1.732 = 5.43848 3 Hence, the max. value of n is 5. p -1 Ê 5 ˆ -1 Ê 12 ˆ 28. Given sin Á ˜ + sin Á ˜ = Ë x¯ Ë x¯ 2 fi

Ê 5ˆ p Ê 12 ˆ sin -1 Á ˜ = - sin -1 Á ˜ Ë x¯ 2 Ë x¯

TR_04a.indd 56

fi

Ê 5ˆ Ê 12 ˆ sin -1 Á ˜ = cos -1 Á ˜ Ë x¯ Ë x¯

fi

Ê 5ˆ Ê 12 ˆ sin -1 Á ˜ = sin -1 1 - Á ˜ Ë x¯ Ë x¯

fi

Ê 5ˆ Ê 12 ˆ ÁË ˜¯ = 1 - ÁË ˜¯ x x

fi

Ê 169 ˆ ÁË 2 ˜¯ = 1 x

2

2

2

fi x2 = 169 fi x = ±13 29. We have x = sin–1(b6 + 1) + cos–1(b4 + 1) + tan–1(a2 + 1) It is possible only when a = 0 Thus, x = sin–1(1) + cos–1(1) + tan–1(1) p p 3p = +0+ = 2 4 4 pˆ pˆ Ê Ê Therefore, sin Á x + ˜ + cos Á x + ˜ Ë ¯ Ë 4 4¯ Ê 3p p ˆ Ê 3p p ˆ = sin Á + ˜ + cos Á + ˜ Ë 4 ¯ Ë 4 4 4¯ = sin p + cos p = 0 – 1 = –1 30. Given, sin–1 x + tan–1 x = 2k + 1 Let g(x) = sin–1 x + tan–1 x Domain of g = [–1, 1] Now, g(–1) = sin–1(–1) + tan–1(–1) p p 3p =- - =2 4 4 –1 g(1) = sin (1) + tan–1(1) p p 3p = + = 2 4 4 È 3p 3p ˘ Range of g = Í , Î 4 4 ˚˙ 3p 3p £ 2k + 1 £ 4 4 3p 3p fi - 1 £ 2k £ -1 4 4 3 ¥ 3.14 3 ¥ 3.14 fi - 1 £ 2k £ -1 4 4 fi –2.35 – 1 £ 2k £ 2.35 – 1 fi –3.35 £ 2k £ 1.35 3.35 1.35 fi £k£ 2 2 fi –1.67 £ k £ 6.7 Thus, the integral values of k are –1 and 0 31. Given, p sin -1 x + sin -1 y = 2 Thus, -

2/10/2017 4:11:40 PM

4.57

Inverse Trigonometric Functions

fi fi fi

p - sin -1 y 2 sin–1 x = cos–1 y sin -1 x = -1

sin x = sin

-1

1– y

fi

2

fi x = 1 – y2 fi x2 = 1 – y2 fi x2 = y2 = 1 Now,

36.

Ê 1 + x 4 + y 4 ˆ Ê 1 + ( x 2 + y 2 ) 2 - 2x 2 y 2 ˆ ˜ Á 2 ˜ =Á Ë x - x2 y 2 + y 2 ¯ Ë ( x2 + y 2 - x2 y 2 ) ¯ Ê 1 + 1 - 2x 2 y 2 ˆ =Á ˜ Ë (1 - x 2 y 2 ) ¯ 2(1 – x 2 y 2 ) (1 - x 2 y 2 ) =2 =

37.

32. Given, cos–1 x + cos–1(2x) + cos–1(3x) = p fi cos–1 2x + cos–1 3x = p – cos–1 x fi cos–1 2x + cos–1 3x = cos–1(–x) fi

cos -1 (2x ◊ 3x – 1 – 4x 2 1 - 9x 2 ) = cos -1 (– x)

fi

(6x 2 – 1 – 4x 2 1 - 9x 2 ) = - x

fi (6x 2 + x) 2 = ( 1 – 4x 2 1 - 9x 2 ) 2 fi 36x4 + 12x3 + x2 = 1 – 4x2 – 9x2 + 36x4 fi 12x3 + 14x2 = 1 Also, ax3 + bx2 + cx = 1 Thus, a = 12, b = 14 and c = 0 Hence, the value of a2 + b2 + c2 + 10 = 144 + 196 + 10 = 350 33. The domain of sin–1 x + tan–1 x is [–1, 1] Now, f (1) = sin–1(1) tan–1(1) + 1 + 4 + 5 3p = + 10 4 and f (–1) = sin–1(–1) + tan–1(–1) + 1 – 4 + 5 3p =+2 4 Therefore, a + b + 5 = 10 + 2 + 5 = 17. p 34. Clearly, x = . 2 Thus, sin x = 1 Ans. (a) 35. As we know that, domain of sin–1 x is [–1, 1] Ê xˆ Therefore, -1 £ log 3 ÁË ˜¯ £ 1 3 x Ê ˆ fi 3-1 £ Á ˜ £ 3 Ë 3¯

TR_04a.indd 57

38.

