adrian yeo trig or treat- an encyclopedia of trigonometric identity proofs with intellectually...

6651 tp.indd 1 9/28/07 4:50:42 PM

T Trigonometric Identity Proofs

(TIPs)

Intellectually challenging games

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N E W J E R S E Y • L O N D O N • S I N G A P O R E • B E I J I N G S H A N G H A I • H O N G K O N G • TA I P E I • C H E N N A I

World Scientific

Y E O AdrianM.A., Ph.D. (Cambridge University)

Honorary Fellow, Christ’s College, Cambridge University

6651 tp.indd 2 9/28/07 4:50:44 PM

An encyclopedia ofTrigonometric Identity Proofs

Intellectually challenging games

(TIPs)

British Library Cataloguing-in-Publication DataA catalogue record for this book is available from the British Library.

For photocopying of material in this volume, please pay a copying fee through the CopyrightClearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, USA. In this case permission tophotocopy is not required from the publisher.

ISBN-13 978-981-277-618-1ISBN-10 981-277-618-4ISBN-13 978-981-277-619-8 (pbk)ISBN-10 981-277-619-2 (pbk)

All rights reserved. This book, or parts thereof, may not be reproduced in any form or by any means,electronic or mechanical, including photocopying, recording or any information storage and retrievalsystem now known or to be invented, without written permission from the Publisher.

Copyright © 2007 by World Scientific Publishing Co. Pte. Ltd.

Published by

World Scientific Publishing Co. Pte. Ltd.

5 Toh Tuck Link, Singapore 596224

USA office: 27 Warren Street, Suite 401-402, Hackensack, NJ 07601

UK office: 57 Shelton Street, Covent Garden, London WC2H 9HE

Printed in Singapore.

TRIG OR TREATAn Encyclopedia of Trigonometric Identity Proofs (TIPs) with IntellectuallyChallenging Games

SC - Trig or Treat.pmd 9/17/2007, 10:43 AM1

September 12, 2007 19:16 Book:- Trig or Treat (9in x 6in) 01˙dedi

dedicated to the memoryof my mother

tan peck hiah

tan A+ tan B+ tan C ≡ tan A tan B tan C

(A+B+C = 180◦)

tan A+ tan(A+120◦)+ tan(A+240◦) = 3 tan(3A+360◦)

tan−1

(

11

)

= tan−1

(

12

)

+ tan−1

(

13

)

1π

=122 tan

π22 +

123 tan

π23 +

124 tan

π24 +

125 tan

π25 + · · ·

September 17, 2007 19:18 Book:- Trig or Treat (9in x 6in) 02˙contents

Contents

Preface ix

Introduction xiii

Trig — Level One 1

The Basics of Trigonometry 1Pythagoras’ Theorem 11

Trig — Level Two 15

Compound Angles, Double Angles and Half Angles

Trig — Level Three 25

Angles in a TriangleSum and Difference of sin and cos

Practical Trig 33

Numerical Values of Special Angles —Graphs of sin, cos and tan for 0◦–360◦

Appendix 45

The Concordance of Trigonometric Identities 47The Encyclopedia of Trigonometric Games or

Trigonometric Identity Proofs (TIPs) 71

vii

September 17, 2007 19:18 Book:- Trig or Treat (9in x 6in) 02˙contents

viii Trig or Treat

Level-One-Games: Easy Proofs 73

Level-Two-Games: Less-Easy Proofs 251

Level-Three-Games: Not-So-Easy Proofs 349

Addenda 393

September 17, 2007 22:30 Book:- Trig or Treat (9in x 6in) 03˙preface

Preface

Many students find “Trigonometry” to be a difficult topic in a diffi-cult subject “Math”. Yet most students have no difficulty with com-puter games, and enjoy playing them even though many of thesegames have lots of pieces to manipulate, and are subjected to com-plex rules. For example, a simple game like “Tetris” has seven differ-ent pieces; and the player has to orientate and manipulate each piecein turn, as it falls. The objective is to construct a solid wall with all therandom pieces — and to do it, racing against the clock.

“Trigonometry”, or “Trig” for short, can be thought of as an intellec-tual equivalent of “Tetris”. There are six main pieces to manipulate. Threeof them, sine, cosine, and tangent, are most important — that is why thisbranch of Math is called Trigonometry; the “tri” refers to three functions,three angles and three sides of a triangle. And there is only one simple rule— Logic. Trig can be thought of as a game that involves the logical manip-ulation of various trig pieces to achieve different identities and equations,and to solve numerical problems.

Trig can also be viewed as a non-numerical equivalent of the numbergame “Sudoku”. The logic and the arrangement of the digits 1 to 9, is nowapplied to the six trig pieces — sine, cosine, tangent, cosecant, secant andcotangent.

Tetris and Sudoku are both simple games that give lots of fun andpleasure. Trig is also a simple game, but with a vital difference —knowledge of it has invaluable applications in Math, surveying, building,

ix


x Trig or Treat

navigation, astronomy, and other branches of science, engineering, andtechnology.

Adults, children and students can play Sudoku and Tetris for hours onend. So they should have little difficulty playing “Trig”, if they derivesimilar fun and pleasure from it.

Albert Einstein said:

“Everything should be made as simple as possible,But not simpler”.

This book seeks to make Trig as simple as possible, by treating it as agame — albeit, an intellectual game — as interesting and stimulating asTetris and Sudoku. Mastering of Trig will not only give mental and intel-lectual satisfaction and pleasure, but it will also lead to beneficial results inone’s future career and life.

This book is the third math book∗ that I have written for my two grand-children, Kathryn and Rebecca, ages 5 and 7, respectively. The challengethat I set for myself here is to explain Trig so simply that my seven-year-old granddaughter, Rebecca, can understand it. Indeed she has been able todo some of the “Level-One-Games”. My hope is that in the coming yearsboth Kathryn and Rebecca would be able to play the “Level-Two-Games”and the “Level-Three-Games” in this book also.

I thank Lim Sook Cheng and her excellent team at World ScientificPublishing for the production of this book; and Zee Jiak Gek for her metic-ulous reading and critique of all the details in the manuscript.

∗ The other two books are “ The Pleasures of Pi,e and Other Interesting Numbers”,and “Are You the King or Are You the Joker?”.


xi


xii

September 12, 2007 19:17 Book:- Trig or Treat (9in x 6in) 04˙introduction

Introduction

The approach taken in this book is to treat Trig as a game. Beginning withonly the definition of sine, the superstar of Trig, the book introduces thereader slowly to the basics of Trig. Then, by applying simple logic, the twoco-stars, cosine and tangent, are introduced.

Thereafter three supporting starlets, named the reciprocals — cosecant,secant and cotangent — are added. With these six pieces, the applicationof simple logic, arithmetic and algebra will give countless Trig equationscalled identities. Played like jigsaw puzzles, Tetris and Sudoku, movingthe Trig pieces around to give different identities can be a lot of fun.

As with other games and puzzles, practice can lead to greater skill andmental agility. About 300 games (proofs) are provided in this book to givefun (and confidence) to readers who want to try their hands (and worktheir brains) on these intellectual games. The numerous games are broadlygrouped into three overlapping levels — Level-One-Games (Easy Proofs),Level-Two-Games, (Less-Easy Proofs) and Level-Three-Games (Not-so-Easy Proofs).

For the first time ever, a “Concordance of Trigonometric Identities”has been created. Trigonometric identities are given a 6-digit code, whichenables readers (and students) to have easy reference to the identity to beproved, and to locate rapidly the proof in the Encyclopedia of Trigonomet-ric Identity Proofs (TIPs) in the Appendix.

Readers are welcome to look at the identities in the Concordance first,and try their hand at proving any of the identities, prior to looking at the

xiii

September 12, 2007 19:17 Book:- Trig or Treat (9in x 6in) 04˙introduction

xiv Trig or Treat

detailed proofs in the Encyclopedia. (Some identities which may appearsimple, may be difficult to prove; conversely, some complex-looking iden-tities may turn out to be relatively easy!)

The games provide the challenge to readers to match their skills, andprogress up the ladder of increasing intellectual agility. If you are reallygood in Trig, then the speed of proving the identities is the speed withwhich you write out the proofs, i.e. your brain works faster than your brawn(hand).

Have fun with Trig!

September 17, 2007 19:21 Book:- Trig or Treat (9in x 6in) 05˙level-1

Trig — Level One

The Basics of Trigonometry


2 Trig or Treat

90°

90°

0° 60°

45° 30°

Acute Angles

120°

150° 135°

180°

90°

180°

Obtuse Angles


The Basics of Trigonometry 3

Measuring Angles

“The sun rises in the east, and sets in the west”. Similarly, the measur-ing of angles begins “in the east” (0◦), goes counterclockwise, up into theoverhead sky at noon (90◦) and sets in the west (180◦).

People in many ancient civilisations (including the Babylonians,Mesopotamians and the Egyptians) used a numbering system based on 60called the sexagesimal system. This resulted in the convention of 360◦

(60◦ × 6) for the angle round a point. This convention for measuringangles continues to the present day, despite the widespread use of themetric system based on decimals (10’s). Another sexagesimal legacy fromthe past is the use of 60 seconds in a minute, and 60 minutes in an hour.


4 Trig or Treat

Sine

A

HypotenuseOpposite side of angle A

90°

sine A =

sin A =

OppositeHypotenuse

OH



Sine

Over the centuries, many civilisations used calculations based on right-angled triangles and the relationships of their sides for various purposes,including the building of monuments such as palaces, temples, and pyra-mids and other tombs for their rulers. Some of these mathematical tech-niques were also applicable to the study of the stars (astronomy) which ledto calender making.

The origins of Trig are lost in the mist of antiquity. One of the earliestrecorded reference to the concept of “the sine of an angle” — “jya” — wasfound in a sixth century Indian math book. This word was later translatedinto “jiba” or “jaib” in Arabic. A further translation into Latin convertedthe word into “sinus”, meaning a bay or curve, the same meaning as “jaib”.This was further simplified in the 17th century into English — “sine”, andabbreviated as “sin” (but always pronounced as “SINE”and not “SIN”.)

Sine is simply the name of a specific ratio:

sine of an angle (A) =length of the opposite side of angle (A)

length of the hypotenuse

This definition is often abbreviated to

sin A =OH

You cannot do Trig if you cannot remember the definition of sin! Thereare many simple ways of remembering. How about:

1. O/H lang SINE?∗

2. O/H , it’s so SINple?

Can you create your own mnemonics?

∗Sounds like “Auld Lang Syne”, the universally popular song sung at the strokeof midnight on New Year’s Day.


6 Trig or Treat

Cosine

Hypotenuse

Adjacent side of angle A

(90−A)°

Adjacent

Hypotenuse

AH

cos A =

=

A 90°

Tangent

Adjacent side of angle A

Opposite side of angle A

Opposite

Adjacent

= OA

tan A =

sin Acos A

tan A =

A 90°



Cosine and Tangent

The complementary angle to the angle A in a right-angled triangle is thethird angle, with the value of (90−A)◦, because the three angles of a tri-angle sum to a total of 180◦. The term “co-sine” was derived from thephrase “the sine of the complementary angle”

co-sine A = sine of complementary angle of A

= sine(90−A)◦

∴ cos A =length of the adjacent side of angle A

length of the hypotenuse

=AH

Tangent∗ is defined as the ratio of sin A/cos A

∴ tan A =sin Acos A

=

OHAH

=OA

∗This ratio (tangent) should be distinguished from the line which touches a circle,which is also called tangent in geometry.


8 Trig or Treat

Reciprocals

cosec A =1

sin A

sec A =1

cos A

cot A =1

tan A

tan AA =sin AAcos AA

cot AA =cos AAsin AA



Reciprocals

The superstar “sin” and its two co-stars (cos and tan) make up the three keyplayers in Trig. Their definitions and their relationships are essential for allproblems in Trig. Hence it is important that they be committed to memory.

Three more trig terms — the supporting cast — are also used. Theseare known as the reciprocals, and are best remembered as the reciprocalsof sin, cos and tan.

1sin A

= cosec A (cosecant)

1cos A

= sec A (secant)

1tan A

= cot A (cotangent)

These reciprocals are rarely used in applications in science, engineeringand technology. But for intellectual gymnastics (and in examinations!),these reciprocals are often used in equations and identities.


10 Trig or Treat

Pythagoras’ Theorem

b

a2 + b2 = c2

a

c

The Famous “3-4-5 Triangle”

5 3

4

32 + 42 = 52



Pythagoras’ Theorem∗

The most well-known theorem in Math, which practically every student haslearnt, is the Pythagoras’ Theorem, named after the Greek mathematicianPythagoras (∼580–500 BC).

This theorem states that in a right-angled triangle, the square of thehypotenuse (the longest side) is equal to the sum of the squares of thetwo other sides. This lengthy statement can be represented accurately inmathematical terms:

a2 +b2 = c2

where a and b are lengths of the two sides, and c is the length of thehypotenuse, the side facing the right-angle.

The most famous right-angled triangle is the “3-4-5 triangle”:

32 +42 = 52

(9+16 = 25)

A less famous sister is the “5-12-13 triangle” (52 + 122 = 132;25 + 144 = 169).

Recent research has shown that many civilisations, including the Baby-lonian, the Egyptian, the Chinese and the Indian civilisations, indepen-dently knew about the relationship between the squares of the three sidesof the right-angled triangle, in some cases, centuries before Pythagoras wasborn. (This illustrates a truism in Math, that often, your discoveries basedon your own efforts, may have been preceded by others. However this doesnot diminish in any way, the pleasure, excitement and sense of achievementthat you experienced — the so-called “eureka effect”. Indeed, it proves thatyou have a mathematical mind, capable of the same deep thoughts as theancient heroes of Math.)

∗A theorem is simply a mathematical statement whose validity has been provenby meticulous mathematical reasoning.


12 Trig or Treat

Trig Equivalent of Pythagoras’ Theorem

b

a

c

A

A

cos A

1sin A

sin2 A + cos2 A = 1

cos2 A

sin2 A = 1 − cos2 A

= 1 − sin2 A



Trig Equivalent of Pythagoras’ Theorem

One of the most important of Trig identities∗ is the trig equivalent of thePythagoras’ Theorem. The proof is simple:

sin A =ac

cos A =bc

sin2 A+ cos2 A =a2

c2 +b2

c2

=a2 +b2

c2

=c2

c2

(

by Pythagoras’ Theorem

a2 +b2 = c2

)

= 1

sin2AA+ cos2AA ≡ 1

A simpler visual proof can be obtained by using a special right-angledtriangle with a hypotenuse of unit length (1).

Then the length of the opposite side is now equal to sin A, and thelength of the adjacent side is equal to cos A (see figure opposite). Thenby Pythagoras’ Theorem:

sin2 A+ cos2 A ≡ 12

sin2 A+ cos2 A ≡ 1 .

This unity trig identity is the simplest and the most important of all trigidentities. It is also extremely useful in helping to solve trig problems.Whenever you see sin2 A or cos2 A, always consider the possibility of usingthis identity to simplify further.

∗An identity is a mathematical equation that is true for all values of the angle A. Itdoes not matter whether A = 30◦, 60◦, 90◦ etc, whether it is acute or obtuse, etc.The symbol (≡) is used to show that the two sides of an equation are identical.


14 Trig or Treat

We can also derive two other identities:

dividingby sin2 A

sin2 A

sin2 A+

cos2 A

sin2 A≡

1

sin2 A

1+ cot2 A ≡ cosec2A

dividingby cos2 A

sin2 Acos2 A

+cos2 Acos2 A

≡

1cos2 A

tan2 A+1 ≡ sec2A.

After this simple introduction, you are now ready to play Level-One-Games (Easy Proofs), some of which seven-year-old Rebecca could play.

The general approach for playing the games (proving the identities)is to:

1. start with the more complex side of the identity (usually the left handside (LHS));

2. eyeball the key terms, and think in terms of sin and cos of the angle;3. engage in some mental gymnastics — rearranging and simplifying;4. whilst at all times, keeping the terms in the right hand side — the final

objective — in mind.

Like a guided missile, your logic and math manipulation of the LHSshould lead you to zoom in to the RHS.

Have Fun!


Trig — Level Two

Compound Angles


16 Trig or Treat

Sine of (the Sum of Two Angles)

sin(A+B) = sin A cos B+ cos A sin B

Substituting B = A,

sin(A+A) = sin A cos A+ cos A sin A

∴ sin 2A = 2 sin A cos A (Double Angle Formula)

Substituting A with (A/2),

sin 2

(

A2

)

= 2 sinA2

cosA2

∴ sin A = 2 sinA2

cosA2

(Half Angle Formula)

Sine of (the Difference of Two Angles)

Substituting B in the first equation by (−B),

sin (A+(−B)) = sin A cos(−B)+ cos A sin(−B)

∴ sin(A−B) = sin A cos B− cos A sin B

since cos(−B) = cos B

and sin(−B) = −sin B

See Proof in Addenda p. 395


Compound Angles 17

Sine of the Sum and the Difference of Angles

This is the beginning of Level-Two-Games. The sin of compound angles(i.e. angles that are the sum or the difference of two other angles) can beexpressed in terms of trig functions of the single angles.

For example, sin of (A + B) can be expressed as a combination of thesum and products of the sin and cos of A and B separately. Similarly, sincewe know that sin(−A) = −sin A, and cos(−A) = cos A, we can derivesin(A−B). By judicious substitution (using B = A), sin(A + B) can bechanged to sin 2A and then into sin A (using A = 2(A/2)).

In earlier days, before calculators and computers were available, knowl-edge of the trig functions of compound angles was invaluable in practicalworkplace calculations. Knowing the basics for 0◦, 30◦, 45◦, 60◦ and 90◦,one could work out the values of trig functions of 15◦ and 22.5◦ and othersuch angles by way of these functions. In those days, Trig was both “puremath” and “applied math”, useful in many professions involving science,engineering and architecture.

Today, where the pressing of a few buttons on a calculator or computerwill give answers for all such calculations, Trig is largely “pure math” —a mental pursuit, an intellectual game.

But what a beautiful game it is, especially when you are immersed inthe proving of the vast number of identities!


18 Trig or Treat

Cosine of (the Sum of Two Angles)

cos(A+B) = cos A cos B− sin A sin B

Substituting B = A,

cos 2A = cos2 A− sin2 A (Double Angle Formula)

= 2 cos2 A−1

= 1−2 sin2 A

Note the minussign on the RHS,even though thesign on the LHSis plus

Remember:-

s2 + c2 = 1

∴ −s2 = c2−1

and c2 = 1− s2


cos 2

(

A2

)

= cos2 A2− sin2 A

2

∴ cos A = cos2 A2− sin2 A

2(Half Angle Formula)

= 2 cos2 A2−1

= 1−2 sin2 A2

Cosine of (the Difference of Two Angles)


cos(A+(−B)) = cos A cos(−B)− sin A sin(−B)

∴ cos(A−B) = cos A cos B+ sin A sin B

since cos(−B) = cos B

and sin(−B) = −sin B

Note the plussign on the RHS,even though thesign on the LHSis minus



Compound Angles 19

Cosine of the Sum and Difference of Angles

Cos functions require special attention as occassionally they act in a con-trarian, counter-intuitive manner — this arises largely from the fact thatcos(−A) = cos A.

This is the first function where such counter-intuitive behaviour of cosshows itself.

Although the LHS of the cos function is for the sum of two angles, theRHS shows a difference of the two products. Students are usually carelessand are not sufficiently sensitive to such minor (???) intricacies in Math.Unfortunately such minor (???) inattention can be very costly in examina-tions because they lead to wrong answers and major (???) losses in marks!


20 Trig or Treat

Tangent of (the sum of Two Angles)

tan(A+B) =tan A+ tan B

1− tan A tan BNote the minussign in thedenominator onthe RHS, eventhough the LHShas a plus sign!

Substituting B = A,

tan 2A =2 tan A

1− tan2 A(Double Angle Forumla)


tan 2

(

A2

)

=2 tan

A2

1− tan2 A2

(1)

∴ tan A =2 tan

A2

1− tan2 A2

(Half Angle Formula) (2)

Tangent of (the Difference of Two Angles)


tan(A+(−B)) =tan A+ tan(−B)

1− tan A tan(−B)

∴ tan(A−B) =tan A− tan B

1+ tan A tan B

since tan(−B) = − tan B

Note the plus signin the denomina-tor on the RHS,even though theLHS has a minussign!



Compound Angles 21

Tangent of the Sum and the Difference of Two Angles

The tan formulas for the sum and difference of two angles derive directlyfrom the sin and cos formulas. The double angle formula, tan(2A), andthe half angle formula, tan A, have proven to be extremely valuable inmany mathematical proofs, and have resulted in sophisticated methods forthe calculation of π to a large number of decimal places. (Would youbelieve that π has been calculated to 1.24 trillion decimal places — yes,1,240,000,000,000 decimals?)


22 Trig or Treat

Summary of Trig Functions for Compound Angles

sin(A+B) = sin A cos B+ cos A sin B

sin 2A = 2 sin A cos A

sin A = 2 sinA2

cosA2

sin(A−B) = sin A cos B− cos A sin B

cos(A+B)= cos A cos B− sin A sin B

cos 2A = cos2 A− sin2 A

cos A = cos2 A2− sin2 A

2

cos(A−B)= cos A cos B+ sin A sin B

tan(A+B) =tan A+ tan B

1− tan A tan B

tan 2A =2 tan A

1− tan2 A

tan A =2 tan

A2

1− tan2 A2

tan(A−B) =tan A− tan B

1+ tan A tan B


Compound Angles 23

The sin, cos and tan of compound angles, and their “double angle” and“half angle” formulas provide the basis for many of the Level-Two-Games.

Together with the most important identity:

sin2 A+ cos2 A ≡ 1 ,

the 12 identities on the opposite page, make up the total of the key Trigfunctions. Most other Trig identities can be derived from these “12 + 1”key identities.

The typical student (or Trig player) is expected to know these “12+1”key functions (very lucky if you know them instinctively, and very un-lucky if you don’t). With these “12+1” key functions, Level-Two-Games(Less-Easy Proofs) should prove to be easy also.

Within Level-Two-Games, the proofs begin with the simpler ones andprogress upwards in difficulty.

Have Fun!


Trig — Level Three

Angles in a Triangle


26 Trig or Treat

Special Trig Identities for AllThree Angles in a Triangle

A + B + C = 90° + +2 2 2A B C

= sin (90° −sin ( ))2C

2A + B

= cos ( )2A + B

= cos (90° −cos ( ))2C

2A + B

= sin2

A + B

= 180°

C = 180° − (A + B )

sin C = sin (180° − (A + B ))

= sin (A + B )

cos C = cos (180° − (A + B ))

= −cos (A + B )

sin (A + B + C ) = sin 180° = 0

cos (A + B + C ) = cos 180° = −1

A

B C

See Graphs on p. 42


Angles in a Triangle 27

Trig Identities Involving All Three Angles in a Triangle

Some special trig identities apply only when all three angles in a triangleare involved. For such identities, the additional constraint of

(A+B+C) = 180◦

is a critical one.For such identities the relationship between the three angles is always

necessary for simplification, and sometimes result in beautiful identities asseen in some of the examples in Level-Three-Games.


28 Trig or Treat

The Sum and Difference of Sine Functions

sin(A+B) = sin Acos B+ cos Asin B

sin(A−B) = sin Acos B− cos Asin B

Adding the two equations:

sin(A+B)+ sin(A−B) = 2 sin Acos B

sin S+ sin T = 2 sin

(

S+T2

)

cos

(

S−T2

)

Let (A+B) = S

and (A−B) = T

∴ A =S+T

2

and B =S−T

2

Subtracting the second equation from the first:

sin(A+B)− sin(A−B) = 2 cos A sin B

sin S− sin T = 2cos

(

S+T2

)

sin

(

S−T2

)



The Sum and Difference of Sine Functions

Often in Math, the addition or subtraction of similar equations, or the re-arrangement of sums and differences can lead to new insights and newequations which may be of special value.

The equations in the earlier pages are the sin and cos formulas of com-pound angles.

Here we are looking at the sum and difference of the trig functionsof such compound angles, and after further simplification, we derive newrelationships. Knowing these new relationships provide greater flexibilityand agility in the manipulation of the trig building blocks, and enables newequations or identities to be proved.


30 Trig or Treat

The Sum and Difference of Cosine Functions

cos(A+B) = cos A cos B− sin A sin B

cos(A−B) = cos A cos B+ sin A sin B

Adding the two equations:

cos(A+B)+ cos(A−B) = 2 cos A cos B

cos S+ cos T = 2 cos

(

S+T2

)

cos

(

S−T2

)

Let (A+B) = S

and (A−B) = T

∴ A =S+T

2

and B =S−T

2

Subtracting the second equation from the first:

cos(A+B)− cos(A−B) = −2 sin A sin B

cos S− cos T = −2 sin

(

S+T2

)

sin

(

S−T2

)

Note theminus signin front ofthe RHSterms



The Sum and Difference of Cosine Functions

Similar addition and subtraction of the cos functions for compound anglesgive similar equations for the sum and difference of cos functions.

While these functions were extremely useful before calculators andcomputers were available, they have fallen into disuse in modern timesexcept for purposes of “examinations”, to test the student’s versatility.

You are now ready to play Level-Three-Games — the “Not-So-Easy”Proofs.∗

Have Fun!

∗There are no simple sum and difference formulas for tan!


Practical Trig

Numerical Values of Special Angles


34 Trig or Treat


11

22

30° 30°

60° 60°

3

3

3

3sin 30° =

2

1tan 30° =

1cos 30° =

2

31

sin 60° =2

tan 60° =cos 60° =2

45°

1

21

sin 45° = 2

1 cos 45° = 2

1 tan 45° = 11

90°


Numerical Values of Special Angles 35


Five angles are of special interest in trigonometry — 0◦, 30◦, 45◦, 60◦, 90◦.Therefore it is important to know them and to remember them. The proofsare visual (resulting from the Pythagoras’ Theorem), and are easy to follow(see figures on the opposite page). Written in the form of square roots, theratios are easy to remember, beginning with sin 0◦ = 0, and cos 0◦ = 1.

sin 0◦ = 0 =

√

04

cos 0◦ = 1 =

√

44

tan 0◦ = 0

sin 30◦ =12

=

√

14

cos 30◦ =

√3

2=

√

34

tan 30◦ =1√

3

sin 45◦ =

√2

2=

√

24

cos 45◦ =

√2

2=

√

24

tan 45◦ = 1

sin 60◦ =

√3

2=

√

34

cos 60◦ =12

=

√

14

tan 45◦ =√

3

sin 90◦ = 1 =

√

44

cos 90◦ = 0 =

√

04

tan 90◦ (indeterminate)

Both sin 0◦ and tan 0◦ are zero because the length for “opposite” is zero.cos 0◦ = 1 because the “adjacent” and the hypotenuse are identical.

Similarly sin 90◦ = 1 because the “opposite” is coincident with the hy-potenuse; and cos 90◦ = 0 because the length of “adjacent” is zero.

It is important to stress than tan 90◦ does not have a value (indetermi-nate); in Math, we refer to it as “tending to infinity” and write it as “→ ∞”.


36 Trig or Treat

All Values of Sin (0◦ − 360◦)

Fourth QuadrantH

A4

O4

Second Quadrant H

A2 O2

Third Quadrant H

A3

O3

First Quadrant H

A1 O1



Values of Obtuse and Negative Angles

Sine

One of the potential obstacles to having fun with Trig games is discoveringthat not all trig functions are positive. Trig functions of angles larger than90◦ (obtuse angles) can sometime have negative values.

One of the easiest ways to overcome these potential obstacles to havefun with Trig is to focus on the values of the sin function. Consider fourangles A1, A2, A3 and A4 in the four quadrants, respectively:

First Quadrant : sin A1 =O1

H∗=

+ve+ve

= +ve

Second Quadrant : sin A2 =O2

H=

+ve+ve

= +ve

Third Quadrant : sin A3 =O3

H=

−ve+ve

= −ve

Fourth Quadrant : sin A4 =O4

H=

−ve+ve

= −ve.

∗The hypotenuse always has a positive value.


38 Trig or Treat

Value of Sine in the Four Quadrants

1

090° 180° 270° 360°

−1



Value of Sine in the Four Quadrants

The values of the sine of angles in the first and second quadrants (i.e.between 0◦ and 180◦) are always positive as seen in the graph on theopposite page, rising from 0 for sin 0◦ to a maximum value of 1 for sin 90◦.

Similarly the values of the sine of angles in the third and fourth quad-rants (i.e. between 180◦ and 360◦) are all negative, going to a minimum of−1 for sin 270◦, and returning to 0 for sine 360◦.

The brain remembers pictures better than equations or words; so com-mit the graph on the opposite page to memory; and this would prove to beextremely valuable in solving trig problems. This graph is well-known inMath as “the sinusoidal curve” or the “sine curve”, for short.

A quick sketch of the sinusoidal curve (done in 10 seconds) will providea good guide for ensuring that the correct values of sine in the differentquadrants are obtained.

It is also important to remember that the sine of a negative angle is thenegative of the sine of the angle.

sin(−A) = −sin A.


40 Trig or Treat

Value of Cosine in the Four Quadrants

1

090° 180° 270° 360°

−1

Value of Tangent in the Four Quadrants

0° 90° 180° 270° 360°



Cosine and Tangent

By similar consideration, the values of the cos and tan of angles in thefour quadrants can be obtained. Again it is easier to remember the pictures(graphs on the opposite page).

Always remember that cos 0◦ = 1, so the graph for cos always beginsat 1, and go to −1 for cos 180◦. (The cos graph for 0◦ to 360◦ looks like ahole in the ground).

Again, remember:

cos(−A) = cos A.

The cos function has this unusual feature and hence cos functions needspecial attention (i.e. be extra careful with cos functions).

For tan, the value goes from tan 0◦ = 0 all the way to the indeterminatevalue for tan 90◦. Interestingly, the third quadrant is an exact replica of thefirst quadrant, and the fourth an exact replica of the second.

Again, remember:

tan(−A) = − tan A


42 Trig or Treat

(A) 90° 0°

0°

0°

cos

180° 270°

(180° + A)

(180° − A)

(180° − A)

(180° − A)

(180° + A)

(180° + A)

360°

90° 180° 270° 360°

90° 180° 270° 360°

(A)

(−A)

(−A)

(−A)

(A)

tan

sin



It is very difficult to remember all the equations for calculating the dif-ferent values of sin, cos and tan of angles in all four quadrants, as well asthe negative angles.

Fortunately, the three pictures (graphs on the opposite page) enable easyderivation of their values based on their relation to their primary values inthe first quadrant). Let us use an arbitrary angle (say 39◦) to see how wecan work out the correct values for angles in all the four quadrants.

sin(−39◦) = −sin 39◦ cos(−39◦) = cos 39◦

2nd Q sin 141◦ = sin(180◦−39◦) cos 141◦ = −cos(180◦−39◦)

= sin 39◦ = −cos 39◦

3rd Q sin(180◦ +39◦) = −sin 39◦ cos(180◦ +39◦) = −cos 39◦

4th Q sin(360◦−39◦) = −sin 39◦ cos(360◦−39◦) = cos 39◦

tan(−39◦) = − tan 39◦

2nd Q tan 141◦ = − tan(180◦−39◦)

= − tan 39◦

3rd Q tan(180◦ +39◦) = tan 39◦

4th Q tan(360◦−39◦) = − tan 39◦

September 14, 2007 1:0 Book:- Trig or Treat (9in x 6in) 08˙appendix

Appendix


Concordance 45


46 Trig or Treat


The Concordanceof

Trigonometric Identities

The six-digit code for the Concordance is based on the number of trig functionson the LHS of the identity e.g. 123 000 means that the LHS has 1 sin, 2 cos, 3 tanand no cosec, sec and cot functions. On the rare occasion when you cannot findthe identity in the Concordance, use the functions on the RHS to determine thecode.


48 Trig or Treat

sin

cos

tan

cose

cse

c

cot

Lev

el

Page

0 0 0 0 0 2 2 cot A cot 2A ≡ cot2 A−1 2 288

(1+ cot A)2 +(1− cot A)2≡

2

sin2 A1 233

cot4 A+ cot2 A ≡ cosec4 A− cosec2 A 1 188

cot A+1cot A−1

≡

1+ tan A1− tan A

1 110

1− cot2 A1+ cot2 A

≡ sin2 A− cos2 A 1 183

cot2 A−12 cot A

≡ cot 2A 2 315

0 0 0 0 0 4cot A cot B−1cot A+ cot B

≡ cot(A+B) 2 323

cot A cot B+1cot A− cot B

≡ cot(A−B) 2 274

0 0 0 0 1 0 2 sec2 A−1 ≡ 1+2tan2 A 1 201

1+2 sec2 A ≡ 2 tan2 A+3 1 128

0 0 0 0 1 1 cot A(sec2 A−1)≡ tan A 1 148

0 0 0 0 2 0 sec4−sec2 A ≡ tan4 A+ tan2 A 1 157

sec A1+ sec A

≡

1− cos A

sin2 A1 203

1+ sec A1− sec A

≡

cos A+1cos A−1

1 79


Concordance 49

sin

cos

tan

cose

cse

c

cot

Lev

el

Page

0 0 0 0 2 0sec2 A

2− sec2 A≡ sec 2A 2 295

1+ sec Asec A

≡

sin2 A1− cos A

1 244

sec A+1sec A−1

≡ cot2A2

2 312

0 0 0 1 0 1 cosec2 A− cot2 A ≡ 1 1 145

cosec A− cot A ≡

sin A1+ cos A

1 219

cosec4 A− cot4 A ≡ cosec2 A+ cot2 A 1 103

cosec 2A− cot 2A ≡ tan A 2 299

cosec A−1cot A

≡

cot Acosec A+1

1 225

0 0 0 1 1 0 cosec A sec A ≡ 2cosec2A 2 270

cosec A sec A ≡ tan A+ cot A 1 135

cosec2 A+ sec2 A ≡ cosec2 A sec2 A 1 186

1sec2 A

+1

cosec2 A≡ 1 1 97

0 0 0 2 0 0cosec A−1cosec A+1

≡

1− sin A1+ sin A

1 109

cosec A1+ cosec A

≡

1− sin Acos2 A

1 215

cosec4A− cosec2A ≡ cot4 A+ cot2 A 1 158


50 Trig or Treat

sin

cos

tan

cose

cse

c

cot

Lev

el

Page

0 0 0 2 0 2cosec A cosec Bcos A cot B−1

≡ sec(A+B) 2 273

0 0 0 2 2 0sec A

cosec2 A−

cosec Asec2 A

≡ (1+ cot A+ tan A)(sin A− cos A) 1 243

sec A− cosec Asec Acosec A

≡ sin A− cos A 1 113

0 0 1 0 0 0 tan2 A+1 ≡ sec2 A 1 146

tan

(

45◦ +A2

)

≡ tan A+ sec A 2 300

tan(

A−

π4

)

≡

tan A−1tan A+1

2 285

0 0 1 0 0 1 tan A+ cot A ≡ 2 cosec 2A 2 306

tan A+ cot A ≡ cosec A sec A 1 144

tan2 A+ cot2 A+2 ≡ cosec2 A sec2 A 1 101

tan A+ cot A ≡

2sin 2A

2 290

1+ cot A1+ tan A

≡ cot A 1 152

2cot A tan 2A

≡ 1− tan2 A 2 252

1tan A+ cot A

≡

sin Asec A

1 163

12(cot A− tan A) ≡ cot 2A 2 269


Concordance 51

sin

cos

tan

cose

cse

c

cot

Lev

el

Page

0 0 1 0 0 11

tan A+ cot A≡ sin A cos A 1 166

0 0 1 0 0 2cot A(1+ tan2 A)

1+ cot2 A≡ tan A 1 180

0 0 1 0 1 0 sec 2A− tan 2A ≡ tan(45◦−A) 2 303

sec A− tan A ≡

cos A1+ sin A

1 218

sec4 A− tan4 A ≡

1+ sin2 Acos2 A

1 236

(sec A− tan A)2≡

1− sin A1+ sin A

1 230

tan Asec A−1

≡

sec A+1tan A

1 222

0 0 1 1 0 0 tan Acosec A ≡ sec A 1 104

tan2 A(cosec2 A−1)≡ 1 1 90

0 0 1 1 0 1 cosec A+ cot A+ tan A ≡

1+ cos Asin A cos A

1 131

cosec Acot A+ tan A

≡ cos A 1 159

0 0 1 1 1 0sec A+ cosec A

1+ tan A≡ cosec A 1 162

0 0 1 1 1 1 (tan A− cosec A)2− (cot A− sec A)2

≡ 2(cosec A− sec A) 2 338

0 0 2 0 0 0 tan A+ tan 2A ≡

sin A(4 cos2 A−1)

cos A cos 2A2 337


52 Trig or Treat

sin

cos

tan

cose

cse

c

cot

Lev

el

Page

0 0 2 0 0 0 tan(45◦ +A) tan(45◦−A)

≡ cot(45◦ +A)cot(45◦−A) 2 326

(1+ tan A)2 +(1− tan A)2≡ 2 sec2 A 1 185

1− tan2 A1+ tan2 A

+1 ≡ 2 cos2 A 1 247


≡ cos 2A 2 317

2 tan A1+ tan2 A

≡ sin 2A 2 316

1+ tan A1− tan A

≡

cot A+1cot A−1

1 110

0 0 2 0 0 1tan A(1+ cot2 A)

1+ tan2 A≡ cot A 1 178

tan A(cot A+ tan A) ≡ sec2 A 1 89

0 0 2 0 0 2 (tan A+ cot A)2− (tan A− cot A)2

≡ 4 1 182

tan A+ tan Bcot A+ cot B

≡ tan A tan B 1 171

tan A− cot Atan A+ cot A

≡ sin2 A− cos2 A 1 189

cot A− tan Acot A+ tan A

≡ cos 2A 2 280


+1 ≡ 2 sin2 A 1 239


≡ 1−2 cos2 A 1 245


Concordance 53

sin

cos

tan

cose

cse

c

cot

Lev

el

Page

0 0 2 0 0 2cot A

1− tan A+

tan A1− cot A

≡ 1+ tan A+ cot A 1 224

cot A1− tan A

+tan A

1− cot A≡ 1+ sec Acosec A 1 220

0 0 2 0 2 0 (sec A− tan A)(sec A+ tan A) ≡ 1 1 194

sec2 A− tan2 A+ tan Asec A

≡ sin A+ cos A 1 192

tan A+ sec A−1tan A− sec A+1

≡ tan A+ sec A 1 208

sec A sec B1+ tan A tan B

≡ sec(A−B) 2 272

0 0 2 1 2 0(sec A− tan A)2 +1

cosec A(sec A− tan A)≡ 2 tan A 1 249

0 0 3 0 0 0 tan A+ tan(A+120◦)+ tan(A+240◦)

≡ 3 tan 3A 3 391

tan A+ tan B+ tan C ≡ tan A tan B tan C(where A+B+C = 180◦) 3 356

3 tan A− tan3 A1−3 tan2 A

≡ tan 3A 2 325

tan Atan 2A− tan A

≡ cos 2A 2 302

0 0 4 0 0 01− tan A tan B1+ tan A tan B

≡

cos(A+B)

cos(A−B)2 277

tan A+ tan Btan A− tan B

≡

sin(A+B)

sin(A−B)2 278


54 Trig or Treat

sin

cos

tan

cose

cse

c

cot

Lev

el

Page

0 0 4 0 0 01+ tan A1− tan A

+1− tan A1+ tan A

≡ 2 sec 2A 2 332

0 0 4 0 0 4 (tan A+ tan B)(1− cot A cot B)

+(cot A+ cot B)(1− tan A tan B) ≡ 0 1 114

0 1 0 0 0 0 2 cos2(45◦−A)≡ 1+ sin 2A 2 307

18(1− cos 4A) ≡ sin2 A cos2 A 2 314

1−2 cos2 A ≡ 2 sin2 A−1 1 149

21+ cos A

≡ sec2 A2

2 292

21− cos A

≡ cosec2 A2

2 291

0 1 0 0 0 21− cot2 A1+ cot2 A

+2 cos2 A ≡ 1 1 240

0 1 0 0 1 0 sec A− cos A ≡ sin A tan A 1 143

(1− cos A)(1+ sec A) ≡ sin A tan A 1 199

1+ sec A1+ cos A

≡ sec A 1 160

0 1 0 0 1 1 (cos A+ cot A)sec A ≡ 1+ cosec A 1 141

0 1 0 1 0 0 cos Acosec A ≡ cot A 1 83

cosec A1− cos A

≡

1+ cos A

sin3 A1 226


Concordance 55

sin

cos

tan

cose

cse

c

cot

Lev

el

Page

0 1 0 1 0 1 cos2 A(cosec2 A− cot2 A) ≡ cos2 A 1 150

cot A cos Acosec2 A−1

≡ sin A 1 123

0 1 1 0 0 0 cos A tan A ≡ sin A 1 82

cos2 A(1+ tan2 A) ≡ 1 1 117

cos2 A tan A ≡ sin A cos A 1 86

(1+ cos A) tanA2≡ sin A 2 293

(cos2 A−1)(tan2 A+1)≡− tan2 A 1 122

1cos2 A(1+ tan2 A)

≡ 1 1 181

0 1 1 0 1 0sec A− cos A

tan A≡ sin A 1 207

0 1 1 0 1 1sec A+ tan Acot A+ cos A

≡ tan A sec A 1 210

0 1 1 1 0 0 cos A tan Acosec A ≡ 1 1 85

cosec Acot A+ tan A

≡ cos A 1 159

cos Acosec Atan A

≡ cot2 A 1 126

0 2 0 0 0 0 4 cos3 A−3 cos A ≡ cos 3A 2 340

cos 3A+ cos 2A ≡ 2 cos5A2

cosA2

3 358

8 cos4 A−4 cos 2A−3≡ cos 4A 2 345


56 Trig or Treat

sin

cos

tan

cose

cse

c

cot

Lev

el

Page

0 2 0 0 0 0cos A+1cos A−1

≡

1+ sec A1− sec A

1 79

11− cos A

+1

1+ cos A≡ 2 cosec2A 1 176

1− cos A1+ cos A

≡ (cosec A− cot A)2 1 231

cos(A+B)cos(A−B)≡ cos2 A− sin2 B 2 308

cos 4A+4 cos 2A+3≡ 8 cos4 A 2 342

cos 2A− cos 10A

≡ tan 4A(sin 2A+ sin 10A) 3 383

0 2 0 0 0 1 cos2 A+ cot2 Acos2 A ≡ cot2 A 1 184

cos A+ cos A cot2 A ≡ cot A cosec A 1 147

0 2 0 0 1 0 cos A(sec A− cos A) ≡ sin2 A 1 121

1− sec2 A(1− cos A)(1+ cos A)

≡−sec2 A 1 179

0 2 0 0 2 0sec A− cos Asec A+ cos A

≡

sin2 A1+ cos2 A

1 190

0 2 0 2 0 0cos A

cosec A−1+

cos Acosec A+1

≡ 2 tan A 1 205

0 2 1 0 0 0 cos2 A+ tan2 A cos2 A ≡ 1 1 119

0 2 1 1 0 0 cosec A tanA2−

cos 2A1+ cos A

≡ 4 sin2 A2

2 328

0 3 0 0 0 0 32 cos6 A−48 cos4 A+18 cos2 A−1

≡ cos 6A 2 341


Concordance 57

sin

cos

tan

cose

cse

c

cot

Lev

el

Page

0 3 0 0 0 0 cos 3A+2 cos 5A+ cos 7A

≡ 4 cos2 A cos 5A 3 385

cos(A−B)

cos A cos B≡ 1+ tan A tan B 2 255

1− cos 2A+ cos 4A− cos 6A

≡ 4 sin A cos 2A sin 3A 3 380

1+ cos 2A+ cos 4A+ cos 6A

≡ 4 cos A cos 2A cos 3A 3 386

cos(A+B)

cos A cos B≡ 1− tan A tan B 2 275

cos A+ cos 3A2 cos 2A

≡ cos A 3 368

cos A+ cos B+ cos C

≡ 4 sinA2

sinB2

sinC2

+1

(where A+B+C = 180◦) 3 357

0 4 0 0 0 0cos 4A− cos 8Acos 4A+ cos 8A

≡ tan 2A tan 6A 3 377

cos 2A− cos 4Acos 2A+ cos 4A

≡ tan 3A tan A 3 381

cos A+ cos Bcos A− cos B

≡−cot

(

A+B2

)

cot

(

A−B2

)

3 362


58 Trig or Treat

sin

cos

tan

cose

cse

c

cot

Lev

el

Page

1 0 0 0 0 0 1−2 sin2 A ≡ 2cos2 A−1 1 155

sin(A+B+C)≡

sin A cos B cos C

+sin B cos C cos A

+sin C cos A cos B

−sin A sin B sin C

3 352

1 0 0 0 0 1 sin A cot A ≡ cos A 1 80

1 0 0 0 1 0 (1− sin2 A) sec2 A ≡ 1 1 88

sec A1− sin A

≡

1+ sin Acos3 A

1 242

sin A sec A ≡ tan A 1 81

1− sin Asec A

≡

cos3 A1+ sin A

1 237

1 0 0 1 0 0 cosec A− sin A ≡ cot A cos A 1 140

(1− sin A)(1+ cosec A) ≡ cos A cot A 1 200

1+ sin A1+ cosec A

≡ sin A 1 75

1 0 1 0 0 0tan Asin A

≡ sec A 1 105

sin A tanA2≡ 1− cos A 2 297

sin 2A tan A ≡ 1− cos 2A 2 298

tan A sin A ≡ sec A− cos A 1 197


Concordance 59

sin

cos

tan

cose

cse

c

cot

Lev

el

Page

1 0 1 0 0 0 tan2 A− sin2 A ≡ tan2 A sin2 A 1 151

1 0 1 0 1 0 sec A− sin A tan A ≡ cos A 1 139

(1− sin A)(sec A+ tan A) ≡ cos A 1 196

sin A sec Atan A

≡ 1 1 91

1+ sec Atan A+ sin A

≡ cosec A 1 204

tan A sin Asec2 A−1

≡ cos A 1 127

1 1 0 0 0 0 3 sin2 A+4 cos2 A ≡ 3+ cos2 A 1 111

cos 2A1+ sin 2A

≡

cot A−1cot A+1

2 322

sin 2A1− cos 2A

≡ cot A 2 267

1−cos2 A

1+ sin A≡ sin A 1 193

cos4 A− sin4 A ≡ cos2 A− sin2 A 1 102

cos4 A− sin4 A ≡

1sec 2A

2 253

cos4 A− sin4 A ≡ cos 2A 2 260

sin A1+ cos A

≡ tanA2

2 262

sin 2A1+ cos 2A

≡ tan A 2 305

1− cos Asin A

≡ tanA2

2 263


60 Trig or Treat

sin

cos

tan

cose

cse

c

cot

Lev

el

Page

1 1 0 0 0 0 cos2 A− sin2 A ≡ 2 cos2 A−1 1 118

1−sin2 A

1− cos A≡−cos A 1 174

1−sin2 A

1+ cos A≡ cos A 1 227

sin4 A+ cos4 A ≡

34

+14

cos 4A 2 344

(cos2 A−2)2−4 sin2 A ≡ cos4 A 1 153

sin(30◦+A)+ cos(60◦ +A)≡ cos A 2 265

cos2 A− sin2 A ≡ 1−2 sin2 A 1 87

cos2 A− sin2 A ≡ cos4 A− sin4 A 1 102

1−8 sin2 A cos2 A ≡ cos 4A 2 346

(

sinA2

+ cosA2

)2

≡ 1+ sin A 2 264

1−cos2 A

1+ sin A≡ sin A 1 168

1+cos2 A

sin A−1≡−sin A 1 232

cos2 A1− sin A

≡ 1+ sin A 1 125

1− cos Asin A

≡

sin A1+ cos A

1 165

1 1 0 0 0 1 sin A+ cos A cot A ≡ cosec A 1 138


Concordance 61

sin

cos

tan

cose

cse

c

cot

Lev

el

Page

1 1 0 1 1 0 sin A cos Acosec A sec A ≡ 1 1 84

sin Acosec A

+cos Asec A

≡ 1 1 124

sec Acosec A

+sin Acos A

≡ 2 tan A 1 95

1 1 1 0 0 0 cos A+ sin A tan A ≡ sec A 1 137

tan A+cos A

1+ sin A≡ sec A 1 223

cos2 A− sin2 A1− tan2 A

≡ cos2 A 1 116

tan A+ cos Asin A

≡ sec A+ cot A 1 92

1 1 1 0 0 1 tan2 A cos2 A+ cot2 A sin2 A ≡ 1 1 112

tan A(sin A+ cot A cos A) ≡ sec A 1 198

sin2 A− tan Acos2 A− cot A

≡ tan2 A 1 241

1− sin2 A1− cos2 A

+ tan A cot A ≡ cosec2 A 1 130

cos A1− tan A

+sin A

1− cot A≡ sin A+ cos A 1 206

1 1 1 1 0 0 tan A cos A+ cosec A sin2 A ≡ 2 sin A 1 120

1 1 1 1 1 1 (cosecA− sin A)(sec A− cos A)