39.

1 Ê xˆ £Á ˜ £3 3 Ë 3¯

fi 1£x£9 Thus the domain of f (x) is [1, 9] As we know that, the range of sin–1 x p p is ÈÍ - , ˘˙ Î 2 2˚ p p Therefore, - £ 2 sin -1a £ 2 2 p p -1 fi - £ sin a £ 4 4 1 1 £a£ fi 2 2 1 | a| £ fi 2 Ans. (c) Now, sin–1(x – 3) is defined for –1 £ (x – 3) £ 1 fi 2£x£4 1 Also, the function is defined for 9 - x2 2 fi 9–x >0 fi x2 – 9 < 0 fi (x + 3)(x – 3) < 0 fi –3 < x < 3 Thus, the solution is x Œ [2, 3) Hence, the domain is [2, 3) Ans. (b) Since the f is ont, so the range of f is co-domain. i.e., range = B Ê p pˆ Clearly, range of f is Á - , ˜ Ë 2 2¯ p pˆ Ê Thus, B = Á - , ˜ Ë 2 2¯ Ans. (c) Ê yˆ Given, cos –1 x - cos -1 Á ˜ = a Ë 2¯ Ê y y2 ˆ cos -1 Á x ◊ + 1 - x 2 1 fi ˜ =a Ë 2 4 ¯ Ê xy y2 ˆ 2 fi Á + 1- x 1˜ = cos a Ë 2 4 ¯ 2 2 Ê xy ˆ y2 ˆ Ê 2 fi ˜ ÁË cos a – ˜¯ = ÁË 1 - x 1 2 4 ¯ 2 Ê 2 Ê xy ˆ ˆ fi Á cos a – xy cos a + ÁË ˜¯ ˜ Ë 2 ¯ 2 2 2 y x y fi = 1 - x2 + 4 4 2 y fi x 2 - xy cos a + = 1 - cos 2a 4 y2 x 2 - xy cos a + = sin 2a fi 4

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4.58

Trigonometry Booster

fi 4x2 – 4xy cos a + y2 = 4 sin2 a Ans. (d) Note: No question asked in 2006. Ê xˆ Ê 5ˆ p 40. Given, sin –1 Á ˜ + cosec -1 Á ˜ = Ë 5¯ Ë 4¯ 2 Ê xˆ Ê 4ˆ p sin –1 Á ˜ + sin -1 Á ˜ = fi Ë 5¯ Ë 5¯ 2 Ê xˆ Ê 3ˆ p sin -1 Á ˜ + cos -1 Á ˜ = Ë 5¯ Ë 5¯ 2 fi x=3 Ans. (d) Ê Ê 5ˆ Ê 2ˆ ˆ 41. Given, cot Á cosec -1 Á ˜ + tan -1 Á ˜ ˜ Ë 3¯ Ë 3¯ ¯ Ë fi

Ê Ê 3ˆ Ê 2ˆ ˆ = cot Á sin -1 Á ˜ + tan -1 Á ˜ ˜ Ë 5¯ Ë 3¯ ¯ Ë Ê Ê 3ˆ Ê 2ˆ ˆ = cot Á tan -1 Á ˜ + tan -1 Á ˜ ˜ Ë ¯ Ë 3¯ ¯ Ë 4 Ê Ê 3 2 ˆˆ Á -1 Á 4 + 3 ˜ ˜ = cot Á tan Á 3 2 ˜˜ Á ÁË 1 - ◊ ˜¯ ˜ Ë 4 3 ¯ Ê Ê 17 ˆ ˆ = cot Á tan -1 Á ˜ ˜ Ë 6 ¯¯ Ë Ê Ê 6 ˆˆ = cot Á cot -1 Á ˜ ˜ Ë 17 ¯ ¯ Ë =

6 17

Ans. (b) Note: No questions asked in between 2009 to 2014.