·(tan A+ cot A) ≡ 1 1 167


62 Trig or Treat

sin

cos

tan

cose

cse

c

cot

Lev

el

Page

1 2 0 0 0 0sin3 A+ cos3 A

1−2 cos2 A≡

sec A− sin Atan A−1

1 191

2 sin(A−B)

cos(A+B)− cos(A−B)≡ cot A− cot B 2 321

(2 cos2 A−1)2

cos4 A− sin4 A≡ 1−2 sin2 A 1 246

sin A+ cos Acos A

≡ 1+ tan A 1 76

cos Acos A− sin A

≡

11− tan A

1 77

1−2 cos2 Asin A cos A

≡ tan A− cot A 1 175

cos(A+B)

sin A cos A≡ cot A− cot B 2 256

cos(A−B)

sin A cos B≡ cot A+ tan B 2 257

sin(A+B)

cos A cos B≡ tan A+ tan B 2 259

cos(A+B)

cos A sin B≡ cot B− tan A 2 271

5−10 cos2 Asin A− cos A

≡ 5(sin A+ cos A) 1 228

1 3 0 0 0 0cos A

1+ cos 2A+

sin A1− cos 2A

≡

sin A+ cos Asin 2A

2 331


Concordance 63

sin

cos

tan

cose

cse

c

cot

Lev

el

Page

1 3 0 0 0 0sin A

(

4 cos2 A−1)

cos A cos 2A≡ tan A+ tan 2A 2 337

2 0 0 0 0 0 sin 5A− sin 3A ≡ 2 sin A cos 4A 3 359

3 sin A−4 sin3 A ≡ sin 3A 2 324

2 sin2 A6− sin2 A

7≡ cos2 A

7− cos

A3

2 304

sin2 A

1− sin2 A≡ tan2 A 1 100

1+ sin A1− sin A

≡ (sec A+ tan A)2 1 229

1+ sin A1− sin A

≡

cosec A+1cosec A−1

1 78

1− sin A1+ sin A

≡ (sec A− tan A)2 1 248

sin(A+B)sin(A−B)≡ cos2 B− cos2 A 2 311

1sin A+1

−

1sin A−1

≡ 2 sec2 A 1 177

2 sin 2A(1−2 sin2 A) ≡ sin 4A 2 339

sin(A+B)sin(A−B)≡ sin2 A− sin2 B 2 310

2 0 0 0 0 1 sin A+ sin Acot2 A ≡ cosec A 1 195

2 0 0 1 0 0 sin A(cosec A− sin A) ≡ cos2 A 1 134


64 Trig or Treat

sin

cos

tan

cose

cse

c

cot

Lev

el

Page

2 0 1 0 0 0 tan 4A(sin 2A+ sin 10A)

≡ cos 2A− cos 10A 3 388

sin A+ tan Asin A

≡ 1+ sec A 1 94

sin A+ sin A tan2 A ≡ tan A sec A 1 154

2 1 0 0 0 0 (4 sin A cos A)(1−2 sin2 A) ≡ sin 4A 2 283

(

2 sin2 A−1)2

sin4 A− cos4 A≡ 1−2 cos2 A 1 136

sin2 A+ cos2 A

sin2 A≡ cosec2 A 1 99

sin Asin A− cos A

≡

11− cot A

1 93

1−2 sin2 Asin A cos A

≡ cot A− tan A 1 169

1+ sin A− sin2 Acos A

≡ cos A+ tan A 1 96

cos(A−B)

sin A sin B≡ cot A cot B+1 1 258

sin(A+B)

sin A cos B≡ cot A tan B+1 2 276

sin(A−B)

sin A cos B≡ 1− cot A tan B 2 254

2 1 0 0 1 0sec Asin A

−

sin Acos A

≡ cot A 1 213


Concordance 65

sin

cos

tan

cose

cse

c

cot

Lev

el

Page

2 1 0 0 2 0 2 sec2 A−2 sec2 A sin2 A

−sin2 A− cos2 A ≡ 1 1 187

2 2 0 0 0 0 (cos A− sin A)2 +2 sin A cos A ≡ 1 1 133

(a sin A+b cos A)2 +(a cos A−b sin A)2

≡ a2 +b2 1 106

(2a sin A cos A)2 +a2(cos2 A− sin2 A)2≡ a2 2 261

sin 3Asin A

−

cos 3Acos A

≡ 2 2 287

1+ sin Acos A

+cos A

1+ sin A≡

2cos A

1 164

cos A

1− sin2 A− cos2 A+ sin A≡ cot A 1 129

sin A cos A

cos2 A− sin2 A≡

tan A1− tan2 A

1 98

sin(A+45◦)cos(A+45◦)

+cos(A+45◦)sin(A+45◦)

≡ 2 sec 2A 2 336

cos 2Asin A

+sin 2Acos A

≡ cosec A 2 320

sin2 2A+2 cos2A−1

sin2 2A+3 cos 2A−3≡

11− sec 2A

2 335

sin A1− cos A

+1− cos A

sin A≡ 2cosec A 1 221

cos A1+ sin A

+1+ sin A

cos A≡ 2 sec A 1 216

1− sin Acos A

+cos A

1− sin A≡ 2 sec A 1 214


66 Trig or Treat

sin

cos

tan

cose

cse

c

cot

Lev

el

Page

2 2 0 0 0 01+ cos A

sin A+

sin A1+ cos A

≡ 2cosec A 1 212

sin A1+ cos A

+1+ cos A

sin A≡

2sin A

1 161

1− cos 2A+ sin Asin 2A+ cos A

≡ tan A 2 347

sin A+ sin 2A2+3 cos A+ cos 2A

≡ tanA2

2 327

sin A+ sin 2A1+ cos A+ cos 2A

≡ tan A 2 268

(sin A+ cos A)2 +(sin A− cos A)2≡ 2 1 156

sin(2A+B)+ sin Bcos(2A+B)+ cos B

≡ tan(A+B) 3 361

sin(A+B)− sin(A−B)

cos(A+B)+ cos(A−B)≡ tan B 3 370

sin 4A− sin 2Acos 4A+ cos 2A

≡ tan A 3 366

sin 2A+ cos 2A+1sin 2A+ cos 2A−1

≡

tan(45◦ +A)

tan A2 333

1+ sin 2A+ cos 2Asin A+ cos A

≡ 2 cos A 2 266

cos2 A

sin2 A+ cos2 A+ sin2 A ≡

1

sin2 A1 132


Concordance 67

sin

cos

tan

cose

cse

c

cot

Lev

el

Page

2 2 0 0 0 01+ sin A+ cos A1+ sin A− cos A

≡

1+ cos Asin A

1 211

1+ cos A+ sin A1+ cos A− sin A

≡ sec A+ tan A 1 217

sin3 A+ cos3 Asin A+ cos A

≡ 1− sin A cos A 1 107

cos3 A− sin3 Acos A− sin A

≡

2+ sin 2A2

2 286

sin A− cos A+1sin A+ cos A−1

≡

sin A+1cos A

1 209


≡ 1−12

sin 2A 2 294

sin 4A+ sin 2Acos 4A+ cos 2A

≡ tan 3A 3 373

cos A− cos 3Asin 3A− sin A

≡ tan 2A 3 374

cos A− cos 3Asin 3A+ sin A

≡ tan A 3 367

cos A− cos 5Asin A+ sin 5A

≡ tan 2A 3 371


≡ tan 6A 3 375

sin 4A− sin 8Acos 4A− cos 8A

≡−cot 6A 3 376


68 Trig or Treat

sin

cos

tan

cose

cse

c

cot

Lev

el

Page

2 2 0 0 0 0sin A+ sin Bcos A+ cos B

≡ tan

(

A+B2

)

3 363

sin A− sin Bcos A− cos B

≡−cot

(

A+B2

)

3 364

2 2 0 0 0 2sin A cos Bcos A sin B

(cot Acot B)+1 ≡

1

sin2 B1 234

3 0 0 0 0 0 sin 5A+2sin 3A+ sin A ≡ 4sin 3Acos2 A 3 379

sin 2A+ sin 4A− sin 6A

≡ 4 sin A sin 2Asin 3A 3 384

sin 2A+ sin 2B+ sin 2C

≡ 4 sin A sin Bsin C

(where A+B+C = 180◦) 3 354

sin A+ sin B+ sin C

≡ 4cosA2

cosB2

cosC2

(where A+B+C = 180◦) 3 350

sin A+ sin 3A2sin 2A

≡ cos A 3 365

sin A(sin 3A+ sin 5A)

≡ cos A(cos 3A− cos 5A) 3 389

sin A(sin A+ sin 3A)

≡ cos A(cos A− cos 3A) 3 390


Concordance 69

sin

cos

tan

cose

cse

c

cot

Lev

el

Page

3 1 0 0 0 0cos A+ sin A− sin3 A

sin A

≡ cot A+ cos2 A 1 115

sin A− sin 3A

sin2 A− cos2 A≡ 2 sin A 3 369

3 1 0 2 0 1 cosec A(cosec A− sin A)

+

(

sin A− cos Asin A

)

+ cot A ≡ cosec2 A 1 235

3 2 0 0 0 0sin 3A cos A− sin A cos 3A

sin 2A≡ 1 2 284

3 3 0 0 0 0 (sin A+ cos B)2

+(cos B+ sin A)(cos B− sin A)

≡ 2cos B(sin A+ cos B) 1 108

(sin A− cos B)2

+(cos B+ sin A)(cos B− sin A)

≡−2cos B(sin A− cos B) 1 202

sin A+ cos Acos A

−

sin A− cos Asin A

≡ sec Acosec A 1 173

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

≡ tan 2A 3 372


70 Trig or Treat

sin

cos

tan

cose

cse

c

cot

Lev

el

Page

3 3 0 0 0 0sin A+ cos A

sin A−

cos A− sin Acos A

≡ cosec A sec A 1 170

3 6 0 0 0 0 sin A cos B cos C + sin B cos A cos C

+sin C cos A cos B

≡ sin A sin B sin C

(A+B+C = 180◦) 3 353

4 0 0 0 0 0sin A+ sin Bsin A− sin B

≡ tan

(

A+B2

)

cot

(

A−B2

)

3 360

sin 4A+ sin 8Asin 4A− sin 8A

≡−

tan 6Atan 2A

3 378

1+ sin A1− sin A

−

1− sin A1+ sin A

≡ 4 tan A sec A 1 172

4 0 2 0 0 0sin 2A+ sin 4Asin 2A− sin 4A

+tan 3Atan A

≡ 0 3 382

4 2 0 0 0 0sin 2A cos A−2cos 2A sin A

2 sin A− sin 2A≡ 2 cos2 A

22 330

4 4 0 0 0 0cos A− cos Bsin A+ sin B

+sin A− sin Bcos A+ cos B

≡ 0 2 281

cos A− sin Acos A+ sin A

+cos A+ sin Acos A− sin A

≡ 2 sec 2A 2 279

cos A+ sin Acos A− sin A

−


≡ 2 tan 2A 2 282

September 12, 2007 19:18 Book:- Trig or Treat (9in x 6in) 09˙tips

The Encyclopedia ofTrigonometric Games or

Trigonometric Identity Proofs(TIPs)


Level-One-GamesEasy Proofs


74 Trig or Treat

An Example for Proving a Trig Identity

1+ sin A1+ cosec A

≡ sin A

Eyeballing and Mental Gymnastics

1. Start with the more complex side, normally the LHS; this is more amenable tosimplification.

2. Consider simplifying tan, cot, and the reciprocal functions to sin and cos.3. Consider the use of common denominators.4. Rearrange and simplify through cancellation of common terms, if available.

Let us explore this first game (proof) together. Eyeballing the identity,we see that the more complex side is indeed the LHS (occassionally theRHS is the more complex; then it may be preferable to begin with theRHS. On rare occassions both the LHS and the RHS are complex; thenone can explore simplifying both side to a common set of terms).

We note the cosec term in the denominator and remember that cosec =

1/sin.Generally speaking it is easier to work with sin and cos than with

their reciprocals as it makes rearrangement, simplification and cancellationeasier.

With a reciprocal term in the denominator, we expect to use commondenominators prior to rearrangement, simplification and cancellation.

If all goes well, and no careless mistakes are made, we should end upwith “sin A” which is the target objective on the RHS.

It is useful to begin a proof by writing down accurately the LHS. If amistake is made here, no amount of effort will give the required identity.


Level-One-Games 75

1+ sin A1+ cosec A

≡ sin A

LHS =1+ sin A

1+ cosec A

=1+ sin A

(

1+1

sin A

)

= (1+ sin A)

1sin A+1

sin A

= (1+ sin A)

(

sin Asin A+1

)

= sin A

≡ RHS.

Use a bracket, ifit helps focus onthe key groups andminimises carelessmistakes.


76 Trig or Treat

sin A+ cos Acos A

≡ 1+ tan A

Eyeballing and Mental Gymnastics∗

1. t = s/c2. rearrange and simplify.

LHS =sin A+ cos A

cos A

= tan A+1

≡ RHS.

∗Some teachers prefer that students always write out the trig functions to includethe angle i.e. sin A instead of sin. In the “Eyeballing and Mental Gymnastics”sections, and in the short explanatory notes, we will use the abbreviations: s = sin,c = cos, and t = tan. Such abbreviations reflect the mental process in action, andconveys a sense of speed with eyeballing and mental gymnastics taking place.


Level-One-Games 77

cos Acos A− sin A

≡

11− tan A


1. divide LHS by cos2. t = s/c

LHS =cos A

cos A− sin A

=1

1− tan A

≡ RHS.


78 Trig or Treat

1+ sin A1− sin A

≡

cosec A+1cosec A−1


1. cosec = 1/sin2. rearrange and simplify.

LHS =1+ sin A1− sin A

=cosec +1cosec−1

≡ RHS.

divide bothnumeratorand denominatorby sin


Level-One-Games 79

cos A+1cos A−1

≡

1+ sec A1− sec A


1. sec = 1/cos2. rearrange and simplify.

LHS =cos A+1cos A−1

=1+ sec A1− sec A

≡ RHS.

divide bothnumeratorand denominatorby cos


80 Trig or Treat

sin A cot A ≡ cos A


1. cot = c/s2. simplify.

LHS = sin A cot A

= sin A ·

cos Asin A

= cos A

≡ RHS.


Level-One-Games 81

sin A sec A ≡ tan A


1. sec = 1/cos, t = s/c2. simplify.

LHS = sin A · sec A

= sin A ·

1cos A

= tan A

≡ RHS.


82 Trig or Treat

cos A tan A ≡ sin A


1. t = s/c2. simplify.

LHS = cos A tan A

= cos Asin Acos A

= sin A

≡ RHS.


Level-One-Games 83

cos A cosec A ≡ cot A


1. cosec = 1/sin, cot = c/s2. simplify.

LHS = cos A cosec A

= cos A ·

1sin A

= cot A

≡ RHS.


84 Trig or Treat

sin A cos A cosec A sec A ≡ 1


1. cosec = 1/sin, sec = 1/cos2. simplify.

LHS = sin A cos A cosec A sec A

= sin A cos A ·

1sin A

·

1cos A

= 1

≡ RHS.


Level-One-Games 85

cos A tan A cosec A ≡ 1


1. t = s/c, cosec = 1/sin2. simplify.

LHS = cos A tan A cosec A

= cos A ·

sin Acos A

·

1sin A

= 1

≡ RHS.


86 Trig or Treat

tan A cos2 A ≡ sin A cos A



LHS = tan A cos2 A

=sin Acos A

· cos2 A

= sin A cos A

≡ RHS.


Level-One-Games 87

cos2 A− sin2 A ≡ 1−2 sin2 A


1. c2, s2 suggest s2 + c2 ≡ 1∗

2. simplify.

LHS = cos2 A− sin2 A

= (1− sin2 A)− sin2 A

= 1−2 sin2 A

≡ RHS.

∗We shall use the abbreviation: s2 + c2 ≡ 1 for the trig equivalent of Pythagoras’Theorem sin2 A+ cos2 A ≡ 1.


88 Trig or Treat

(1− sin2 A)sec2 A ≡ 1


1. s2 suggests s2 + c2 ≡ 12. sec = 1/cos3. simplify.

LHS = (1− sin2 A)sec2 A

= cos2 A ·

1cos2 A

= 1

≡ RHS.


Level-One-Games 89

tan A(cot A+ tan A) ≡ sec2 A


1. t = s/c, cot = c/s2. sec2 suggests s2 + c2 = 1∗

3. simplify.

LHS = tan A(cot A+ tan A)

= 1+ tan2 A

= sec2 A

≡ RHS.

∗The identity s2 + c2 ≡ 1 should also trigger off possible reference to:

tan2 A+1 ≡ sec2 A (divide s2 + c2 ≡ 1 by c2)

and1+ cot2 A ≡ cosec2A (divide s2 + c2 ≡ 1 by s2).


90 Trig or Treat

tan2 A(cosec2 A−1) ≡ 1


1. tan2, cosec2 suggest s2 + c2 ≡ 12. rearrange and simplify.

LHS = tan2 A(cosec2 A−1)

= tan2 A(cot2 A)

≡ 1

≡ RHS.


Level-One-Games 91

sin Asec Atan A

≡ 1



LHS =sin A sec A

tan A

= sin A ·

1cos A

·

cos Asin A

≡ 1

≡ RHS.


92 Trig or Treat

tan A+ cos Asin A

≡ sec A+ cot A


1. t = s/c, sec = 1/cos, cot = c/s2. simplify.

LHS =tan A+ cos A

sin A

=sin Acos A

·

1sin A

+cos Asin A

= sec A+ cot A

≡ RHS.


Level-One-Games 93

sin Asin A− cos A

≡

11− cot A


1. cot = c/s2. rearrange and simplify.

LHS =sin A

sin A− cos A

=1

1−cos Asin A

=1

1− cot A

≡ RHS.

divide bothnumeratorand denominatorby sin A


94 Trig or Treat

sin A+ tan Asin A

≡ 1+ sec A


1. t = s/c, sec = 1/cos2. simplify.

LHS =sin A+ tan A

sin A

=sin Asin A

+sin Acos A

·

1sin A

= 1+ sec A

≡ RHS.


Level-One-Games 95

sec Acosec A

+sin Acos A

≡ 2 tan A


1. sec = 1/cos, cosec = 1/sin, t = s/c2. rearrange and simplify.

LHS =sec A

cosec A+

sin Acos A

=1

cos A·

sin A1

+sin Acos A

= tan A+ tan A

= 2 tan A.


96 Trig or Treat

1+ sin A− sin2 Acos A

≡ cos A+ tan A


1. s2 suggests s2 + c2 ≡ 12. t = s/c3. rearrange and simplify.

LHS =1+ sin A− sin2 A

cos A

=cos2 A+ sin A

cos A

= cos A+ tan A

= RHS.


Level-One-Games 97

1sec2 A

+1

cosec2 A≡ 1


1. sec = 1/cos, cosec = 1/sin2. sec2, cosec2 suggest s2 + c2 ≡ 13. simplify.

LHS =1

sec2 A+

1cosec2 A

= cos2 A+ sin2 A

= 1

≡ RHS.


98 Trig or Treat

sin A cos A

cos2 A− sin2 A≡

tan A1− tan2 A


1. t = s/c2. divide both numerator and denominator by cos2 A to give cos2 A/cos2 A = 13. rearrange and simplify.

LHS =sin Acos A

cos2 A− sin2 A

=

sin Acos Acos2 A

cos2 A− sin2 Acos2 A

=tan A

1− tan2 A

≡ RHS.


Level-One-Games 99

sin2 A+ cos2 A

sin2 A≡ cosec2 A


1. s2, c2 suggest s2 + c2 ≡ 12. cosec = 1/sin3. simplify.

LHS =sin2 A+ cos2 A

sin2 A

=1

sin2 A

= cosec2 A

≡ RHS.


100 Trig or Treat

sin2 A

1− sin2 A≡ tan2 A


1. s2, t2 suggest s2 + c2 ≡ 12. t = s/c.

LHS =sin2 A

1− sin2 A

=sin2 Acos2 A

= tan2 A

≡ RHS.


Level-One-Games 101

tan2 A+ cot2 A+2 ≡ cosec2 A+ sec2 A


1. tan2, cot2, cosec2, sec2 suggest s2 + c2 ≡ 12. rearrange and simplify.

LHS = tan2 A+ cot2 A+2

= (tan2 A+1)+(cot2 A+1)

= sec2 A+ cosec2 A

≡ RHS.


102 Trig or Treat

cos4 A− sin4 A ≡ cos2 A− sin2 A


1. cos4−sin4 suggest a4 −b4 = (a2 −b2)(a2 +b2)2. c2, s2 suggest s2 + c2 ≡ 13. rearrange and simplify.


= (cos2 A− sin2 A)(cos2 A+ sin2 A)

= (cos2 A− sin2 A)(1)

= cos2 A− sin2 A

≡ RHS.


Level-One-Games 103

cosec4 A− cot4 A ≡ cosec2 A+ cot2 A


1. (cosec4 − cot4) suggest a4 −b4 = (a2 −b2)(a2 +b2)2. squares suggest s2 + c2 ≡ 13. rearrange and simplify.

LHS = cosec4 A− cot4 A

= (cosec2 A− cot2 A)(cosec2 A+ cot2 A)

= (1)(cosec2 A+ cot2 A)

≡ RHS.


104 Trig or Treat

tan A cosec A ≡ sec A


1. t = s/c, cosec = 1/sin, sec = 1/cos2. simplify.

LHS = tan A cosec A

=sin Acos A

·

1sin A

=1

cos A

= sec A

≡ RHS.


Level-One-Games 105

tan Asin A

≡ sec A



LHS =tan Asin A

=sin Acos A

·

1sin A

=1

cos A

= sec A

≡ RHS.


106 Trig or Treat

(a sin A+b cos A)2 +(a cos A−b sin A)2≡ a2 +b2


1. ( )2 suggests expansion2. s2 + c2 ≡ 13. rearrange and simplify.

LHS = (a sin A+b cos A)2 +(a cos A−b sin A)2

= a2 sin2 A+2ab sin A cos A+b2 cos2 A

+a2 cos2 A−2ab sin Acos A+b2 sin2 A

= a2(sin2 A+ cos2 A)+b2(sin2 A+ cos2 A)

= a2 +b2

≡ RHS.


Level-One-Games 107


≡ 1− sin A cos A


1. s3 + c3 suggests (s+ c)(s2 − cs+ c2)2. s2 + c2 ≡ 13. rearrange and simplify.

LHS =sin3 A+ cos3 Asin A+ cos A

=(sin A+ cos A)(sin2 A− sin A cos A+ cos2 A)

(sin A+ cos A)

= sin2 A− sin A cos A+ cos2 A

= 1− sin A cos A

≡ RHS.


108 Trig or Treat

(sin A+cos B)2+(cos B+sin A)(cos B−sin A)≡2 cos B(sin A+cos B)


1. Although the identity looks lengthy and complicated, closer inspection showsthat the term (sin A+ cos B) is common to both sides of the equation.

2. rearrange and simplify.

LHS = (sin A+ cos B)2 +(cos B+ sin A)(cos B− sin A)

= (sin A+ cos B)[(sin A+ cos B)+(cos B− sin A)]

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

= 2cos B(sin A+ cos B)

≡ RHS.


Level-One-Games 109

cosec A−1cosec A+1

≡

1− sin A1+ sin A


1. cosec = 1/sin2. rearrange and simplify.

LHS =cosec A−1cosec A+1

=

1sin A

−1

1sin A

+1

=1− sin A

sin A·

sin A1+ sin A

=1− sin A1+ sin A

≡ RHS.


110 Trig or Treat

1+ tan A1− tan A

≡

cot A+1cot A−1


1. t = s/c, cot = c/s2. rearrange and simplify.

LHS =1+ tan A1− tan A

=1+

sin Acos A

1−sin Acos A

=cos A+ sin A

cos A·

cos Acos A− sin A

=cos A+ sin Acos A− sin A

=cot A+1cot A−1

≡ RHS.



Level-One-Games 111

3 sin2 A+4 cos2 A ≡ 3+ cos2 A


1. s2 + c2 ≡ 12. rearrange and simplify.

LHS = 3 sin2 A+4 cos2 A

= 3 sin2 A+3 cos2 A+ cos2 A

= 3(sin2 A+ cos2 A)+ cos2 A

= 3+ cos2 A

≡ RHS.


112 Trig or Treat

tan2 A cos2 A+ cot2 A sin2 A ≡ 1


1. t = s/c, cot = c/s2. s2 + c2 ≡ 13. rearrange and simplify.

LHS = tan2 A cos2 A+ cot2 A sin2 A

=sin2 Acos2 A

· cos2 A+cos2 A

sin2 A· sin2 A

= sin2 A+ cos2 A

= 1

≡ RHS.


Level-One-Games 113

sec A− cosec Asec A cosec A

≡ sin A− cos A


1. sec = 1/cos, cosec = 1/sin2. rearrange and simplify.

LHS =sec A− cosec Asec A cosec A

=sec A

sec A cosec A−

cosec Asec A cosec A

=1

cosec A−

1sec A

= sin A− cos A

≡ RHS.


114 Trig or Treat

(tan A+ tan B)(1− cot A cot B)+(cot A+ cot B)(1− tan A tan B) ≡ 0


1. cot = 1/ tan∗

2. expansion of factors3. rearrange and simplify.

LHS = (tan A+ tan B)(1− cot A cot B)+(cot A+ cot B)(1− tan A tan B)

= tan A+ tan B− tan A cot A cot B− tan B cot A cot B

+ cot A+ cot B− cot A tan A tan B− cot B tan A tan B

= tan A+ tan B− cot B− cot A

+ cot A+ cot B− tan B− tan A

= 0

≡ RHS.

∗Since the terms in the identity to be proved are all tan and cot, it is easier to thinkof cot as 1/tan rather than to convert tan and cot into sin and cos.


Level-One-Games 115

cos A+ sin A− sin3 Asin A

≡ cot A+ cos2 A


1. sin A− sin3 A suggests sin A(1− sin2 A)2. s2 + c2 ≡ 13. cot = c/s4. rearrange and simplify.

LHS =cos A+ sin A− sin3 A

sin A

=cos A+ sin A(1− sin2 A)

sin A

=cos A+ sin A(cos2 A)

sin A

= cot A+ cos2 A

≡ RHS.


116 Trig or Treat

cos2 A− sin2 A1− tan2 A

≡ cos2 A


1. t = s/c2. s2 + c2 ≡ 13. rearrange and simplify.

LHS =cos2 A− sin2 A

1− tan2 A

=(cos2 A− sin2 A)

1−sin2 Acos2 A

= (cos2 A− sin2 A)

(

cos2 A

cos2 A− sin2 A

)

= cos2 A

≡ RHS.


Level-One-Games 117

cos2 A(1+ tan2 A) ≡ 1


1. c2, t2 suggest s2 + c2 = 12. t = s/c3. rearrange and simplify.

LHS = cos2 A(1+ tan2 A)

= cos2 A+ cos2 A ·

sin2 Acos2 A

= cos2 A+ sin2 A

= 1

≡ RHS.


118 Trig or Treat

cos2 A− sin2 A ≡ 2cos2 A−1


1. c2, s2 suggest s2 + c2 ≡ 12. simplify.


= cos2 A− (1− cos2 A)

= cos2 A−1+ cos2 A

= 2cos2 A−1

≡ RHS.


Level-One-Games 119

cos2 A+ tan2 Acos2 A ≡ 1


1. c2, t2 suggest s2 + c2 ≡ 12. t = s/c3. rearrange and simplify.

LHS = cos2 A+ tan2 Acos2 A

= cos2 A+sin2 Acos2 A

· cos2 A

= cos2 A+ sin2 A

= 1

≡ RHS.


120 Trig or Treat

tan A cos A+ cosec A sin2 A ≡ 2 sin A


1. t = s/c, cosec = 1/sin2. rearrange and simplify.

LHS = tan A cos A+ cosec A sin2 A

=sin Acos A

· cos A+1

sin A· sin2 A

= sin A+ sin A

= 2 sin A

= RHS.


Level-One-Games 121

cos A(sec A− cos A) ≡ sin2 A



LHS = cos A(sec A− cos A)

= cos A

(

1cos A

− cos A

)

= 1− cos2 A

= sin2 A

≡ RHS.


122 Trig or Treat

(cos2 A−1)(tan2 A+1) ≡− tan2 A


1. c2, t2 suggest s2 + c2 ≡ 12. t = s/c3. rearrange and simplify.

LHS = (cos2 A−1)(tan2 A+1)

= −sin2 A · sec2A

= −

sin2 Acos2 A

= − tan2 A

≡ RHS.


Level-One-Games 123

cot A cos Acosec2

−1≡ sin A


1. cosec2 suggests s2 + c2 ≡ 12. cot = c/s3. rearrange and simplify.

LHS =cot A cos A(cosec2

−1)

=cos Asin A

· cos A

(

1cot2 A

)

=cos Asin A

· cos A ·

sin2 Acos2 A

= sin A

≡ RHS.


124 Trig or Treat

sin Acosec A

+cos Asec A

≡ 1


1. cosec = 1/sin; sec = 1/cos2. rearrange and simplify.

LHS =sin A

cosec A+

cos AsecA

= sin A · sin A+ cos A · cos A

= sin2 A+ cos2 A

= 1

≡ RHS.


Level-One-Games 125

cos2 A1− sin A

≡ 1+ sin A


1. c2 suggests s2 + c2 ≡ 12. rearrange and simplify.

LHS =cos2 A

1− sin A

=1− sin2 A1− sin A

=(1+ sin A)(1− sin A)

(1− sin A)

= 1+ sin A

≡ RHS.


126 Trig or Treat

cos A cosec Atan A

≡ cot2 A


1. cosec = 1/sin, t = s/c, cot = c/s2. simplify.

LHS =cos A cosec A

tan A

= cos A ·

1sin A

·

cos Asin A

=cos2 A

sin2 A

= cot2 A

≡ RHS.


Level-One-Games 127

tan A sin Asec2 A−1

≡ cos A


1. sec2 suggest s2 + c2 ≡ 12. simplify.

LHS =tan Asin Asec2 A−1

=sin Acos A

· sin A ·

1tan2 A

=sin2 Acos A

·

cos2 A

sin2 A

= cos A

≡ RHS.


128 Trig or Treat

1+2 sec2 A ≡ 2 tan2 A+3


1. sec2, tan2 suggests s2 + c2 ≡ 12. rearrange and simplify.

LHS = 1+2 sec2 A

= 1+2(1+ tan2 A)

= 1+2+2 tan2 A

= 2 tan2 A+3

= RHS.


Level-One-Games 129

cos A

1− sin2 A− cos2 A+ sin A≡ cot A


1. s2, c2 suggest s2 + c2 ≡ 12. cot = c/s3. rearrange and simplify.

LHS =cos A

1− sin2 A− cos2 A+ sin A

=cos A

1− (sin2 A+ cos2 A)+ sin A

=cos Asin A

= cotA

≡ RHS.


130 Trig or Treat

1− sin2 A1− cos2 A

+ tan A cot A ≡ cosec2 A


1. s2, c2 suggest s2 + c2 ≡ 12. tan ·cot = 13. rearrange and simplify.

LHS =1− sin2 A1− cos2 A

+ tan A cot A

=cos2 A

sin2 A+1

= cot2 A+1

= cosec2 A

≡ RHS.


Level-One-Games 131

cosec A+ cot A+ tan A ≡

1+ cos Asin Acos A


1. cosec = 1/sin, cot = c/s, t = s/c2. rearrange and simplify.

LHS = cosec A+ cot A+ tan A

=1

sin A+

cos Asin A

+sin Acos A

=cos A+ cos2 A+ sin2 A

sin A cos A

=1+ cos A

sin Acos A

≡ RHS.


132 Trig or Treat

cos2 A

sin2 A+ cos2 A+ sin2 A ≡

1

sin2 A


1. s2, c2 suggests s2 + c2 ≡ 12. rearrange and simplify.

LHS =cos2 A

sin2 A+ cos2 A+ sin2 A

=cos2 A

sin2 A+1

=cos2 A+ sin2 A

sin2 A

=1

sin2 A

≡ RHS.


Level-One-Games 133

(cos A− sin A)2 +2 sin A cos A ≡ 1


1. ( )2 suggests s2 + c2 ≡ 12. rearrange and simplify.

LHS = (cos A− sin A)2 +2 sin A cos A

= (cos2 A−2 sin A cos A+ sin2 A)+2 sin A cos A

= cos2 A+ sin2 A

= 1

≡ RHS.


134 Trig or Treat

sin A(cosec A− sin A) ≡ cos2 A


1. cosec = 1/sin2. c2 suggests s2 + c2 = 13. simplify.

LHS = sin A(cosec A− sin A)

= sin A

(

1sin A

− sin A

)

= 1− sin2 A

= cos2 A

≡ RHS.


Level-One-Games 135

cosec A sec A ≡ tan A+ cot A


1. cosec = 1/sin, sec = 1/cos, t = s/c, cot = c/s2. rearrange and simplify.

LHS = cosec Asec A

=1

sin A·

1cos A

=(sin2 A+ cos2 A)

sin Acos A

= tan A+ cot A

≡ RHS.


136 Trig or Treat

(2 sin2 A−1)2

sin4 A− cos4 A≡ 1−2cos2 A


1. sin4−cos4 suggests (a2 −b2)(a2 +b2)2. 2 sin2 A−1 equals 1−2 cos2 A3. rearrange and simplify.

LHS =(2 sin2 A−1)2

sin4 A− cos4 A

=(1−2 cos2 A)2

(sin2 A− cos2 A)(sin2 A+ cos2 A)

=(1−2 cos2 A)2

(1−2 cos2 A)(1)

= 1−2 cos2 A

≡ RHS.


Level-One-Games 137

cos A+ tan A sin A ≡ sec A


1. t = s/c, sec = 1/cos2. simplify.

LHS = cos A+ tan A sin A

= cos A+sin Acos A

· sin A

=cos2 A+ sin2 A

cos A

=1

cos A

= sec A

≡ RHS.


138 Trig or Treat

sin A+ cos A cot A ≡ cosec A


1. cot = c/s, cosec = 1/sin2. rearrange and simplify.

LHS = sin A+ cos A cot A

= sin A+ cos A ·

cos Asin A

=sin2 A+ cos2 A

sin A

=1

sin A

= cosec A

≡ RHS.


Level-One-Games 139

sec A− sin A tan A ≡ cos A


1. sec = 1/cos, t = s/c2. rearrange and simplify.

LHS = sec A− sin A tan A

=1

cos A−

sin A · sin Acos A

=1− sin2 A

cos A

=cos2 Acos A

= cos A

≡ RHS.


140 Trig or Treat

cosec A− sin A ≡ cot A cos A


1. cosec = 1/sin, cot = c/s2. rearrange and simplify.

LHS = cosec A− sin A

=1

sin A− sin A

=1− sin2 A

sin A

=cos2 Asin A

= cot A · cos A

≡ RHS.


Level-One-Games 141

(cos A+ cot A)sec A ≡ 1+ cosec A


1. cot = c/s, sec = 1/cos, cosec = 1/sin2. rearrange and simplify.

LHS = (cos A+ cot A)sec A

=

(

cos A+cos Asin A

)

·

1cos A

= cos A

(

1+1

sin A

)

1cos A

= 1+1

sin A

= 1+ cosec A

≡ RHS.


142 Trig or Treat

cosec A− sin A ≡ cos A cot A



LHS = cosec A− sin A

=1

sin A− sin A

=1− sin2 A

sin A

=cos2 Asin A

= cos A cot A

≡ RHS.


Level-One-Games 143

sec A− cos A ≡ sin A tan A



LHS = sec A− cos A

=1

cos A− cos A

=1− cos2 A

cos A

=sin2 Acos A

= sin A tan A

≡ RHS.


144 Trig or Treat

tan A+ cot A ≡ sec A cosec A


1. t = s/c, cot = c/s, sec = 1/cos, cosec = 1/sin2. s2 + c2 ≡ 13. rearrange and simplify.

LHS = tan A+ cot A

=sin Acos A

+cos Asin A

=sin2 A+ cos2 A

sin A cos A

=1

sin A cos A

= sec A cosec A

≡ RHS.


Level-One-Games 145

cosec2 A− cot2 A ≡ 1


1. cosec = 1/sin, cot = c/s2. cosec2, cot2 suggest s2 + c2 ≡ 13. simplify.

LHS = cosec2 A− cot2 A

=1

sin2 A−

cos2 A

sin2 A

=1− cos2 A

sin2 A

=sin2 A

sin2 A

= 1

≡ RHS.


146 Trig or Treat

tan2 A+1 ≡ sec2 A


1. tan2, sec2 suggest s2 + c2 ≡ 12. simplify.

LHS = tan2 A+1

=sin2 Acos2 A

+1

=sin2 A+ cos2 A

cos2 A

=1

cos2 A

= sec2 A

≡ RHS.


Level-One-Games 147

cos A+ cos A cot2 A = cot A cosec A


1. c2 suggests s2 + c2 ≡ 12. cosec = 1/sin3. rearrange and simplify.

LHS = cos A+ cos A cot2 A

= cos A(1+ cot2 A)

= cos A · cosec2A

= cos A ·

1

sin2 A

= cot A cosec A

= RHS.


148 Trig or Treat

cot A(sec2 A−1) ≡ tan A


1. cot = c/s, sec = 1/cos, t = s/c2. sec2 suggests s2 + c2 ≡ 13. rearrange and simplify.

LHS = cot A(sec2 A−1)

=1

tan A(tan2 A)

= tan A

≡ RHS.


Level-One-Games 149

1−2 cos2 A ≡ 2 sin2 A−1


1. c2, s2, 1 suggest s2 + c2 ≡ 12. rearrange and simplify.

LHS = 1−2 cos2 A

= 1−2(1− sin2 A)

= 1−2+2 sin2 A

= 2 sin2 A−1

≡ RHS.


150 Trig or Treat

cos2 A(cosec2 A− cot2 A) ≡ cos2 A


1. cosec = 1/sin, cot = c/s2. cos2, cosec2, cot2 suggest s2 + c2 ≡ 13. rearrange and simplify.

LHS = cos2 A(cosec2 A− cot2 A)

= cos2 A

(

1

sin2 A−

cos2 A

sin2 A

)

= cos2 A

(

1− cos2 A

sin2 A

)

= cos2 A

(

sin2 A

sin2 A

)

= cos2 A

≡ RHS.


Level-One-Games 151

tan2 A− sin2 A ≡ tan2 A sin2 A


1. t = s/c2. t2, s2 suggest s2 + c2 ≡ 13. rearrange and simplify.

LHS = tan2 A− sin2 A

=sin2 Acos2 A

− sin2 A

=sin2 A− sin2 A cos2 A

cos2 A

=sin2 A(1− cos2 A)

cos2 A

= tan2 A · sin2 A

≡ RHS.


152 Trig or Treat

1+ cot A1+ tan A

≡ cot A


1. cot = c/s, t = s/c2. rearrange and simplify.

LHS =1+ cot A1+ tan A

=

(

1+cos Asin A

)

1

1+sin Acos A

=

(

sin A+ cos Asin A

)

·

(

cos Acos A+ sin A

)

=cos Asin A

= cot A

≡ RHS.


Level-One-Games 153

(cos2 A−2)2−4 sin2 A ≡ cos4 A


1. c2, s2 suggest s2 + c2 ≡ 12. rearrange and simplify.

LHS = (cos2 A−2)2−4 sin2 A

= (cos4 A−4 cos2 A+4)−4 sin2 A

= cos4 A−4(cos2 A+ sin2 A)+4

= cos4 A−4(1)+4

= cos4 A

≡ RHS.


154 Trig or Treat

sin A+ sin A tan2 A ≡ tan A sec A


1. t2 suggests s2 + c2 ≡ 12. sec = 1/cos3. rearrange and simplify.

LHS = sin A+ sin A tan2 A

= sin A(1+ tan2 A)

= sin A · sec2 A

= sin A ·

1cos2 A

= tan A sec A

≡ RHS.


Level-One-Games 155

1−2 sin2 A ≡ 2 cos2 A−1


1. s2, c2 suggest s2 + c2 ≡ 12. rearrange and simplify.

LHS = 1−2 sin2 A

= 1− sin2 A− sin2 A

= cos2 A− sin2 A

= cos2 A+(cos2 A−1)

= 2 cos2 A−1

≡ RHS.


156 Trig or Treat

(sin A+ cos A)2 +(sin A− cos A)2≡ 2


1. ( )2 suggests expansion2. s2 + c2 ≡ 13. rearrange and simplify.

LHS = (sin A+ cos A)2 +(sin A− cos A)2

= (sin2 A+2 sin A cos A+ cos2 A)

+(sin2 A−2 sin A cos A+ cos2 A)

= 2(sin2 A+ cos2 A)

= 2(1)

= 2

≡ RHS.


Level-One-Games 157

sec4 A− sec2 A ≡ tan4 A+ tan2 A


1. sec = 1/cos, t = s/c2. s2 + c2 ≡ 13. rearange and simplify.

LHS = sec4 A− sec2 A

= sec2 A(sec2 A−1)

= sec2 A tan2 A

= (tan2 A+1) tan2 A

= tan4 A+ tan2 A

≡ RHS.


158 Trig or Treat

cosec4 A− cosec2 A ≡ cot4 A+ cot2 A


1. cosec = 1/sin, cot = c/s2. s2 + c2 ≡ 13. rearrange and simplify.

LHS = cosec4 A− cosec2 A

= cosec2 A(cosec2 A−1)

= cosec2 A(cot2A)

= (1+ cot2 A)(cot2 A)

= cot2 A+ cot4 A

≡ RHS.


Level-One-Games 159

cosec Acot A+ tan A

≡ cos A


1. cosec = 1/sin, cot = c/s, t = s/c2. rearrange and simplify.

LHS =cosec A

cot A+ tan A

=1

sin A

1cos Asin A

+sin Acos A

=1

sin A

(

sin Acos A

cos2 A+ sin2 A

)

=1

sin A

(

sin A cos A1

)

= cos A

≡ RHS.


160 Trig or Treat

1+ sec A1+ cos A

≡ sec A



LHS =1+ sec A1+ cos A

=

(

1+1

cos A

)(

11+ cos A

)

=

(

cos A+1cos A

)(

11+ cos A

)

=1

cos A

= sec A

≡ RHS.


Level-One-Games 161

sin A1+ cos A

+1+ cos A

sin A≡

2sin A


1. common denominator2. s2 + c2 ≡ 13. rearrange and simplify.

LHS =sin A

1+ cos A+

1+ cos Asin A

=sin2 A+(1+ cos A)2

(1+ cos A)(sin A)

=sin2 A+(1+2 cos A+ cos2 A)

(1+ cos A)(sin A)

=2+2 cos A

(1+ cos A)(sin A)

=2

sin A

≡ RHS.


162 Trig or Treat

sec A+ cosec A1+ tan A

≡ cosec A


1. sec = 1/cos, cosec = 1/sin, t = s/c2. rearrange and simplify.

LHS =sec A+ cosec A

1+ tan A

=

(

1cos A

+1

sin A

)

1

1+sin Acos A

=(sin A+ cos A)

cos A sin A

(

cos Acos A+ sin A

)

=1

sin A

= cosec A

≡ RHS.


Level-One-Games 163

1tan A+ cot A

≡

sin Asec A


1. t = s/c, cot = c/s, sec = 1/cos2. common denominator3. rearrange and simplify.

LHS =1

tan A+ cot A

=1

sin Acos A

+cos Asin A

=cos A sin A

sin2 A+ cos2 A

= cos A sin A

=sin Asec A

≡ RHS.


164 Trig or Treat

1+ sin Acos A

+cos A

1+ sin A≡

2cos A


1. common denominator2. s2 + c2 ≡ 13. rearrange and simplify.

LHS =1+ sin A

cos A+

cos A1+ sin A

=(1+ sin A)2 + cos2 A

cos A(1+ sin A)

=1+2 sin A+ sin2 A+ cos2 A

cos A(1+ sin A)

=2+2 sin A

cos A(1+ sin A)

=2

cos A

≡ RHS.


Level-One-Games 165

1− cos Asin A

≡

sin A1+ cos A


1. 1 suggests s2 + c2 ≡ 12. rearrange and simplify.

LHS =1− cos A

sin A

=1− cos A

sin A·

(

1+ cos A1+ cos A

)

=1− cos2 A

sin A(1+ cos A)

=sin2 A

sin A(1+ cos A)

=sin A

1+ cos A

≡ RHS.


166 Trig or Treat

1tan A+ cot A

≡ sin A cos A


1. t = s/c, cot = c/s2. rearrange and simplify.

LHS =1

tan A+ cot A

=1

sin Acos A

+cos Asin A

=cos A sin A

sin2 A+ cos2 A

=sin A cos A

1

= sin A cos A

≡ RHS.


Level-One-Games 167

(cosec A− sin A)(sec A− cos A)(tan A+ cot A) ≡ 1


1. cosec = 1/sin, sec = 1/cos, t = s/c, cot = c/s2. common denominator3. rearrange and simplify.

LHS = (cosec A− sin A)(sec A− cos A)(tan A+ cot A)

=

(

1sin A

− sin A

)(

1cos A

− cos A

)(

sin Acos A

+cos Asin A

)

=

(

(1− sin2 A)

sin A

)(

1− cos2 Acos A

)(

sin2 A+ cos2 Acos A sin A

)

=

(

cos2 Asin A

)(

sin2 Acos A

)(

1cos A sin A

)

= 1

≡ RHS.


168 Trig or Treat

1−cos2 A

1+ sin A≡ sin A


1. Common denominator2. c2 suggests s2 + c2 = 13. rearrange and simplify.

LHS = 1−cos2 A

1+ sin A

=1+ sin A− cos2 A

1+ sin A

=sin2 A+ sin A

1+ sin A

=sin A(1+ sin A)

(1+ sin A)

= sin A

≡ RHS.


Level-One-Games 169

1−2 sin2 Asin A cos A

≡ cot A− tan A


1. cot = c/s, t = s/c2. s2 suggests s2 + c2 ≡ 13. rearrange and simplify.

LHS =1−2 sin2 Asin A cos A

=1− sin2 A− sin2 A

sin A cos A

=cos2 A− sin2 A

sin A cos A

=cos Asin A

−

sin Acos A

= cot A− tan A

≡ RHS.


170 Trig or Treat

sin A+ cos Asin A

−

cos A− sin Acos A

≡ cosec A sec A


1. cosec = 1/sin, sec A = 1/cos A2. common denominator3. s2 + c2 ≡ 14. rearrange and simplify.

LHS =sin A+ cos A

sin A−

cos A− sin Acos A

=(sin A cos A+ cos2 A)− (sin A cos A− sin2 A)

sin A cos A

=cos2 A+ sin2 A

sin A cos A

=1

sin A cos A

= cosec A sec A

≡ RHS.


Level-One-Games 171

tan A+ tan Bcot A+ cot B

≡ tan A tan B


1. cot = 1/ tan2. since both LHS and RHS are in terms of tan and cot, it may be easier to leave

the terms in tan rather than convert them to sin and cos.3. common denominator4. rearrange and simplify.

LHS =tan A+ tan Bcot A+ cot B

=tan A+ tan B

1tan A

+1

tan B

=(tan A+ tan B)

(

tan B+ tan Atan A tan B

)

= (tan A+ tan B)

(

tan A tan Btan A+ tan B

)

= tan A tan B

≡ RHS.


172 Trig or Treat

1+ sin A1− sin A

−

1− sin A1+ sin A

≡ 4 tan A sec A


1. t = s/c, sec = 1/cos2. s2 + c2 ≡ 13. rearrange and simplify.


−

1− sin A1+ sin A

=(1+ sin A)2

− (1− sin A)2

(1− sin A)(1+ sin A)

=(1+2 sin A+ sin2 A)− (1−2 sin A+ sin2 A)

1− sin2 A

=4 sin Acos2 A

= 4 tan A sec A

≡ RHS.


Level-One-Games 173

sin A+ cos Acos A

−

sin A− cos Asin A

≡ sec A cosec A


1. sec = 1/cos, cosec = 1/sin2. common denominator3. s2 + c2 ≡ 14. rearrange and simplify.