LEVEL III | x| - 2 ˆ 1. Let D1 : -1 £ ÊÁ £1 Ë 3 ˜¯ fi –3 £ (|x| – 2) £ 3 fi –1 £ |x| £ 5 fi –5 £ x £ 5 Ê 1 - | x| ˆ £1 and D2: -1 £ Á Ë 4 ˜¯ fi –4 £ 1 – |x| £ 4 fi –5 £ –|x| £ 3 fi –3 £ |x| £ 5 fi –5 £ x £ 5 Thus, Df = D1 « D2 = [–5, 5] 2. The function f is defined for fi 5p sin–1 x – 6 (sin–1 x)2 £ 0

TR_04a.indd 58

fi fi

6 (sin–1 x)2 – 5p sin–1 x £ 0 (sin–1 x) (6(sin–1 x) – 5p) £ 0 5p fi 0 £ sin -1 x £ 6 1 fi 0£ x£ 2 Also, sin–1 x is defined for [–1, 1] È 1˘ Thus, D f = Í0, ˙ Î 2˚ 3. f is defined for –1 £ log2(x2 + 3x + 4) £ 1 fi 2–1 £ (x2 + 3x + 4) £ 2 when (x2 + 3x + 4) £ 2 fi (x2 + 3x + 2) £ 0 fi (x + 1) (x + 2) £ 0 fi –2 £ x £ –1 1 when x 2 + 3x + 4 ≥ 2 fi 2x2 + 6x + 7 ≥ 0 fi xŒR Thus, Df = [–2, –1] 4. Given, cos–1 x + cos–1 x2 = 2p It is possible only when cos–1 x = p and cos–1 x2 = p x = cos p = –1 and x2 = –1 and x2 = –1 Thus, no such value of x is exist. 1 5. It is true only when 2 = x2 - 1 x -1 fi (x2 – 1)2 = 1 fi (x2 – 1) = ±1 fi x2 = 1 ± 1 = 2, 0 x = 0, ± 2 fi Ê x 2 - 1ˆ Ê 2x ˆ 2p 6. Given, cot -1 Á + tan -1 Á 2 ˜ = Ë 2x ˜¯ 3 Ë x - 1¯ fi

Ê 2x ˆ Ê 2x ˆ 2p tan -1 Á + tan -1 Á = 2˜ 3 Ë1 - x ¯ Ë 1 - x 2 ˜¯

fi

Ê 2x ˆ 2p 2 tan -1 Á = 3 Ë 1 - x 2 ˜¯

fi

Ê 2x ˆ p tan -1 Á = Ë 1 - x 2 ˜¯ 3

fi

Ê 2x ˆ Êpˆ ÁË 1 - x 2 ˜¯ = tan ÁË 3 ˜¯ = 3

fi fi

2x = 1 - x2 3 2 x2 + x -1= 0 3 2

fi

1 ˆ 1 4 Ê ÁË x + ˜¯ = 1 + 3 = 3 3

2/10/2017 4:11:41 PM

4.59

Inverse Trigonometric Functions

fi

1 ˆ 2 Ê ÁË x + ˜¯ = ± 3 3

fi

x=-

fi

x=

1 2 ± 3 3

1 ,- 3 3

1 ¸ Ï Hence, the solution set is Ì - 3, ˝ 3˛ Ó Ê Ê 2x 2 + 4 ˆ ˆ 7. Given, sin -1 Á sin Á 2 ˜˜ < p - 3 Ë Ë x +1 ¯¯

3 ˆˆ Ê Ê Á -1 Á x + 2 ˜ ˜ = cos Á cos Á Ë 1 ˜¯ ˜¯ Ë Ê =Áx+ Ë

3ˆ ˜ 2¯

Thus, the given equation reduces to

fi

Ê Ê 2x 2 + 4 ˆ ˆ sin -1 Á sin Á p - 2 ˜˜ < p - 3 x +1 ¯¯ Ë Ë

3ˆ Ê -1 ÁË x + ˜¯ + tan(sec x) = 0 2

fi

Ê 2x 2 + 4 ˆ p Á ˜ 3 x2 + 1

fi

2 3ˆ Ê x - 1ˆ Ê x + + Á ˜ =0 ÁË ˜ 2¯ Ë 1 ¯

fi

2x 2 + 4 -3> 0 x2 + 1

fi

fi

2x 2 + 4 – 3x 2 - 3 >0 x2 + 1

Ê ÁË x +

fi

– x2 + 1 >0 x2 + 1

x2 - 1