LHS =sin A+ cos A

cos A−

sin A− cos Asin A

=(sin2 A+ sin A cos A)− (sin A cos A− cos2 A)

cos A sin A

=sin2 A+ cos2 A

cos A sin A

=1

cos A sin A

= sec A cosec A

≡ RHS.


174 Trig or Treat

1−sin2 A

1− cos A≡−cos A



LHS = 1−sin2 A

1− cos A

= 1−(1− cos2 A)

(1− cos A)

= 1−(1− cos A)(1+ cos A)

(1− cos A)

= 1− (1+ cos A)

= −cos A

≡ RHS.


Level-One-Games 175

1−2 cos2 Asin A cos A

≡ tan A− cot A


1. c2 suggests s2 + c2 ≡ 12. t = s/c, cot = c/s3. rearrange and simplify.

LHS =1−2 cos2 Asin A cos A

=(sin2 A+ cos2 A)−2 cos2 A

sin A cos A

=sin2 A− cos2 A

sin A cos A

=sin Acos A

−

cos Asin A

= tan A− cot A

≡ RHS.


176 Trig or Treat

11− cos A

+1

1+ cos A≡ 2 cosec2 A


1. common denominator2. cosec = 1/sin3. cosec2 suggests s2 + c2 ≡ 14. rearrange and simplify.

LHS =1

1− cos A+

11+ cos A

=1+ cos A+1− cos A(1− cos A)(1+ cos A)

=2

1− cos2 A

=2

sin2 A

= 2 cosec2 A

≡ RHS.


Level-One-Games 177

1sin A+1

−

1sin A−1

≡ 2 sec2 A


1. sec = 1/cos2. sec2 suggests s2 + c2 ≡ 13. rearrange and simplify.

LHS =1

sin A+1−

1sin A−1

=(sin A−1)− (sin A+1)

(sin A+1)(sin A−1)

=−2

sin2 A−1

=−2

−cos2 A

= 2 sec2 A

≡ RHS.


178 Trig or Treat

tan A(1+ cot2 A)

1+ tan2 A≡ cot A


1. cot2, tan2 suggest s2 + c2 ≡ 12. cot = c/s, t = s/c3. rearrange and simplify.

LHS =tan A(1+ cot2 A)

1+ tan2 A

=sin Acos A

(cosec2 A)

(sec2 A)

=sin Acos A

(

1

sin2 A

)

·

(

cos2 A1

)

=cos Asin A

= cot A

≡ RHS.


Level-One-Games 179

1− sec2 A(1− cos A)(1+ cos A)

≡−sec2 A


1. sec2 suggests s2 + c2 ≡ 12. sec = 1/cos3. rearrange and simplify.

LHS =1− sec2 A

(1− cos A)(1+ cos A)

=− tan2 A

(1− cos2 A)

= −

sin2 Acos2 A

·

(

1

sin2 A

)

= −

1cos2 A

= −sec2 A

≡ RHS.


180 Trig or Treat

cot A(1+ tan2 A)

1+ cot2 A≡ tan A


1. tan2, cot2 suggest s2 + c2 ≡ 12. cot = c/s, t = s/c3. rearrange and simplify.

LHS =cot A(1+ tan2 A)

(1+ cot2 A)

=cos Asin A

(sec2 A)

(

1cosec2 A

)

=cos Asin A

·

(

1cos2 A

)

·

(

sin2 A1

)

=sin Acos A

= tan A

≡ RHS.


Level-One-Games 181

1cos2 A(1+ tan2 A)

≡ 1


1. c2, t2 suggest s2 + c2 = 12. t = s/c3. common denominator4. rearrange and simplify.

LHS =1

cos2 A(1+ tan2 A)

=1

cos2 A(sec2 A)

=1

cos2 A

(

1cos2 A

)

= 1

≡ RHS.


182 Trig or Treat

(tan A+ cot A)2− (tan A− cot A)2

≡ 4


1. cot = 1/ tan2. expansion of squares3. ( )2 suggests s2 + c2 ≡ 14. rearrange and simplify.

LHS = (tan A+ cot A)2− (tan A− cot A)2

= (tan2 A+2 tan A cot A+ cot2 A)− (tan2 A−2 tan A cot A+ cot2 A)

= 2tan A cot A+2 tan A cot A

= 2+2

= 4

≡ RHS.


Level-One-Games 183


= sin2 A− cos2 A


1. cot2, s2, c2 suggest s2 + c2 ≡ 12. cot = c/s3. rearrange and simplify.

LHS =1− cot2 A1+ cot2 A

=1− cot2 Acosec2 A

=

(

1−cos2 A

sin2 A

)(

sin2 A1

)

=(sin2 A− cos2 A)(sin2 A)

sin2 A

= sin2 A− cos2 A

≡ RHS.


184 Trig or Treat

cos2 A+ cot2 A cos2 A ≡ cot2 A


1. cot = c/s2. c2, cot2 suggest s2 + c2 ≡ 13. rearrange and simplify.

LHS = cos2 A+ cot2 A cos2 A

= cos2 A(1+ cot2 A)

= cos2 A(cosec2 A)

= cos2 A ·

1

sin2 A

= cot2 A

≡ RHS.


Level-One-Games 185

(1+ tan A)2 +(1− tan A)2≡ 2sec2A


1. t2, sec2 suggest s2 + c2 ≡ 12. rearrange and simplify.

LHS = (1+ tan A)2 +(1− tan A)2

= (1+2 tan A+ tan2 A)+(1−2 tan A+ tan2 A)

= 2+2 tan2 A

= 2(1+ tan2 A)

= 2 sec2 A

≡ RHS.


186 Trig or Treat

cosec2 A+ sec2 A ≡ cosec2 A sec2 A


1. cosec2, sec2 suggest s2 + c2 ≡ 12. cosec = 1/sin, sec = 1/cos3. rearrange and simplify.

LHS = cosec2 A+ sec2 A

=1

sin2 A+

1cos2 A

=cos2 A+ sin2 A

sin2 A cos2 A

=1

sin2 A cos2 A

= cosec2 A sec2 A

≡ RHS.


Level-One-Games 187

2 sec2 A−2 sec2 A sin2 A− sin2 A− cos2 A ≡ 1


1. sec2, s2, c2 suggest s2 + c2 ≡ 12. sec = 1/cos3. rearrange and simplify.

LHS = 2 sec2 A−2 sec2 A sin2 A− sin2 A− cos2 A

= 2 sec2 A(1− sin2 A)− (sin2 A+ cos2 A)

= 21

cos2 A· (cos2 A)− (1)

= 2−1

= 1

≡ RHS.


188 Trig or Treat

cot4 A+ cot2 A ≡ cosec4 A− cosec2 A


1. cot = c/s, cosec = 1/sin2. cot2, cosec2 suggest s2 + c2 ≡ 13. rearrange and simplify.

LHS = cot4 A+ cot2 A

= cot2 A(cot2 A+1)

= cot2 A(cosec2 A)

= (cosec2 A−1)(cosec2 A)

= cosec4 A− cosec2 A

≡ RHS.


Level-One-Games 189


≡ sin2 A− cos2 A


1. t = s/c, cot = c/s2. common denominator3. s2 + c2 ≡ 14. rearrange and simplify.

LHS =tan A− cot Atan A+ cot A

=

sin Acos A

−

cos Asin A

sin Acos A

+cos Asin A

=

sin2 A− cos2 Acos A sin A


=sin2 A− cos2 A

sin2 A+ cos2 A

= sin2 A− cos2 A

≡ RHS.


190 Trig or Treat

sec A− cos Asec A+ cos A

≡

sin2 A1+ cos2 A


1. sec = 1/cos2. common denominator3. s2 + c2 ≡ 14. rearrange and simplify.

LHS =sec A− cos Asec A+ cos A

=

1cos A

− cos A

1cos A

+ cos A

=

1− cos2 Acos A

1+ cos2 Acos A

=1− cos2 A1+ cos2 A

=sin2 A

1+ cos2 A

≡ RHS.


Level-One-Games 191

sin3 A+ cos3 A1−2 cos2 A

≡

sec A− sin Atan A−1


1. (s3 + c3) suggests (s+ c)(s2 − sc+ c2)2. (1−2 cos2 A) suggests (s2 − c2) and (s+ c)(s− c)3. sec = 1/cos, t = s/c4. rearrange and simplify.

LHS =sin3 A+ cos3 A

1−2 cos2 A


sin2 A− cos2 A

=(sin A+ cos A)(1− sin A cos A)

(sin A+ cos A)(sin A− cos A)

=1− sin A cos Asin A− cos A

=sec A− sin A

tan A−1

≡ RHS.

divide bothnumeratorand denominatorby cos A


192 Trig or Treat

sec2 A− tan2 A+ tan Asec A

≡ sin A+ cos A


1. sec = 1/cos, t = s/c2. s2 + c2 ≡ 13. rearrange and simplify.

LHS =sec2

− tan2 A+ tan Asec A

=(1+ tan2 A)− tan2 A+ tan A

sec A

=1+ tan A

sec A

=1

sec A+

sin Acos A

·

cos A1

= cos A+ sin A

≡ RHS.


Level-One-Games 193

1−cos2 A

1+ sin A≡ sin A



LHS = 1−cos2 A

1+ sin A

= 1−(1− sin2 A)

1+ sin A

= 1−(1+ sin A)(1− sin A)

(1+ sin A)

= 1− (1− sin A)

= sin A

≡ RHS.


194 Trig or Treat

(sec A− tan A)(sec A+ tan A) ≡ 1


1. (sec− tan)(sec+ tan) suggests (a−b)(a+b) = a2 −b2


LHS = (sec A− tan A)(sec A+ tan A)

= sec2− tan2 A

= (1+ tan2 A)− tan2 A

= 1

≡ RHS.


Level-One-Games 195

sin A+ sin A cot2 A ≡ cosec A


1. cot2 suggests s2 + c2 ≡ 12. cosec = 1/sin3. rearrange and simplify.

LHS = sin A+ sin A cot2 A

= sin A(1+ cot2 A)

= sin Acosec2 A

= sin A ·

1

sin2 A

=1

sin A

= cosec A

≡ RHS.


196 Trig or Treat

(1− sin A)(sec A+ tan A) ≡ cos A


1. sec = 1/cos, t = s/c2. common denominator3. rearrange and simplify.

LHS = (1− sin A)(sec A+ tan A)

= (1− sin A)

(

1cos A

+sin Acos A

)

= (1− sin A)

(

1+ sin Acos A

)

=1− sin2 A

cos A

=cos2 Acos A

= cos A

≡ RHS.


Level-One-Games 197

tan A sin A ≡ sec A− cos A


1. t = s/c, sec = 1/cos2. rearrange and simplify.

LHS = tan A sin A

=sin Acos A

· sin A

=sin2 Acos A

=(1− cos2 A)

cos A

=1

cos A− cos A

= sec A− cos A

≡ RHS.


198 Trig or Treat

tan A(sin A+ cot A cos A) ≡ sec A


1. t = s/c, cot = c/s, cosec = 1/sin2. simplify.

LHS = tan A(sin A+ cot A cos A)

=sin Acos A

(

sin A+cos Asin A

· cos A

)

=sin2 Acos A

+ cos A

=sin2 A+ cos2 A

cos A

=1

cos A

= sec A

≡ RHS.


Level-One-Games 199

(1− cos A)(1+ sec A) ≡ sin A tan A



LHS = (1− cos A)(1+ sec A)

= (1− cos A)

(

1+1

cos A

)

= (1− cos A)

(

1+ cos Acos A

)

=1− cos2 A

cos A

=sin2 Acos A

= sin A tan A

≡ RHS.


200 Trig or Treat

(1− sin A)(1+ cosec A) ≡ cos A cot A



LHS = (1− sin A)(1+ cosec A)

= 1− sin A+ cosec A− sin A · cosec A

= cosec A− sin A

=1

sin A− sin A

=1− sin2 A

sin A

=cos2 Asin A

= cos A cot A

≡ RHS.

sin A · cosec A= 1


Level-One-Games 201

2 sec2 A−1 ≡ 1+2 tan2 A


1. sec2, t2 suggest s2 + c2 ≡ 12. rearrange and simplify.

LHS = 2 sec2 A−1

= 2(1+ tan2 A)−1

= 2+2 tan2 A−1

= 1+2 tan2 A

≡ RHS.


202 Trig or Treat

(sinA−cosB)2+(cosB+sinA)(cosB−sinA)=−2cosB(sinA−cosB)


1. Although the identity looks lengthy and complicated, closer inspection showsthat the term (sin A− cos B) is common on both sides of the equation.

2. Rearrange and simplify.

LHS = (sin A− cos B)2 +(cos B+ sin A)(cos B− sin A)

= (sin A− cos B)2− (sin A− cos B)(cos B+ sin A)

= (sin A− cos B)((sin A− cos B)− (cos B− sin A))

= (sin A− cos B)(−2 cos B)

= −2 cos B(sin A− cos B)

≡ RHS.


Level-One-Games 203

sec A1+ sec A

≡

1− cos A

sin2 A



LHS =sec A

1+ sec A

=1

cos A·

1

1+1

cos A

=1

cos A·

cos A(cos A+1)

=1

cos A+1

=

(

1− cos A1− cos A

)

·

1(1+ cos A)

=1− cos A1− cos2 A

=1− cos A

sin2 A

≡ RHS.


204 Trig or Treat

1+ sec Atan A+ sin A

≡ cosec A


1. sec = 1/cos, t = s/c, cosec = 1/sin2. rearrange and simplify.

LHS =1+ sec A

tan A+ sin A

=

(

1+1

cos A

)

1sin Acos A

+ sin A

=

(

cos A+1cos A

)(

cos Asin A+ sin A cos A

)

=cos A+1

sin A(1+ cos A)

=1

sin A

= cosec A

≡ RHS.


Level-One-Games 205

cos Acosec A−1

+cos A

cosec +1≡ 2 tan A


1. common denominator2. t = s/c, cosec = 1/sin3. rearrange and simplify.

LHS =cos A

cosec A−1+

cos Acosec A+1

=cos A(cosec A+1)+ cos A(cosec A−1)

(cosec A−1)(cosec A+1)

=(cot A+ cos A)+(cot A− cos A)

cosec2 A−1

=2 cot Acot2 A

|cosec2−1 = cot2

=2

cot A

= 2 tan A

≡ RHS.


206 Trig or Treat

cos A1− tan A

+sin A

1− cot A≡ sin A+ cos A


1. t = s/c, cot = c/s2. common denominator3. rearrange and simplify.

LHS =cos A

1− tan A+

sin A1− cot A

=cos A

(

1−sin Acos A

) +sin A

(

1−cos Asin A

)

= cos A ·

cos A(cos A− sin A)

+ sin A ·

sin A(sin A− cos A)

=cos2 A− sin2 Acos A− sin A

=(cos A+ sin A)(cos A− sin A)

(cos A− sin A)

= cos A+ sin A

≡ RHS.


Level-One-Games 207

sec A− cos Atan A

≡ sin A



LHS =sec A− cos A

tan A

=

(

1cos A

− cos A

)(

cos Asin A

)

=

(

1− cos2 Acos A

)(

cos Asin A

)

=1− cos2 A

sin A

=sin2 Asin A

= sin A

≡ RHS.


208 Trig or Treat

tan A+ sec A−1tan A− sec A+1

≡ tan A+ sec A


1. s2 + c2 ≡ 1tan2 +1 ≡ sec2

1 ≡ sec2− tan2


LHS =tan A+ sec A−1tan A− sec A+1

=tan A+ sec A− (sec2 A− tan2 A)

tan A− sec A+1

=(tan A+ sec A)− (sec A+ tan A)(sec A− tan A)

tan A− sec A+1

=(sec A+ tan A)(1− sec A+ tan A)

tan A− sec A+1

=(sec A+ tan A)(tan A− sec A+1)

(tan A− sec A+1)

= tan A+ sec A

≡ RHS.


Level-One-Games 209

sin A− cos A+1sin A+ cos A−1

≡

sin A+1cos A


1. s2 + c2 ≡ 12. multiple terms in denominator; may have to use s2 + c2 = 1 to arrive at com-

mon factor with numerator3. rearrange and simplify.

LHS =sin A− cos A+1sin A+ cos A−1

=sin A− cos A+1sin A+ cos A−1

·

(

sin A+1sin A+1

)

=(sin A− cos A+1)(sin A+1)

(sin2 A+ sin A cos A− sin A)+(sin A+ cos A−1)


−cos2 A+ sin A cos A+ cos A


cos A(−cos A+ sin A+1)

=sin A+1

cos A

≡ RHS.


210 Trig or Treat

sec A+ tan Acot A+ cos A

≡ tan A sec A


1. sec = 1/cos, t = s/c, cot = c/s2. common denominator3. rearrange and simplify.

LHS =sec A+ tan Acot A+ cos A

=

1cos A

+sin Acos A

cos Asin A

+ cos A

=

1+ sin Acos A

cos A+ cos A sin Asin A

=(1+ sin A)

cos A·

sin Acos A(1+ sin A)

=sin Acos A

·

1cos A

= tan A sec A

≡ RHS.


Level-One-Games 211

1+ sin A+ cos A1+ sin A− cos A

≡

1+ cos Asin A


1. multiple term in denominator; may have to find common factor forcancellation.


LHS =1+ sin A+ cos A1+ sin A− cos A

=1+ sin A+ cos A1+ sin A− cos A

·

(

1+ cos A1+ cos A

)

=(1+ sin A+ cos A)(1+ cos A)

(1+ sin A− cos A)+(cos A+ sin A cos A− cos2 A)


sin2 A+ sin A+ sin A cos A


sin A(sin A+1+ cos A)

=1+ cos A

sin A

≡ RHS.


212 Trig or Treat

1+ cos Asin A

+sin A

1+ cos A≡ 2 cosec A


1. common denominator2. cosec = 1/sin3. rearrange and simplify.

LHS =1+ cos A

sin A+

sin A1+ cos A

=(1+ cos A)2 + sin2 A

sin A(1+ cos A)

=1+2 cos A+ cos2 A+ sin2 A

sin A(1+ cos A)

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

=2(1+ cos A)

sin A(1+ cos A)

=2

sin A

= 2 cosec A

≡ RHS.


Level-One-Games 213

sec Asin A

−

sin Acos A

≡ cot A


1. sec = 1/cos, cot = c/s2. common denominator3. rearrange and simplify.

LHS =sec Asin A

−

sin Acos A

=1

cos A·

1sin A

−

sin Acos A

=1− sin2 A

cos A sin A

=cos2 A

cos A sin A

=cos Asin A

= cot A

= RHS.


214 Trig or Treat

1− sin Acos A

+cos A

1− sin A≡ 2 sec A



LHS =1− sin A

cos A+

cos A1− sin A

=(1− sin A)2 + cos2 A

cos A(1− sin A)

=(1− sin A)2 +(1− sin2 A)

cos A(1− sin A)

=(1− sin A)2 +(1− sin A)(1+ sin A)

cos A(1− sin A)

=(1− sin A)(1− sin A+1+ sin A)

cos A(1− sin A)

=2

cos A

= 2 sec A

≡ RHS.


Level-One-Games 215

cosec A1+ cosec A

≡

1− sin Acos2 A


1. cosec = 1/sin2. common denominator3. s2 + c2 ≡ 14. rearrange and simplify.

LHS =cosec A

1+ cosec A

=1

sin A

1

1+1

sin A

=1

sin A

(

sin Asin A+1

)

=1

1+ sin A

=1

(1+ sin A)

(

1− sin A1− sin A

)

=1− sin A

1− sin2 A

=1− sin A

cos2 A

≡ RHS.


216 Trig or Treat

cos A1+ sin A

+1+ sin A

cos A≡ 2 sec A



LHS =cos A

1+ sin A+

1+ sin Acos A

=cos2 A+(1+ sin A)2

cos A(1+ sin A)

=(1− sin2 A)+(1+ sin A)2

cos A(1+ sin A)

=(1+ sin A)(1− sin A)+(1+ sin A)2

cos A(1+ sin A)

=(1+ sin A)(1− sin A+1+ sin A)

cos A(1+ sin A)

=2

cos A

= 2 sec A

≡ RHS.


Level-One-Games 217

1+ cos A+ sin A1+ cos A− sin A

≡ sec A+ tan A


1. multiple terms in denominator suggest use of common factor for cancellation2. sec+ tan in RHS suggest 1/cos+ sin/cos which gives (1+ sin)/cos3. s2 + c2 ≡ 14. rearrange and simplify.

LHS =1+ cos A+ sin A1+ cos A− sin A

=1+ cos A+ sin A1+ cos A− sin A

·

(

1+ sin A1+ sin A

)

=(1+ cos A+ sin A)(1+ sin A)

1+ cos A− sin A+ sin A+ sin A cos A− sin2 A


cos2 A+ cos A+ sin A cos A


cos A(cos A+1+ sin A)

=1+ sin A

cos A

= sec A+ tan A

≡ RHS.


218 Trig or Treat

sec A− tan A ≡

cos A1+ sin A


1. sec = 1/cos, t = s/c2. s2 + c2 ≡ 13. rearrange and simplify.

LHS = sec A− tan A

=1

cos A−

sin Acos A

=1− sin A

cos A

=1− sin A

cos A

(

1+ sin A1+ sin A

)

=1− sin2 A

cos A(1+ sin A)

=cos2 A

cos A(1+ sin A)

=cos A

1+ sin A

≡ RHS.


Level-One-Games 219

cosec A− cot A ≡

sin A1+ cos A



LHS = cosec A− cot A

=1

sin A−

cos Asin A

=1− cos A

sin A

=1− cos A

sin A

(

1+ cos A1+ cos A

)

=1− cos2 A

sin A(1+ cos A)

=sin2 A

sin A(1+ cos A)

=sin A

1+ cos A

≡ RHS.


220 Trig or Treat

cot A1− tan A

+tan A

1− cot A≡ 1+ sec A cosec A


1. cot = c/s, t = s/c, sec = 1/cos, cosec = 1/sin2. rearrange and simplify.

LHS =cot A

1− tan A+

tan A1− cot A

=cos Asin A

1(

1−sin Acos A

) +sin Acos A

·

1(

1−cos Asin A

)

=cos Asin A

·

cos Acos A− sin A

+sin Acos A


=cos2 A

sin A(cos A− sin A)−

sin2 Acos A(cos A− sin A)

=cos3 A− sin3 A

(cos A− sin A)(sin A cos A)

=(cos A− sin A)(cos2 A+ cos A sin A+ sin2 A)

(cos A− sin A)(sin A cos A)

=(1+ cos A sin A)

sin A cos A

= cosec A sec A+1

≡ RHS.


Level-One-Games 221

sin A1− cos A

+1− cos A

sin A≡ 2 cosec A


1. cosec = 1/sin2. common denominator3. rearrange and simplify.

LHS =sin A

1− cos A+

1− cos Asin A

=sin2 A+(1− cos A)2

sin A(1− cos A)

=sin2 A+(1−2 cos A+ cos2 A)

sin A(1− cos A)

=sin2 A+ cos2 A+1−2 cos A

sin A(1− cos A)

=2−2 cos A

sin A(1− cos A)

=2(1− cos A)

sin A(1− cos A)

=2

sin A

= 2 cosec A

≡ RHS.


222 Trig or Treat

tan Asec A−1

≡

sec A+1tan A


1. t = s/c, sec = 1/cos c2. s2 + c2 ≡ 13. common denominator4. rearrange and simplify.

LHS =tan A

sec A−1

=sin Acos A

11

cos A−1

=sin Acos A

(

cos A1− cos A

)

=sin A

1− cos A

=sin A

(1− cos A)

(

1+ cos A1+ cos A

)

=sin A+ sin A cos A

1− cos2 A

=sin A+ sin A cos A

sin2 A

=1+ cos A

sin A

=sec A+1

tan A

≡ RHS.


Level-One-Games 223

tan A+cos A

1+ sin A≡ sec A


1. t = s/c, sec = 1/cos2. common denominator3. rearrange and simplify.

LHS = tan A+cos A

1+ sin A

=sin Acos A

+cos A

1+ sin A

=sin A(1+ sin A)+ cos2 A

cos A(1+ sin A)

=sin A+ sin2 A+ cos2 A

cos A(1+ sin A)

=(1+ sin A)

cos A(1+ sin A)

=1

cos A

= sec A

≡ RHS.


224 Trig or Treat

cot A1− tan A

+tan A

1− cot A≡ 1+ tan A+ cot A


1. cot = c/s, t = s/c2. common denominator3. s2 + c2 ≡ 14. rearrange and simplify.

LHS =cot A

1− tan A+

tan A1− cot A

=

cos Asin A

1−sin Acos A

+

sin Acos A

1−cos Asin A

=cos Asin A

·

cos A(cos A− sin A)

+sin Acos A

·


=cos3 A− sin3 A

sin Acos A(cos A− sin A)

=(cos A− sin A)(cos2 A+ sin Acos A+ sin2 A)

(cos A− sin A)(sin Acos A)

= cot A+1+ tan A

= 1+ tan A+ cot A

≡ RHS.


Level-One-Games 225

cosec A−1cot A

≡

cot Acosec A+1



LHS =cosec A−1

cot A

=

1sin A

−1

cos Asin A

=1− sin A

sin A·

sin Acos A

=1− sin A

cos A

=1− sin A

cos A

(

1+ sin A1+ sin A

)

=1− sin2 A

cos A(1+ sin A)

=cos2 A

cos A(1+ sin A)

=cos A

1+ sin A

=cot A

cosec A+1

≡ RHS.



226 Trig or Treat

cosec A1− cos A

≡

1+ cos A

sin3 A


1. cosec = 1/sin2. sin3 suggests s · s2, s2 + c2 = 13. rearrange and simplify.

LHS =cosec A

1− cos A

=1

sin A·

1(1− cos A)

=1

sin A·

1(1− cos A)

·

(

1+ cos A1+ cos A

)

=1+ cos A

sin A(1− cos2 A)

=1+ cos A

sin A(sin2 A)

=1+ cos A

sin3 A

≡ RHS.


Level-One-Games 227

1−sin2 A

1+ cos A≡ cos A


1. s2 suggests s2 + c2 ≡ 12. rearrange and simplify.

LHS = 1−sin2 A

1+ cos A

=(1+ cos A)− sin2 A

1+ cos A

=cos2 A+ cos A

1+ cos A

=cos A(cos A+1)

1+ cos A

= cos A

≡ RHS.


228 Trig or Treat

5−10 cos2 Asin A− cos A

≡ 5(sin A+ cos A)



LHS =5−10 cos2 Asin A− cos A

=5(1−2 cos2 A)

sin A− cos A

=5(cos2 A+ sin2 A−2 cos2 A)

sin A− cos A

=5(sin2 A− cos2 A)

sin A− cos A

=5(sin A+ cos A)(sin A− cos A)

(sin A− cos A)

= 5(sin A+ cos A)

≡ RHS.


Level-One-Games 229

1+ sin A1− sin A

≡ (sec A+ tan A)2


1. (sec+ tan)2 suggests s2 + c2 ≡ 12. sec = 1/cos, t = s/c3. rearrange and simplify.


=1+ sin A1− sin A

(

1+ sin A1+ sin A

)

=(1+ sin A)2

1− sin2 A

=(1+ sin A)2

cos2 A

=

(

1+ sin Acos A

)2

= (sec A+ tan A)2

≡ RHS.


230 Trig or Treat

(sec A− tan A)2≡

1− sin A1+ sin A


1. sec = 1/cos, t = s/c2. ( )2 suggests s2 + c2 ≡ 13. common denominator4. rearrange and simplify.

LHS = (sec A− tan A)2

=

(

1cos A

−

sin Acos A

)2

=(1− sin A)2

cos2 A

=(1− sin A)2

(1− sin2 A)

=(1− sin A)(1− sin A)


=1− sin A1+ sin A

≡ RHS.


Level-One-Games 231

1− cos A1+ cos A

≡ (cosec A− cot A)2


1. cosec = 1/sin, cot = c/s2. ( )2 suggests s2 + c2 ≡ 13. rearrange and simplify.

LHS =1− cos A1+ cos A

=(1− cos A)

(1+ cos A)·

(


)

=(1− cos A)2

(1− cos2 A)

=(1− cos A)2

sin2 A

=

(

1sin A

−

cos Asin A

)2

= (cosec A− cot A)2

≡ RHS.


232 Trig or Treat

1+cos2 A

sin A−1≡−sin A



LHS = 1+cos2 A

sin A−1

=(sin A−1)+ cos2 A

sin A−1

=(sin A− sin2 A− cos2 A)+ cos2 A

sin A−1

=sin A− sin2 A

sin A−1

=sin A(1− sin A)

sin A−1

= −sin A

≡ RHS.


Level-One-Games 233

(1+ cot A)2 +(1− cot A)2≡

2

sin2 A


1. ( )2, s2 suggest s2 + c2 ≡ 12. cot = c/s3. rearrange and simplify.

LHS = (1+ cot A)2 +(1− cot A)2

= (1+2 cot A+ cot2 A)+(1−2 cot A+ cot2 A)

= 2+2 cot2 A

= 2(1+ cot2 A)

= 2 cosec2 A

=2

sin2 A

≡ RHS.


234 Trig or Treat

sin A cos Bcos A sin B

(cot A cot B)+1 ≡

1

sin2 B


1. cot = c/s2. s2 suggests s2 + c2 ≡ 13. rearrange and simplify.

LHS =sin A cos Bcos A sin B

(cot A cot B)+1

=sin A cos Bcos A sin B

·

cos Asin A

·

cos Bsin B

+1

=cos2 B

sin2 B+1

=cos2 B+ sin2 B

sin2 B

=1

sin2 B

≡ RHS.


Level-One-Games 235

cosec A(cosec A− sin A)+sin A− cos A

sin A+ cot A ≡ cosec2 A



LHS = cosec A(cosec A− sin A)+sin A− cos A

sin A+ cot A

=1

sin A

(

1sin A

− sin A

)

+sin A− cos A

sin A+

cos Asin A

=1− sin2 A

sin A · sin A+

sin A− cos Asin A

+cos Asin A

=1− sin2 A+ sin2 A− cos A sin A+ cos A sin A

sin2 A

=1

sin2 A

= cosec2 A

≡ RHS.


236 Trig or Treat

sec4 A− tan4 A ≡

1+ sin2 Acos2 A


1. sec = 1/cos, t = s/c2. sec4− tan4 suggests a4 −b4 = (a2 −b2)(a2 +b2)3. s2, c2 suggest s2 + c2 ≡ 14. rearrange and simplify.

LHS = sec4 A− tan4 A

=1

cos4 A−

sin4 Acos4 A

=1− sin4 A

cos4 A

=(1− sin2 A)(1+ sin2 A)

cos4 A

=(cos2 A)(1+ sin2 A)

cos4 A

=1+ sin2 A

cos2 A

≡ RHS.


Level-One-Games 237

1− sin Asec A

≡

cos3 A1+ sin A


1. sec = 1/cos2. cos3 suggests c · c2, s2 + c2 ≡ 13. rearrange and simplify.

LHS =1− sin A

sec A

= (1− sin A)cos A

= (1− sin A)cos A

(

1+ sin A1+ sin A

)

=(1− sin2 A)cos A

(1+ sin A)

=(cos2 A)cos A

1+ sin A

=cos3 A

1+ sin A

≡ RHS.


238 Trig or Treat

(1− cos A)

(1+ cos A)≡ (cosec A− cot A)2


1. cosec = 1/sin, cot = c/s2. square on RHS suggests s2 + c2 ≡ 13. rearrange and simplify.

LHS =1− cos A1+ cos A

=1− cos A1+ cos A

·

(


)

=(1− cos A)2

1− cos2 A

=(1− cos A)2

sin2 A

=

(

1sin

−

cos Asin A

)2

= (cosec A− cot A)2

≡ RHS.


Level-One-Games 239


+1 ≡ 2 sin2 A


1. t = s/c, cot = c/s2. common denominators3. s2 + c2 ≡ 14. rearrange and simplify.

LHS =tan A− cot Atan A+ cot A

+1

=

sin Acos A

−

cos Asin A

sin Acos A

+cos Asin A

+1

=



+1

= sin2 A− cos2 A+(sin2 A+ cos2 A)

= 2 sin2 A

≡ RHS.


240 Trig or Treat


+2 cos2 A ≡ 1


1. cot = c/s2. s2 + c2 ≡ 13. rearrange and simplify.

LHS =1− cot2 A1+ cot2 A

+2 cos2 A

=1−

cos2 A

sin2 A

1+cos2 A

sin2 A

+2 cos2 A

=

sin2 A− cos2 A

sin2 Asin2 A+ cos2 A

sin2 A

+2 cos2 A

=sin2 A− cos2 A

1+2 cos2 A

= sin2 A+ cos2 A

= 1

≡ RHS.


Level-One-Games 241

sin2 A− tan Acos2 A− cot A

≡ tan2 A


1. t = s/c, cot = c/s2. s2 + c2 ≡ 13. rearrange and simplify.

LHS =sin2 A− tan Acos2 A− cot A

=sin2 A−

sin Acos A

cos2 A−

cos Asin A

=sin2 A cos A− sin A

cos A·

sin Acos2 A sin A− cos A

=sin A(sin A cos A−1)

cos A·

sin Acos A(sin A cos A−1)

=sin2 Acos2 A

= tan2 A

≡ RHS.


242 Trig or Treat

sec A1− sin A

≡

1+ sin Acos3 A


1. sec = 1/cos2. s2 + c2 ≡ 13. rearrange and simplify.

LHS =sec A

1− sin A

=1

cos A·

1(1− sin A)

=1

cos A(1− sin A)·

(

1+ sin A1+ sin A

)

=1+ sin A

cos A(1− sin2 A)

=1+ sin A

cos A(cos2 A)

=1+ sin A

cos3 A

≡ RHS.


Level-One-Games 243

sec Acosec2 A

−

cosec Asec2 A

≡ (1+ cot A+ tan A)(sin A− cos A)


1. cosec2, sec2 suggest s2 + c2 ≡ 12. cosec = 1/sin, sec = 1/cos, cot = c/s, t = s/c3. rearrange and simplify.

LHS =sec A

cosec2 A−

cosec Asec2 A

=1

cos A·

(

sin2 A1

)

−

1sin A

(

cos2 A1

)

=sin2 Acos A

−

cos2 Asin A

=sin3 A− cos3 A∗

sin A cos A

=(sin A− cos A)(sin2 A+ sin A cos A+ cos2 A)

sin A cos A

= (sin A− cos A)(tan A+1+ cot A)

= (1+ cot A+ tan A)(sin A− cos A)

≡ RHS.

∗(a3 −b3) = (a−b)(a2 +ab+b2).


244 Trig or Treat

1+ sec Asec A

≡

sin2 A1− cos A


1. sec = 1/cos2. s2 suggests s2 + c2 = 13. rearrange and simplify.

LHS =1+ sec A

sec A

=1+

1cos A1

cos A

=1+ cos A

cos A·

cos A1

= 1+ cos A

= (1+ cos A)

(


)

=1− cos2 A1− cos A

=sin2 A

1− cos A

≡ RHS.


Level-One-Games 245

tan A− cot Atan A + cot A

≡ 1−2 cos2 A


1. t = s/c, cot = c/s2. common denominators3. s2 + c2 ≡ 14. rearrange and simplify.

LHS =tan A− cot Atan A + cot A

=

sin Acos A

−

cos Asin A

sin Acos A

+cos Asin A

=


sin2 A + cos2 Acos A sin A

=sin2 A− cos2 A

sin2 A + cos2 A

= sin2 A− cos2 A

= (1− cos2 A)− cos2 A

= 1−2 cos2 A

≡ RHS.


246 Trig or Treat

(2 cos2 A−1)2

cos4 A− sin4 A≡ 1−2 sin2 A


1. c4 − s4 suggests (c2 − s2)(c2 + s2)2. s2 + c2 ≡ 13. rearrange and simplify.

LHS =(2 cos2 A−1)2

cos4 A− sin4 A

=(cos2 A− sin2 A)2

(cos2 A− sin2 A)(cos2 A+ sin2 A)

= cos2 A− sin2 A

= (1− sin2 A)− sin2 A

= 1−2 sin2 A

≡ RHS.


Level-One-Games 247


+1 ≡ 2 cos2 A


1. t = s/c2. s2 + c2 ≡ 13. rearrange and simplify.

LHS =1− tan2 A1+ tan2 A

+1

=1−

sin2 Acos2 A

1+sin2 Acos2 A

+1

=


cos2 A+ sin2 Acos2 A

+1

=cos2 A− sin2 A

cos2 A+ sin2 1+1

= cos2 A− sin2 A+1

= cos2 A+ cos2 A

= 2 cos2 A

≡ RHS.


248 Trig or Treat

1− sin A1+ sin A

≡ (sec A− tan A)2


1. sec = 1/cos, t = s/c2. the RHS appears to be more complex and hence should be expanded first3. s2 + c2 ≡ 14. rearrange and simplify.

RHS = (sec A− tan A)2

=

(

1cos A

−

sin Acos A

)2

=

(

1− sin Acos A

)2

=(1− sin A)2

cos2 A

=(1− sin A)2

1− sin2 A

=(1− sin A)2


=1− sin A1+ sin A

≡ LHS.


Level-One-Games 249

(sec A− tan A)2 +1cosec A(sec A− tan A)

≡ 2 tan A


1. sec = 1/cos, t = s/c, cosec = 1/sin2. ( )2 suggests expansion3. s2 + c2 ≡ 14. rearrange and simplify.

LHS =(sec A− tan A)2 +1

cosec A(sec A− tan A)

=sec2 A−2 sec A tan A+ tan2 A+1


=sec2 A−2 sec A tan A+ sec2 A


=2 sec2 A−2 sec A tan Acosec A(sec A− tan A)

=2 sec A(sec A− tan A)


=2 sec Acosec A

= 21

cos A·

sin A1

= 2 tan A

≡ RHS.

September 12, 2007 19:18 Book:- Trig or Treat (9in x 6in) 10˙less-easy

Level-Two-GamesLess-Easy Proofs


252 Trig or Treat

2cot A tan2A

≡ 1− tan2 A


1. 2A suggests expansion of “double angle”2. cot = 1/ tan3. rearrange and simplify.

LHS =2

cot A tan 2A

= 2 tan A ·

(1− tan2 A)

2 tan A

= 1− tan2 A

≡ RHS.


Level-Two-Games 253

cos4 A− sin4 A ≡

1sec 2A


1. (cos4−sin4) suggests (a2 −b2)(a2 +b2)2. sec = 1/cos3. 2A suggests “double angle”4. rearrange and simplify.



= (cos 2A)(1)

=1

sec 2A

≡ RHS.


254 Trig or Treat

sin(A−B)

sin A cos B≡ 1− cot A tan B


1. sin(A−B) suggests expansion2. cot = c/s, t = s/c3. rearrange and simplify.

LHS =sin(A−B)

sin A cos B

=sin A cos B− cos A sin B

sin A cos B

= 1− cot A tan B

≡ RHS.


Level-Two-Games 255

cos(A−B)

cos A cos B≡ 1+ tan A tan B


1. cos(A−B) suggests expansion2. t = s/c3. rearrange and simplify.

LHS =cos(A−B)

cos A cos B

=cos A cos B+ sin A sin B

cos A cos B

= 1+ tan A tan B

≡ RHS.


256 Trig or Treat

cos(A+B)

sin A cos B≡ cot A− tan B


1. cos(A+B) suggests expansion2. cot = c/s, t = s/c3. rearrange and simplify.

LHS =cos(A+B)

sin A cos B

=cos A cos B− sin A sin B

sin A cos B

= cot A− tan B

≡ RHS.


Level-Two-Games 257

cos(A−B)

sin A cos B≡ cot A+ tan B


1. cos(A−B) suggests expansion2. cot = c/s, t = s/c3. rearrange and simplify.

LHS =cos(A−B)

sin A cos B


sin A cos B

= cot A+ tan B

≡ RHS.


258 Trig or Treat

cos(A−B)

sin A sin B≡ 1+ cot A cot B


1. cos(A−B) suggests expansion2. cot = c/s3. rearrange and simplify.

LHS =cos(A−B)

sin A sin B


sin A sin B

=cos A cos Bsin A sin B

+1

= 1+ cot Acot B

≡ RHS.


Level-Two-Games 259

sin(A+B)

cos A cos B≡ tan A+ tan B


1. sin(A+B) suggests expansion2. t = s/c3. rearrange and simplify.

LHS =sin(A+B)

cos A cos B

=sin A cos B+ cos A sin B

cos A cos B

=sin Acos A

+sin Bcos B

= tan A+ tan B

≡ RHS.


260 Trig or Treat

cos4 A− sin4 A ≡ cos 2A


1. c4 − s4 suggests (a2 −b2)(a2 +b2)2. s2 + c2 ≡ 13. rearrange and simplify.



= (cos 2A)(1)

= cos 2A

≡ RHS.


Level-Two-Games 261

(2a sin A cos A)2 +a2(cos2 A− sin2 A)2≡ a2


1. 2 sin A cos A suggests sin 2A2. cos2 A− sin2 A suggests cos 2A3. s2 + c2 ≡ 14. rearrange and simplify.

LHS = (2a sin A cos A)2 +a2(cos2 A− sin2 A)2

= a2(sin 2A)2 +a2(cos 2A)2

= a2(sin2 2A+ cos2 2A)

= a2

≡ RHS.


262 Trig or Treat

sin A1+ cos A

≡ tanA2


1. A/2 suggests use of “half angle formulas” for sin A, cos A2. t = s/c3. rearrange and simplify.

LHS =sin A

1+ cos A

=2 sin

A2

cosA2

1+

(

2 cos2 A2−1

)

=2 sin

A2

cosA2

2 cos2 A2

= tanA2

≡ RHS.


Level-Two-Games 263

1− cos Asin A

≡ tanA2


1. A/2 suggests “half angle formulas” for cos A, sin A2. rearrange and simplify.

LHS =1− cos A

sin A

=

1−

(

1−2 sin2 A2

)

2 sinA2

cosA2

=2 sin2 A

2

2 sinA2

cosA2

=sin

A2

cosA2

= tanA2

≡ RHS.


264 Trig or Treat

(

sinA2

+ cosA2

)2

≡ 1+ sin A


1. A/2 suggests “half angle formulas”2. ( )2 suggests s2 + c2 ≡ 13. rearrange and simplify.

LHS =

(

sinA2

+ cosA2

)2

= sin2 A2

+2 sinA2

cosA2

+ cos2 A2

= sin2 A2

+ cos2 A2

+2 sinA2

cosA2

= 1+ sin A

≡ RHS.


Level-Two-Games 265

cos(60◦ +A)+ sin(30◦ +A) ≡ cos A


1. (60◦+A), (30◦ +A) suggest expansion of “compound angles”2. numerical values for sin, cos of 60◦, 30◦


LHS = cos(60◦ +A)+ sin(30◦ +A)

= cos 60◦ cos A− sin 60◦ sin A

+ sin 30◦ cos A+ cos 30◦ sin A

=12

cos A−

√

32

sin A+12

cos A+

√

32

sin A

= cos A

≡ RHS.

sin cos

0◦ 0 =

√

04

√

44

= 1

30◦12

=

√

14

√

34

=

√

32

45◦√

22

=

√

24

√

24

=

√

22

60◦√

32

=

√

34

√

14

=12

90◦ 1 =

√

44

√

04

= 0


266 Trig or Treat

1+ sin 2A+ cos 2Asin A+ cos A

≡ 2 cos A


1. 2A suggests expansion of “double angle”2. rearrange and simplify.

LHS =1+ sin 2A+ cos 2A

sin A+ cos A

=1+2sin A cos A+(2 cos2 A−1)

sin A+ cos A

2cos A(sin A+ cos A)

sin A+ cos A

= 2 cos A

≡ RHS.


Level-Two-Games 267

sin 2A1− cos 2A

≡ cot A


1. 2A suggests expansion of “double angle”2. s2 + c2 ≡ 1, cot = c/s3. rearrange and simplify.

LHS =sin 2A

1− cos 2A

=2 sin A cos A

1− (1−2 sin2 A)

=2 sin A cos A

2 sin2 A

=cos Asin A

= cot A

≡ RHS.


268 Trig or Treat

sin A+ sin 2A1+ cos A+ cos 2A

≡ tan A


1. 2A suggests expansion of “double angle”2. t = s/c3. rearrange and simplify.

LHS =sin A+ sin 2A

1+ cos A+ cos 2A

=sin A+2sin A cos A

1+ cos A+(2 cos2 A−1)

=sin A(1+2 cos A)

cos A+2 cos2 A

=sin A(1+2 cos A)

cos A(1+2 cos A)

= tan A

≡ RHS.


Level-Two-Games 269

12(cot A− tan A) ≡ cot 2A


1. cot = c/s, t = s/c2. cot 2A = cos 2A/sin 2A3. rearrange and simplify.

LHS =12(cot A− tan A)

=12

(

cos Asin A

−

sin Acos A

)

=12

(

cos2−sin2 A

sin A cos A

)

=cos 2Asin 2A

= cot 2A

≡ RHS.


270 Trig or Treat

cosec A sec A ≡ 2 cosec 2A


1. cosec = 1/sin, sec = 1/cos2. cosec 2A suggests 1/sin 2A3. rearrange and simplify.

LHS = cosec A sec A

=1

sin A·

1cos A

=1

sin A cos A·

(

22

)

=2

2sin A cos A

=2

sin 2A

= 2 cosec 2A

≡ RHS.


Level-Two-Games 271

cos(A+B)

cos Asin B≡ cot B− tan A


1. cos(A+B) suggests expansion2. cot = c/s3. rearrange and simplify.

LHS =cos(A+B)

cos A sin B


cos A sin B

= cot B− tan A

≡ RHS.


272 Trig or Treat

sec A sec B1+ tan A tan B

≡ sec(A−B)


1. sec(A−B) suggests expansion2. sec = 1/cos, t = s/c3. begin with RHS since sec(A − B) = 1/cos(A − B) which is standard

“compound angle” function4. rearrange and simplify.

RHS = sec(A−B)

=1

cos(A−B)

=1

cos A cos B+ sin A sin B

=sec A sec B

1+ tan A tan B

≡ LHS.

divide bothnumeratorand denominatorby cos A cos B


Level-Two-Games 273

cosec Acosec Bcot A cot B−1

≡ sec(A+B)


1. sec(A+B) suggests expansion2. cosec = 1/sin, sec = 1/cos, cot = c/s3. begin with RHS since sec(A + B) = 1/cos(A + B) which is a standard

“compound angle” function.4. rearrange and simplify.

RHS = sec(A+B)

=1

cos(A+B)

=1

cos A cos B− sin A sin B

=cosec Acosec Bcot A cot B−1

≡ LHS.

divide bothnumeratorand denominatorby sin A sin B


274 Trig or Treat

cot A cot B+1cot B− cot A

≡ cot(A−B)


1. cot(A−B) suggests expansion2. cot = 1/ tan3. begin with RHS since cot(A−B) = 1/ tan(A−B), and tan(A−B) is a standard

“compound angle” function4. rearrange and simplify.

RHS = cot(A−B)

=1

tan(A−B)

=1+ tan A tan Btan A− tan B

=cot A cot B+1cot B− cot A

≡ LHS.

divide bothnumeratorand denominatorby tan A tan B


Level-Two-Games 275

cos(A+B)

cos A cos B≡ 1− tan A tan B


1. cos(A+B) suggests expansion2. t = s/c3. rearrange and simplify.

LHS =cos(A+B)

cos A cos B


cos A cos B

= 1−sin A sin Bcos A cos B

= 1− tan A tan B

≡ RHS.


276 Trig or Treat

sin(A+B)

sin A cos B≡ 1+ cot A tan B


1. sin(A+B) suggests expansion2. cot = c/s, t = s/c3. rearrange and simplify.

LHS =sin(A+B)

sin A cos B

=sin A cos B+ cos A sin B

sin A cos B

= 1+cos A sin Bsin A cos B

= 1+ cot A tan B

≡ RHS.


Level-Two-Games 277

1− tan A tan B1+ tan A tan B

≡

cos(A+B)

cos(A−B)


1. cos(A+B), cos(A−B) suggest expansion2. t = s/c3. begin with RHS which are standard “compound angle” functions4. rearrange and simplify.

RHS =cos(A+B)

cos(A−B)

=cos A cos B− sin A sin Bcos A cos B+ sin A sin B

=1− tan A tan B1+ tan A tan B

≡ LHS.



278 Trig or Treat

tan A+ tan Btan A− tan B

≡

sin(A+B)

sin(A−B)


1. sin(A+B), sin(A−B) suggest expansion2. t = s/c3. begin with RHS which are standard “compound angle” functions4. rearrange and simplify.

RHS =sin(A+B)

sin(A−B)

=sin A cos B+ cos A sin Bsin A cos B− cos A sin B

=tan A+ tan Btan A− tan B

≡ LHS.



Level-Two-Games 279



≡ 2 sec 2A


1. sec = 1/cos2. 2A suggests “double angle”3. common denominator4. rearrange and simplify.

LHS =cos A− sin Acos A+ sin A


=(cos A− sin A)2 +(cos A+ sin A)2

(cos A+ sin A)(cos A− sin A)

=(cos2 A−2 sin A cos A+ sin2 A)+(cos2 A+2 sin A cos A+ sin2 A)

cos2 A− sin2 A

=(1)+(1)

cos 2A

= 2 sec 2A

≡RHS.


280 Trig or Treat

cot A− tan Acot A+ tan A

≡ cos 2A


1. cot = c/s, t = s/c2. common denominators3. rearrange and simplify.

LHS =cot A− tan Acot A+ tan A

=

cos Asin A

−

sin Acos A

cos Asin A

+sin Acos A

=cos2 A− sin2 A

sin A cos A·

sin A cos A

cos2 A+ sin2 A

=cos2 A− sin2 A

cos2 A+ sin2 A

=cos 2A

1

= cos 2A

≡ RHS.


Level-Two-Games 281

cos A− cos Bsin A+ sin B

+sin A− sin Bcos A+ cos B

≡ 0


1. common denominators2. rearrange and simplify.

LHS =

(

cos A− cos Bsin A+ sin B

)

+

(

sin A− sin Bcos A+ cos B

)

=(cos A− cos B)(cos A+ cos B)+(sin A− sin B)(sin A+ sin B)

(sin A+ sin B)(cos A+ cos B)

=(cos2 A− cos2 B)+(sin2 A− sin2 B)


=(cos2 A+ sin2 A)− (cos2 B+ sin2 B)


=(1)− (1)


= 0

≡ RHS.


282 Trig or Treat


−


≡ 2 tan 2A


1. common denominators2. tan 2A suggest “double angle” formula3. lots of cos2, sin2 suggest s2 + c2 ≡ 14. rearrange and simplify.

LHS =

(


)

−

(


)

=(cos A+ sin A)(cos A+ sin A)− (cos A− sin A)(cos A− sin A)

(cos A− sin A)(cos A+ sin A)

=(cos2 A+2 sin A cos A+ sin2 A)− (cos2 A−2 sin A cos A+ sin2 A)

cos2 A− sin2 A

=4 sin A cos A

cos 2A

= 2sin 2Acos 2A

= 2 tan 2A

≡RHS.


Level-Two-Games 283

(4 sin A cos A)(1−2 sin2 A) ≡ sin 4A


1. 4 sin A cos A = 2(2 sin A cos A) = 2 sin 2A2. (1−2 sin2 A) = cos 2A3. rearrange and simplify.

LHS = 4 sin A cos A(1−2 sin2 A)

= 2(2 sin A cos A)(cos 2A)

= 2(sin 2A)(cos 2A)

= sin 4A

≡ RHS.


284 Trig or Treat

sin 3A cos A− sin A cos 3Asin 2A

≡ 1


1. sin A cos B− cos A sin B = sin(A−B)2. rearrange and simplify.

LHS =sin 3A cos A− sin A cos 3A

sin 2A

=sin(3A−A)

sin 2A

=sin 2Asin 2A

= 1

≡ RHS.


Level-Two-Games 285

tan(

A−

π4

)

≡

tan A−1tan A+1


1. t = s/c2. (A− (π/4)) suggests expansion of “compound angle”3. rearrange and simplify.

LHS = tan(

A−

π4

)

=tan A− tan

π4

1+ tan A tanπ4

=tan A−11+ tan A

=tan A−1tan A+1

≡ RHS.

tanπ4

= 1


286 Trig or Treat

cos3 A− sin3 Acos A− sin A

≡

2+ sin 2A2


1. cos3−sin3 suggests (a3 −b3) = (a−b)(a2 +ab+b2)2. sin 2A suggests “double angle”3. rearrange and simplify.

LHS =cos3 A− sin3 Acos A− sin A

=(cos A− sin A)(cos2 A+ cos A sin A+ sin2 A)

(cos A− sin A)

= (1+ cos A sin A)

=12(2+2 cos A sin A)

=2+ sin 2A

2

≡ RHS.


Level-Two-Games 287

sin 3Asin A

−

cos 3Acos A

≡ 2


1. common denominator2. sin(A−B) expansion3. rearrange and simplify.

LHS =sin 3Asin A

−

cos 3Acos A

=sin 3A cos A− cos 3A sin A

sin A cos A

=sin(3A−A)

sin A cos A

=sin 2A

12(2 sin A cos A)

= 2sin 2Asin 2A

= 2

≡ RHS.


288 Trig or Treat

2 cot A cot 2A ≡ cot2 A−1


1. cot = c/s2. cos 2A, sin 2A suggest expansion3. rearrange and simplify.

LHS = 2 cot A cot 2A

= 2cos Asin A

·

cos 2Asin 2A

= 2cos Asin A

·

(cos2 A− sin2 A)

2 sin A cos A

=1

sin2 A(cos2 A− sin2 A)

= cot2 A−1

≡ RHS.


Level-Two-Games 289

12

sec Acosec A ≡ cosec 2A


1. sec = 1/cos, cosec = 1/sin2. cosec 2A = 1/sin 2A3. rearrange and simplify.

LHS =12

sec Acosec A

=12

1cos A

·

1sin A

=1

2 sin A cos A

=1

sin 2A

= cosec 2A

≡ RHS.


290 Trig or Treat

tan A+ cot A ≡

2sin 2A


1. t = s/c, cot = c/s2. sin 2A suggests “double angle”3. rearrange and simplify.

LHS = tan A+ cot A

=sin Acos A

+cos Asin A

=sin2 A+ cos2 A

sin A cos A

=1

sin A cos A

=2

2 sin A cos A

=2

sin 2A

≡ RHS.


Level-Two-Games 291

21− cos A

≡ cosec 2 A2


1. A/2 suggests expansion of cos A2. cosec = 1/sin3. rearrange and simplify.

LHS =2

1− cos A

=2

1−

(

1−2 sin2 A2

)

=2

2 sin2 A2

=1

sin2 A2

= cosec 2 A2

≡ RHS.


292 Trig or Treat

21+ cos A

≡ sec2 A2


1. A/2 suggests expansion of cos A2. sec2 = 1/cos2


LHS =2

1+ cos A

=2

1+

(

2 cos2 A2−1

)

=2

2 cos2 A2

=1

cos2 A2

= sec2 A2

≡ RHS.


Level-Two-Games 293

(1+ cos A) tanA2≡ sin A


1. tan(A/2) suggests expansion of cos A as cos 2(A/2) and sin A as sin 2(A/2)2. rearrange and simplify.

LHS = (1+ cos A) tanA2

=

(

1+ cos 2

(

A2

))

tanA2

=

(

1+

(

2 cos2 A2−1

))

tanA2

= 2 cos2 A2·

sinA2

cosA2

= 2 cosA2

sinA2

= sin A

≡ RHS.


294 Trig or Treat


≡ 1−sin 2A

2


1. s3 + c3 = (s+ c)(s2 − sc+ c2)2. s2 + c2 ≡ 13. sin 2A = 2 sin A cos A4. rearrange and simplify.

LHS =sin3 A+ cos3 Asin A+ cos A


(sin A+ cos A)

= sin2 A− sin A cos A+ cos2 A

= 1− sin A cos A

= 1−12(2 sin A cos A)

= 1−sin 2A

2

≡ RHS.


Level-Two-Games 295

sec2 A2− sec2 A

≡ sec 2A


1. sec = 1/cos, sec 2A = 1/cos 2A2. sec2 suggests s2 + c2 ≡ 13. rearrange and simplify.

LHS =sec2 A

2− sec2 A

=1

cos2 A·

1

2−1

cos2 A

=1

cos2 A·

cos2 A(2 cos2 A−1)

=1

2 cos2 A−1

=1

cos 2A

= sec 2A

≡ RHS.


296 Trig or Treat

cosec A− cot A ≡ tanA2


1. cosec = 1/sin, cot = c/s, t = s/c2. A/2 suggests expansion of sin A, cos A3. rearrange and simplify.

LHS = cosec A− cot A

=1

sin A−

cos Asin A

=1− cos A

sin A

=

1−

(

1−2 sin2 A2

)

2 sinA2

cosA2

=2 sin2 A

2

2 sinA2

cosA2

=sin

A2

cosA2

= tanA2

≡ RHS.


Level-Two-Games 297

sin A tanA2≡ 1− cos A


1. tan(A/2) suggests expansion of sin A as sin 2(A/2), and cos A as cos 2(A/2)2. t = s/c3. rearrange and simplify.

LHS = sin A tanA2

= sin 2

(

A2

)

tanA2

= 2 sinA2· cos

A2·

sinA2

cosA2

= 2 sin2 A2

=

(

2 sin2 A2−1

)

+1

= 1− cos A

≡ RHS.


298 Trig or Treat

sin 2A tan A ≡ 1− cos 2A


1. Expand “double angle”2. t = s/c3. rearrange and simplify.

LHS = sin 2A tan A

= 2 sin A cos A ·

sin Acos A

= 2 sin2 A

= (2 sin2 A−1)+1

= 1− cos 2A

≡ RHS.


Level-Two-Games 299

cosec 2A− cot 2A ≡ tan A


1. 2A suggests expansion of “double angle”2. cosec = 1/sin, cot = c/s, t = s/c3. rearrange and simplify.

LHS = cosec 2A− cot 2A

=1

sin 2A−

cos 2Asin 2A

=1− cos 2A

sin 2A

=1− (1−2 sin2 A)

2 sin A cos A

=sin Acos A

= tan A

≡ RHS.


300 Trig or Treat

tan

(

45◦ +A2

)

≡ tan A+ sec A


1. (45◦+A/2) suggests expansion of “compound angle”2. A/2 suggests “half angle” expansion3. both sides are complex; hence explore simplification on both sides to achieve

identity.

LHS = tan

(

45◦ +A2

)

=tan 45◦ + tan

A2

1− tan 45◦ tanA2

=1+ tan

A2

1− tanA2

RHS = tan A+ sec A

=sin Acos A

+1

cos A

=sin A+1

cos A

=

2 sinA2

cosA2

+

(

cos2 A2

+ sin2 A2

)

cos2 A2− sin2 A

2

=

(

sinA2

+ cosA2

)2

(

cosA2

+ sinA2

)(

cosA2− sin

A2

) =

(

sinA2

+ cosA2

)

(

cosA2− sin

A2

)

=tan

A2

+1

1− tanA2

≡ simplified form of LHS.

divide numeratorand denominator

by cosA2


Level-Two-Games 301

A more elegant proof comes from the use of the application of Pythago-ras Theorem with respect to the “half angle formula” of tan A.

tan A =2 tan

A2

1− tan2 A2

.

Let t = tan(A/2).Then from Pythagoras Theorem

the hypotenuse =√

(2t)2 +(1− t2)2

=√

1+2(2t)+ t4

= 1+ t2

∴ cos A =1− t2

1+ t2

∴ RHS = tan A+ sec A

= tan A+1

cos A

=2t

1− t2 +1+ t2

1− t2 where t = tanA2

=1+2t + t2

(1− t2)

=(1+ t)2

(1+ t)(1− t)

=1+ t1− t

=1+ tan

A2

1− tanA2


1 − t2

1 + t2

2t


302 Trig or Treat

tan Atan 2A− tan A

≡ cos 2A


1. t = s/c2. 2A suggests expansion of “compound angle”3. rearrange and simplify.

LHS =tan A

tan 2A− tan A

=sin Acos A

·

1(

sin 2Acos 2A

−

sin Acos A

)

=sin Acos A

·

cos Acos 2A(sin 2A cos A− sin A cos 2A)

=sin A cos 2A

(sin 2A cos A− cos 2A sin A)

=sin A cos 2Asin(2A−A)

= cos 2A

≡ RHS.


Level-Two-Games 303

sec 2A− tan 2A ≡ tan(45◦−A)


1. 2A, (45◦−A) suggest expansion of “compound angles”2. since both LHS and RHS have complex functions, explore simplification of

both side to achieve identity.3. sec = 1/cos, t = s/c4. rearrange and simplify.

LHS = sec 2A− tan 2A

=1

cos 2A−

sin 2Acos 2A

=1− sin 2A

cos 2A

=(cos2 A+ sin2 A−2 sin A cos A)

cos2 A− sin2 A

=(cos A− sin A)2

(cos A+ sin A)(cos A− sin A)

=(cos A− sin A)

(cos A+ sin A)

=1− tan A1+ tan A

RHS = tan(45◦−A)

=tan 45◦− tan A

1+ tan 45◦ tan A

=1− tan A1+ tan A

≡ LHS.

dividing all termsby cos A to preparefor comparisonwith RHS.


304 Trig or Treat

2 sin2 A6− sin2 A

7≡ cos2 A

7− cos

A3


1. sin2(A/7), cos2(A/7) suggest s2 + c2 ≡ 12. A/3 suggests “double angle” formula to give A/63. rearrange and simplify.

LHS = 2 sin2 A6− sin2 A

7

= 2 sin2 A6−

(

1− cos2 A7

)

=

(

2 sin2 A6−1

)

+ cos2 A7

= −cos 2

(

A6

)

+ cos2 A7

= cos2 A7− cos

A3

≡ RHS.


Level-Two-Games 305

sin 2A1+ cos 2A

≡ tan A


1. 2A suggests expansion of “double angle”2. rearrange and simplify.

LHS =sin 2A

1+ cos 2A

=2 sin A cos A

1+(2 cos2 A−1)

=2 sin A cos A

2 cos2 A

=sin Acos A

= tan A

≡ RHS.


306 Trig or Treat

tan A+ cot A ≡ 2cosec 2A


1. t = s/c, cot = c/s, cosec = 1/sin2. 2A suggests “double angle”3. rearrange and simplify.

LHS = tan A+ cot A

=sin Acos A

+cos Asin A

=sin2 A+ cos2 A

cos A sin A

=1

cos A sin A

=22·

1cos A sin A

=2

sin 2A

= 2cosec 2A

≡ RHS.


Level-Two-Games 307

2 cos2(45◦−A) ≡ 1+ sin 2A


1. c2 suggests s2 + c2 ≡ 12. (45◦−A), 2A suggest expansion of “compound angles”3. rearrange and simplify.

LHS = 2 cos2(45◦−A)

= 2(cos 45◦ cos A+ sin 45◦ sin A)2

= 2

(√

22

cos A+

√

22

sin A

)2

= 2

(√

22

)2

(cos A+ sin A)2

= cos2 A+2 sin A cos A+ sin2 A

= 1+ sin 2A

≡ RHS.

sin cos

45◦√

24

√

24


308 Trig or Treat

cos(A+B)cos(A−B) ≡ cos2 B− sin2 A


1. Expand “compound angles”2. c2, s2 suggest s2 + c2 ≡ 13. rearrange and simplify.

LHS = 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 B(1− sin2 A)− sin2 A sin2 B

= cos2 B− cos2 B sin2 A− sin2 A sin2 B

= cos2 B− sin2 A(cos2 B+ sin2 B)

= cos2 B− sin2 A

≡ RHS.


Level-Two-Games 309

cos(A+B)cos(A−B) ≡ cos2 A− sin2 B


1. Expand “compound angles”2. c2, s2 suggest s2 + c2 ≡ 13. rearrange and simplify.

LHS = 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)− sin2 A sin2 B

= cos2 A− cos2 A sin2 B− sin2 A sin2 B

= cos2 A− sin2 B(cos2 A+ sin2 A)

= cos2 A− sin2 B

≡ RHS.


310 Trig or Treat

sin(A+B)sin(A−B)≡ sin2 A− sin2 B


1. Expand “compound angles”2. s2 suggests s2 + c2 ≡ 13. rearrange and simplify.

LHS = 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)− cos2 A sin2 B

= sin2 A− sin2 A sin2 B− cos2 A sin2 B

= sin2 A− sin2 B(sin2 A+ cos2 A)

= sin2 A− sin2 B

≡ RHS.


Level-Two-Games 311

sin(A+B)sin(A−B) ≡ cos2 B− cos2 A


1. Expand “compound angles”2. c2 suggests s2 + c2 ≡ 13. rearrange and simplify.

LHS = 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

= cos2 B(1− cos2 A)− cos2 A sin2 B

= cos2 B− cos2 A cos2 B− cos2 A sin2 B

= cos2 B− cos2 A(cos2 B+ sin2 B)

= cos2 B− cos2 A

≡ RHS.


312 Trig or Treat

sec A+1sec A−1

≡ cot2A2


1. sec = 1/cos, cot = c/s2. A/2 suggests expansion of cos A3. cot2 suggests s2 + c2 ≡ 14. rearrange and simplify.

LHS =sec A+1sec A−1

=

(

1cos A

+1

)

(

1cos A

−1

)

=

(

1+ cos Acos A

)

·

(

cos A1− cos A

)

=1+ cos A1− cos A

=

1+

(

2 cos2 A2−1

)

1−

(

1−2 sin2 A2

)

=2 cos2 A

2

2 sin2 A2

= cot2A2

≡ RHS.


Level-Two-Games 313

sec2 A2− sec2 A

≡ sec 2A


1. sec = 1/cos2. 2A suggests “compound angle”3. rearrange and simplify.

LHS =sec2 A

2− sec2 A

=1

cos2 A·

1(

2−1

cos2 A

)

=1

cos2 A·

cos2 A(2 cos2 A−1)

=1

cos 2A

= sec2A

≡ RHS.


314 Trig or Treat

18(1− cos 4A) ≡ sin2 A cos2 A


1. cos 4A suggests expansion twice: cos 4A → cos 2A → cos A2. s2, c2 suggest s2 + c2 ≡ 13. rearrange and simplify.

LHS =18(1− cos 4A)

=18(1− (2 cos2 2A−1))

=18(2−2(2 cos2 A−1)2)

=18·2(1− (4 cos4 A−4 cos2 A+1))

=14(4 cos2 A(cos2 A−1))

= cos2 A(sin2 A)

= sin2 A cos2 A

≡ RHS.


Level-Two-Games 315

cot2 A−12 cot A

≡ cot 2A


1. cot = c/s, t = s/c2. cot 2A = cos 2A/sin 2A3. rearrange and simplify.

LHS =cot2 A−1

2 cot A

=12

(

cot A−

1cot A

)

=12(cot A− tan A)

=12

(

cos Asin A

−

sin Acos A

)

=12

(

cos2 A− sin2 Asin A cos A

)

=cos 2Asin 2A

= cot 2A

≡ RHS.


316 Trig or Treat

2 tan A1+ tan2 A

≡ sin 2A


1. t = s/c2. 2A suggests “double angle”3. rearrange and simplify.

LHS =2 tan A

1+ tan2 A

=

2

(

sin Acos A

)

1+

(

sin Acos A

)2

= 2

(

sin Acos A

)(

cos2 A

cos2 A+ sin2 A

)

= 2sin Acos A

(

cos2 A1

)

= 2 sin A cos A

= sin 2A

≡ RHS.


Level-Two-Games 317


≡ cos 2A


1. t = s/c2. 2A suggests “double angle”3. rearrange and simplify.

LHS =1− tan2 A1+ tan2 A

=

(

1−

(

sin Acos A

)2)

1(

1+

(

sin Acos A

)2)

=

(


)(

cos2 A

cos2 A+ sin2 A

)

=(cos2 A− sin2 A)

cos2 A·

cos2 A1

= cos2 A− sin2 A

= cos 2A

≡ RHS.


318 Trig or Treat

The previous two identities are part of the series of tan 2A, sin 2A andcos 2A in terms of tan A.

tan 2A =2 tan A

1− tan2 A

sin 2A =2 tan A

1+ tan2 A

cos 2A =1− tan2 A1+ tan2 A

The “double angle” formula for tan is well known to generations ofstudents. But few are those who know about these special formulas forsin2A and cos2A in terms of tan A. You are among the very few. Test itout yourself with your friends!

An easy way to remember the three identities is to write tan A as t.

then: tan 2A =2t

1− t2

From Pythagoras Theorem, the hypotenuse is given by:

√

(2t)2 +(1− t2)2

=√

4t2 +1−2t2 + t4

=√

1+2t2 + t4

= 1+ t2

Then:

sin 2A =OH

=2t

1+ t2 , and

cos 2A =AH

=1− t2

1+ t2 .

1 − t2

1 + t2

2t

2A


Level-Two-Games 319

This sin 2A equation is one of the amazing equations in Trigonometrywhere a slight difference (from minus to plus sign in the denominator)changes the tangent identity for the double angle to the sine identity for thesame double angle.


320 Trig or Treat

cos 2Asin A

+sin 2Acos A

≡ cosec A


1. 2A suggests expansion of “double angle”2. cosec = 1/sin3. rearrange and simplify.

LHS =cos 2Asin A

+sin 2Acos A

=cos2 A− sin2 A

sin A+

2 sin A cos Acos A

=1−2 sin2 A

sin A+2 sin A

=1−2 sin2 A+2 sin2 A

sin A

=1

sin A

= cosec A

≡ RHS.


Level-Two-Games 321

2 sin(A−B)

cos(A+B)− cos(A−B)≡ cot A− cot B


1. (A+B), (A−B) suggest expansion of “compound angles”2. cot = c/s3. rearrange and simplify.

LHS =2 sin(A−B)

cos(A+B)− cos(A−B)

=2(sin A cos B− cos A sin B)

(cos A cos B− sin A sin B)− (cos A cos B+ sin A sin B)

=2(sin A cos B− cos A sin B)

−2 sin A sin B

=(cos A sin B− sin A cos B)

sin A sin B

= cot A− cot B

≡ RHS.


322 Trig or Treat

cos 2A1+ sin 2A

≡

cot A−1cot A+1


1. cos 2A, sin 2A suggest expansion2. cot = c/s3. rearrange and simplify.

LHS =cos 2A

1+ sin 2A

=(cos2 A− sin2 A)

1+2 sin A cos A

=cos2 A− sin2 A

cos2 A+ sin2 A+2 sin A cos A

=(cos A− sin A)(cos A+ sin A)

(cos A+ sin A)2

=cos A− sin Acos A+ sin A

=cot A−1cot A+1

≡ RHS.

dividing both thenumerator andthe denominatorby sin.


Level-Two-Games 323

cot Acot B−1cot A+ cot B

≡ cot(A+B)


1. (A+B) on RHS suggests expansion2. cot = 1/ tan3. begin with RHS expansion of tan(A+B)4. rearrange and simplify.

RHS = cot(A+B)

=1

tan(A+B)

=1− tan A tan Btan A+ tan B

=cot A cot B−1cot B+ cot A

=cot A cot B−1cot A+ cot B

≡ LHS.

divide bothnumeratorand denominatorby tan A tan B


324 Trig or Treat

3 sin A−4 sin3 A ≡ sin 3A


1. 3A suggests expansion of “compound angle” twice2. rearrange and simplify3. the RHS is a standard function, easy to expand, twice.

RHS = sin 3A

= sin(2A+A)

= sin 2A cos A+ cos 2A sin A

= 2 sin A cos A · cos A+ sin A(cos2 A− sin2 A)

= 2 sin A cos2 A+ sin A cos2 A− sin3 A

= 3 sin A cos2 A− sin3 A

= 3 sin A(1− sin2 A)− sin3 A

= 3 sin A−3 sin3 A− sin3 A

= 3 sin A−4 sin3 A

≡ LHS.


Level-Two-Games 325

3 tan A− tan3 A1−3 tan2 A

≡ tan 3A


1. RHS tan 3A is standard double expansion of tan(2A+A) and tan 2A.2. begin with RHS (an exception to normal practice)3. rearrange and simplify.

RHS = tan 3A

= tan(2A+A)

=tan 2A+ tan A

1− tan 2A tan A

=

2 tan A1− tan2 A

+ tan A

1−2 tan A

1− tan2 Atan A

=

2 tan A+ tan A− tan3 A1− tan2 A

1− tan2 A−2tan2 A1− tan2 A

=3 tan A− tan3 A

1−3 tan2 A

≡ LHS.


326 Trig or Treat

tan(45◦ +A) tan(45◦−A) ≡ cot(45◦ +A)cot(45◦−A)


1. tan 45◦ = 1, cot 45◦ = 12. both sides equally complex; therefore maybe easier to work on both sides to

reduce to common terms3. ( ) suggests expansion4. rearrange and simplify.

LHS = tan(45◦ +A) tan(45◦−A)

=(tan 45◦ + tan A)

(1− tan 45◦ tan A)·

(tan 45◦− tan A)

(1+ tan 45◦ tan A)

=(1+ tan A)

(1− tan A)·

(1− tan A)

(1+ tan A)

= 1

RHS = cot(45◦ +A)cot(45◦−A)

=1

tan(45◦ +A)·

1tan(45◦−A)

=(1− tan 45◦ tan A)

(tan 45◦ + tan A)·

(1+ tan 45◦ tan A)

(tan 45◦− tan A)

=(1− tan A)

(1+ tan A)·

(1+ tan A)

(1− tan A)

= 1


tan 45◦ = 1cot 45◦ = 1


Level-Two-Games 327

sin A+ sin 2A2+3 cos A+ cos 2A

≡ tanA2


1. sin 2A, cos 2A suggest “double angle” expansion2. (A/2) on RHS suggests “half-angle formula”3. rearrange and simplify.

LHS =sin A+ sin 2A

2+3 cos A+ cos 2A

=sin A+2 sin A cos A

2+3 cos A+(2 cos2 A−1)

=sin A(1+2 cos A)

1+3 cos A+2 cos2 A

=sin A(1+2 cos A)

(1+ cos A)(1+2 cos A)

=2 sin

A2

cosA2

1+

(

2 cos2 A2−1

)

=2 sin

A2

cosA2

2 cos2 A2

=sin

A2

cosA2

= tanA2

≡ RHS.


328 Trig or Treat

cosec A tanA2−

cos 2A1+ cos A

≡ 4 sin2 A2


1. cosec = 1/sin2. 2A, A/2 suggest expansion using “double angle”, “half angle” formulas3. common denominator4. rearrange and simplify.

LHS = cosec A tanA2−

cos 2A1+ cos A

=1

2 sinA2

cosA2

·

sinA2

cosA2

−

(2 cos2 A−1)

1+ cos A

=1

2 cos2 A2

−

(

2 cos2 A−1)

1+

(

2 cos2 A2−1

)

=1

2 cos2 A2

−

(

2 cos2 A−1)

2 cos2 A2

=1− (2 cos2 A−1)

2 cos2 A2

=2−2 cos2 A

2 cos2 A2

=1− cos2 A

cos2 A2


Level-Two-Games 329

=sin2 A

cos2 A2

=

(

2 sinA2

cosA2

)2

cos2 A2

=4 sin2 A

2cos2 A

2

cos2 A2

= 4 sin2 A2

≡ RHS.


330 Trig or Treat

sin 2A cos A−2 cos 2A sin A2 sin A− sin 2A

≡ 2 cos2 A2


1. 2A suggests “double angle” expansion2. 2 cos2(A/2) on RHS suggest cos A ≡ 2 cos2(A/2)−13. rearrange and simplify.

LHS =sin 2A cos A−2 cos 2A sin A

2 sin A− sin 2A

=2 sin A cos A · cos A−2 cos 2A sin A

2 sin A−2 sin A cos A

=2 sin A(cos2 A− cos 2A)

2 sin A(1− cos A)

=cos2 A− (2 cos2 A−1)

1− cos A

=1− cos2 A1− cos A

=(1− cos A)(1+ cos A)

(1− cos A)

= (1+ cos A)

= 1+

(

2 cos2 A2−1

)

= 2 cos2 A2

≡ RHS.


Level-Two-Games 331

cos A1+ cos 2A

+sin A

1− cos 2A≡

sin A+ cos Asin 2A


1. 2A suggests expansion of “compound angles”2. rearrange and simplify.

LHS =cos A

1+ cos 2A+

sin A1− cos 2A

=cos A(1− cos 2A)+ sin A(1+ cos 2A)

(1+ cos 2A)(1− cos 2A)

=cos A(1− (2 cos2 A−1))+ sin A(2 cos2 A)

(1− cos2 2A)

=cos A(2−2 cos2 A)+2 sin A cos2 A

sin2 2A

=2 cos A−2 cos3 A+2 sin A cos2 A

sin2 2A

=2 cos A(1− cos2 A+ sin A cos A)

sin2 2A

=2 cos A(sin2 A+ sin A cos A)

sin2 2A

=2 cos A sin A(sin A+ cos A)

sin2 2A

=sin 2A(sin A+ cos A)

sin2 2A

=(sin A+ cos A)

sin 2A

≡ RHS.


332 Trig or Treat

1+ tan A1− tan A

+1− tan A1+ tan A

≡ 2 sec 2A


1. t = s/c, sec = 1/cos2. 2A suggests “double angle”3. common denominator4. rearrange and simplify.

LHS =1+ tan A1− tan A

+1− tan A1+ tan A

=(1+ tan A)2 +(1− tan A)2

(1− tan A)(1+ tan A)

=(1+2 tan A+ tan2 A)+(1−2 tan A+ tan2 A)

1− tan2 A

=2+2 tan2 A1− tan2 A

=2(1+ tan2 A)

1− tan2 A

= 2 · (sec2 A)1

1−sin2 Acos2 A

= 2 · sec2 A ·

cos2 A

(cos2 A− sin2 A)

= 21

cos 2A

= 2 sec 2A

≡ RHS.


Level-Two-Games 333

sin 2A+ cos 2A+1sin 2A+ cos 2A−1

≡

tan(45◦ +A)

tan A


1. 2A suggests expansion of “compound angle”2. tan(45◦ +A) suggests expansion of “compound angle”3. both sides to be simplified before identity is established4. t = s/c5. rearrange and simplify.

LHS =sin 2A+ cos 2A+1sin 2A+ cos 2A−1

=(2 sin A cos A)+(cos2 A− sin2 A)+(cos2 A+ sin2 A)

(2 sin A cos A)+(cos2 A− sin2 A)− (cos2 A+ sin2 A)

=2 sin A cos A+2 cos2 A

2 sin A cos A−2 sin2 A

=2 cos A(sin A+ cos A)

2 sin A(cos A− sin A)

=1

tan A

(

tan A+11− tan A

)

dividing allterms by cos A.

RHS =tan(45◦ +A)

tan A

=

(

tan 45◦ + tan A1− tan 45◦ tan A

)(

1tan A

)

=1

tan A

(

1+ tan A1− tan A

)

≡ simplified form of LHS

tan 45◦ = 1


334 Trig or Treat

√

(

1− sin A1+ sin A

)

≡ sec A− tan A


1. square the sq root function2. rearrange and simplify.

LHS =

√

1− sin A1+ sin A

(LHS)2 =1− sin A1+ sin A

=1− sin A1+ sin A

·

(

1− sin A1− sin A

)

=(1− sin A)2

1− sin2 A

=(1− sin A)2

cos2 A

=

(

1cos A

−

sin Acos A

)2

= (sec A− tan A)2

≡ (RHS)2

∴ LHS ≡ RHS.


Level-Two-Games 335

sin2 2A+2 cos 2A−1

sin2 2A+3 cos 2A−3≡

11− sec 2A


1. s2 suggests s2 + c2 ≡ 12. factorisation of LHS3. rearrange and simplify.

LHS =sin2 2A+2 cos 2A−1

sin2 2A+3 cos 2A−3

=(1− cos2 2A)+2 cos 2A−1(1− cos2 2A)+3 cos 2A−3

=cos 2A(2− cos 2A)

3 cos 2A− cos2 2A−2

=cos 2A(2− cos 2A)

(cos 2A−1)(2− cos 2A)

=cos 2A

(cos 2A−1)

=1

1− sec 2A

≡ RHS.

divide numeratorand denominatorby cos 2A


336 Trig or Treat

sin(A+45◦)cos(A+45◦)

+cos(A+45◦)sin(A+45◦)

≡ 2 sec 2A


1. (A+45◦) suggests expansion of “compound angle”2. 2A suggests “double angle”3. sec = 1/cos4. common denominator5. rearrange and simplify.

LHS =sin(A+45◦)cos(A+45◦)

+cos(A+45◦)sin(A+45◦)

=sin2(A+45◦)+ cos2(A+45◦)

cos(A+45◦)sin(A+45◦)

=1

cos(A+45◦)sin(A+45◦)

=2

2 cos(A+45◦)sin(A+45◦)

=2

sin 2(A+45◦)

=2

sin(2A+90◦)

=2

cos 2A

= 2 sec 2A

≡ RHS.

since sin(x+90)= cos x

(turns out that the problem is easier than expected, as the use of commondenominator resulted in s2 +c2

≡ 1; hence there is no need to expand (A+

45◦)!)


Level-Two-Games 337

tan A+ tan 2A ≡

sin A(4 cos2 A−1)

cos A cos 2A


1. t = s/c2. 2A suggests expansion of “double angle”3. rearrange and simplify.

LHS = tan A+ tan 2A

=sin Acos A

+sin 2Acos 2A

=sin A cos 2A+ sin 2A cos A

cos A cos 2A

=sin A(2 cos2 A−1)+ cos A(2 sin A cos A)

cos A cos 2A

=sin A(2 cos2 A−1+2 cos2 A)

cos A cos 2A

=sin A(4 cos2 A−1)

cos A cos 2A

≡ RHS.


338 Trig or Treat

(tan A− cosec A)2− (cot A− sec A)2

≡ 2(cosec A− sec A)


1. t = s/c, cosec = 1/sin, cot = c/s, sec = 1/cos2. ( )2 suggests s2 + c2 ≡ 13. rearrange and simplify.

LHS = (tan A− cosec A)2− (cot A− sec A)2

= (tan2 A−2 tan Acosec A+ cosec 2A)

− (cot2 A−2 cot A sec A+ sec2 A)

= (tan2 A− sec2 A)+(cosec 2A− cot2 A)

−2sin Acos A

·

1sin A

+2cos Asin A

·

1cos A

= (−1)+(1)−2

cos A+

2sin A

= 2

(

1sin A

−

1cos A

)

= 2(cosec A− sec A)

≡ RHS.


Level-Two-Games 339

2 sin 2A(1−2 sin2 A) ≡ sin 4A


1. sin 4A suggests sin 2(2A)2. (1−2 sin2 A) equals cos 2A3. rearrange and simplify.

LHS = 2 sin 2A(1−2 sin2 A)

= 2 sin 2A cos 2A

= sin 4A

≡ RHS.

Alternatively since sin 4A is a standard expression we can proceed with theRHS.

RHS = sin 4A

= 2 sin 2A cos 2A

= 2 sin 2A(1−2 sin2 A)

≡ LHS.


340 Trig or Treat

cos 3A ≡ 4 cos3 A−3 cos A


1. 3A suggests double expansion of “compound angle”2. rearrange and simplify.

LHS = cos 3A

= cos(2A+A)

= cos 2A cos A− sin 2A sin A

= cos A(cos2 A− sin2 A)− sin A(2 sin A cos A)

= cos3 A− sin2 A cos A−2 sin2 A cos A

= cos3 A−3 sin2 A cos A

= cos3 A−3 cos A(1− cos2 A)

= cos3 A−3 cos A+3 cos3 A

= 4 cos3 A−3 cos A

≡ RHS.


Level-Two-Games 341

32 cos6 A−48 cos4 A+18 cos2 A−1 ≡ cos 6A


1. cos 6A on RHS suggest “double angle” formula for (3A)2. 3A suggests expansion of “triple angle”3. this is one of the rare occasions where it may be easier to start with the simpler

RHS4. rearrange and simplify.

RHS = cos 6A

= cos 2(3A)

= 2 cos2(3A)−1

= 2(4 cos3 A−3 cos A)2−1

= 2 cos2 A(4 cos2 A−3)2−1

= 2 cos2 A(16 cos4 A+24 cos2 A+9)−1

= 32 cos6 A+48 cos4 A+18 cos2 A−1

≡ LHS.

see previousproof


342 Trig or Treat

cos 4A+4 cos 2A+3 ≡ 8 cos4 A


1. cos 4A, cos 2A suggest expansion of “double angle”2. rearrange and simplify.

LHS = (cos 4A)+4 cos 2A+3

= (2 cos2 2A−1)+4 cos 2A+3

= 2 cos2 2A+4 cos 2A+2

= 2(cos2 2A+2 cos 2A+1)

= 2(cos 2A+1)2

= 2(2 cos2 A−1+1)2

= 2(4 cos4 A)

= 8 cos4 A

≡ RHS.


Level-Two-Games 343

sin 3Asin A

−

cos 3Acos A

≡ 2


1. sin 3A, cos 3A suggest expansion of “compound angle”2. rearrange and simplify.

LHS =sin 3Asin A

−

cos 3Acos A

=(3 sin A−4 sin3 A)

sin A−

4(cos3 A−3 cos A)

cos A

=sin A(3−4 sin2 A)

sin A−

cos A(4 cos2 A−3)

cos A

= 3−4 sin2 A−4 cos2 A+3

= 6−4(sin2 A+ cos2 A)

= 6−4

= 2

≡ RHS.

see proofson p. 324and 340


344 Trig or Treat

sin4 A+ cos4 A ≡

34

+14

cos 4A


1. The cos 4A on the RHS, a standard function, suggests that it may be easier tostart with the RHS through a double expansion of cos 2(2A)


RHS =34

+14

cos 4A

=34

+14

(2 cos2 2A−1)

=34

+12

cos2 2A−

14

=12

+12

(2 cos2 A−1)2

=12

+12

(4 cos4 A−4 cos2 A+1)

= 1+2 cos4 A−2 cos2 A

= cos4 A+ cos4 A−2 cos2 A+1

= cos4 A+(cos2 A−1)2

= cos4 A+(sin2 A)2

= cos4 A+ sin4 A

≡ LHS.


Level-Two-Games 345

8 cos4 A−4 cos 2A−3 ≡ cos 4A


1. cos 4A on the RHS is a standard expression and suggests the expansion ofcos 4A → cos 2A → cos A

2. hence, easier to begin from the RHS3. rearrange and simplify.

RHS = cos 4A

= cos 2(2A)

= 2 cos2(2A)−1

= 2(2 cos2 A−1)2−1

= 2(4 cos4 A−4 cos2 A+1)−1

= 8 cos4 A−8 cos2 A+2−1

= 8 cos4 A−4(2 cos2 A−1)−2−1

= 8 cos4 A−4 cos 2A−3

≡ LHS.


346 Trig or Treat

1−8 sin2 A cos2 A ≡ cos 4A


1. cos 4A — standard expression; therefore easier to begin with RHS and expandtwice cos 4A → cos 2A → cos A

2. s2, c2 suggest s2 + c2 ≡ 13. rearrange and simplify.

RHS = cos 4A

= 2 cos2(2A)−1

= 2(2 cos2 A−1)2−1

= 2(4 cos4 A−4 cos2 A+1)−1

= (8 cos4 A−8 cos2 A+2)−1

= 8 cos2 A(cos2 A−1)+1

= 8 cos2 A(−sin2 A)+1

= 1−8 sin2 A cos2 A

≡ LHS.


Level-Two-Games 347

1− cos 2A+ sin Asin 2A+ cos A

≡ tan A


1. 2A suggests expansion of “double angle”2. t = s/c3. rearrange and simplify.

LHS =1− cos 2A+ sin A

sin 2A+ cos A

=1− (cos2 A− sin2 A)+ sin A

2 sin A cos A+ cos A

=(cos2 A+ sin2 A)− (cos2 A− sin2 A)+ sin A

cos A(2 sin A+1)

=2 sin2 A+ sin A

cos A(2 sin A+1)

=sin Acos A

(2 sin A+1)

(2 sin A+1)

= tan A

≡ LHS.

September 12, 2007 19:18 Book:- Trig or Treat (9in x 6in) 11˙not-so-easy

Level-Three-GamesNot-So-Easy Proofs


350 Trig or Treat


(A+B+C) = 180◦

sin A+ sin B+ sin C ≡ 4 cosA2

cosB2

cosC2


1. (sin A+ sin B) suggests sin S+ sin T formula2. C = 180◦− (A+B)3. A/2, B/2, C/2 suggest “half angle” formula4. rearrange and simplify.

see next page


Level-Three-Games 351

LHS = sin A+sin B+sin C

= 2 sinA+B

2cos

A−B2

+sin C

= 2 sin

(

180◦−C2

)

cos(A−B)

2+2 sin

C2

cosC2

= 2 cosC2

cos(A−B)

2+2 sin

C2

cosC2

= 2 cosC2

(

cosA−B

2+sin

C2

)

= 2 cosC2

(

cosA−B

2+cos

A+B2

)

= 2 cosC2

(

2 cos(A−B)+(A+B)

4cos

(A−B)−(A+B)

4

)

= 2 cosC2

(

2 cos2A4

cos(−2B)

4

)

= 4 cosC2

cosA2

cosB2

= 4 cosA2

cosB2

cosC2

≡RHS.

sin

(

90◦−C2

)

= cosC2

sinC2

= cos

(

90◦−C2

)

= cosA+B

2

|cos(−B) = cos B


352 Trig or Treat

sin(A+B+C)

≡

sin A cos B cos C+ sin B cos C cos A+ sin C cos A cos B− sin A sin B sin C

∗


1. (A+B+C) suggests expansion of “compound angle” twice2. rearrange and simplify.

LHS = sin(A+B+C)

= sin(A+B)cos C + cos(A+B)sin C

= cos C(sin A cos B+ cos A sin B)

+ sin C(cos A cos B− sin A sin B)

= sin A cos B cos C + sin B cos C cos A

+ sin C cos A cos B− sin A sin B sin C

≡ RHS.

∗Note that:

this general identity is for any three angles A, B and C; they need not necessarilybe for the three angles of a triangle.




(A+B+C = 180◦)

sin A cos B cos C + sin B cos C cos A+ sin C cos A cos B

≡ sin A sin B sin C


1. This identity is easiest proved as a follow-up from the previous proof of thegeneral identity:

sin(A+B+C)≡

sin A cos B cos C

+ sin B cos C cos A

+ sin C cos A cos B

− sin A sin B sin C

2. sin(A+B+C) = sin 180◦ = 0.

Since sin(A+B+C) = sin 180◦ = 0,

∴ sin A cos B cos C

+ sin B cos C cos A

+ sin C cos A cos B

− sin A sin B sin C

= 0

∴ sin A cos B cos C

+ sin B cos C cos A

+ sin C cos A cos B

= sin A sin B sin C

∴ LHS ≡ RHS.

(This identity is extremely valuable for proving many subsequent identities in-volving the three angles of a triangle. Without knowledge of the preceding generalidentity, this identity is more difficult to prove. Try it out for yourself)


354 Trig or Treat


(A+B+C) = 180◦

sin 2A+ sin 2B+ sin 2C ≡ 4 sin A sin B sin C


1. sin 2A+ sin 2B suggests sin S+ sin T formula2. sin 2C suggests “double angle” formula3. (A+B+C) = 180◦ therefore express values for (180◦−C) in terms of A+B4. rearrange and simplify.

LHS = sin 2A+ sin 2B+ sin 2C

= 2 sin

(

2A+2B2

)

cos

(

2A−2B2

)

+2 sin C cos C

= 2 sin(A+B)cos(A−B)+2 sin C cos C

= 2 sin C cos(A−B)−2 sin C cos(A+B)

= 2 sin C(cos(A−B)− cos(A+B))

= 2 sin C

(

−2 sin(A−B)+(A+B)

2sin

(A−B)− (A+B)

2

)

= 2 sin C(−2 sin A sin(−B))

= 4 sin C sin A sin B

= 4 sin A sin B sin C

≡ RHS.

see identitiesbelow

sin(A+B) = sin(180◦− (A+B))

= sin C

cos C = −cos(180◦−C)

= −cos(A+B)




(A+B+C) = 180◦

sin 2A+ sin 2B+ sin 2C ≡ 4 sin A sin B sin C

For this beautiful identity let’s explore a second approach.


1. 2A, 2B, 2C suggest expansion of “double angles”2. A+B+C = 180◦


LHS = sin 2A+ sin 2B+ sin 2C

= 2 sin A cos A

+2 sin B cos B

+2 sin C cos C

= 2 sin A(sin B sin C− cos B cos C)

+2 sin B(sin C sin A− cos C cos A)

+2 sin C(sin A sin B− cos A cos B)

= 6 sin A sin B sin C−2 sin A cos B cos C

−2 sin B cos C cos A

−2 sin C cos A cos B

see identitiesbelow

= 6 sin A sin B sin C−2(sin A sin B sin C)

= 4 sin A sin B sin C

≡ RHS.

from previousproof on p. 353

cos A = cos[180◦− (B+C)]= −cos(B+C)= sin B sin C− cos B cos C

A = 180◦− (B+C)

sin A cos B cos Csin B cos C cos Asin C cos A cos B

=sin(A+B+C)

+ sin A sin B sin Csin(A+B+C)

= sin(180◦)= 0


356 Trig or Treat


(A+B+C) = 180◦

tan A+ tan B+ tan C ≡ tan A tan B tan C


1. t = s/c2. (A+B+C) = 180◦


LHS = tan A+ tan B+ tan C

=sin Acos A

+sin Bcos B

+sin Ccos C

=sin A cos B cos C + sin B cos C cos A+ sin C cos A cos B

cos A cos B cos C

=sin A sin B sin C

cos A cos B cos C

= tan A tan B tan C

≡ RHS.

see identity∗

below

∗For (A+B+C) = 180◦

sin A cos B cos C

+sin B cos C cos A

+sin C cos A cos B

= sin A sin B sin Cfrom previousproof on p. 353

An alternative proof is to begin with tan C = − tan(A + B) with expansion oftan(A + B) so that tan A + tan B can be expressed in terms of tan A, tan B andtan C. Try it.




(A+B+C) = 180◦

cos A+ cos B+ cos C ≡ 4 sinA2

sinB2

sinC2

+1


1. cos A+ cos B suggests cos S+ cos T formula2. since (A + B +C) = 180◦, then C/2 is complementary angle of (A + B)/2;

i.e. (90− (A+B)/2)3. rearrange and simplify.

LHS=(cos A+cos B)+(cos C)

=

(

2 cosA+B

2cos

A−B2

)

+

(

1−2 sin2 C2

)

= 2 sinC2

cosA−B

2+1−2 sin2 C

2

= 2 sinC2

(

cosA−B

2−sin

C2

)

+1

= 2 sinC2

(

cosA−B

2−cos

A+B2

)

+1

= 2 sinC2

(

−2 sin12

(

A−B2

+A+B

2

)

sin12

(

A−B2

−A+B2

))

+1

= 2 sinC2

(

−2 sinA2

sin

(−B2

))

+1

= 4 sinC2

sinA2

sinB2

+1

= 4 sinA2

sinB2

sinC2

+1

≡RHS.


358 Trig or Treat

cos 3A+ cos 2A ≡ 2 cos5A2

cosA2


1. cos S+ cos T ≡ 2 cos(S+T)/2 cos(S−T)/22. rearrange and simplify.

LHS = cos 3A+ cos 2A

= 2 cos3A+2A

2cos

3A−2A2

= 2 cos5A2

cosA2

≡ RHS.



sin 5A− sin 3A ≡ 2 sin A cos 4A


1. sin S− sin T = 2 cos(S+T )/2 sin(S−T )/22. rearrange and simplify.

LHS = sin 5A+ sin 3A

= 2 cos5A+3A

2sin

5A−3A2

= 2 cos 4A sin A

= 2 sin A cos 4A

≡ RHS.


360 Trig or Treat

sin A+ sin Bsin A− sin B

≡ tanA+B

2cot

A−B2


1. sin S+ sin T ≡ 2 sin(S+T )/2 cos(S−T )/22. sin S− sin T ≡ 2 cos(S+T )/2 cos(S−T )/23. rearrange and simplify.

LHS =sin A+ sin Bsin A− sin B

=2 sin

A+B2

cosA−B

2

2 cosA+B

2sin

A−B2

= tanA+B

2cot

A−B2

≡ RHS.



sin(2A+B)+ sin Bcos(2A+B)+ cos B

≡ tan(A+B)


1. sin S+ sin Tcos S+ cos T


LHS =sin(2A+B)+ sin Bcos(2A+B)+ cos B

=2 sin

(2A+B)+B2

cos(2A+B)−B

2

2cos(2A+B)+B

2cos

(2A+B)−B2

=2 sin(A+B)cos A2cos(A+B)cos A

= tan(A+B)

≡ RHS.


362 Trig or Treat

cos A+ cos Bcos A− cos B

≡−cotA+B

2cot

A−B2


1. cos S+ cos T2. cos S− cos T3. rearrange and simplify.

LHS =cos A+ cos Bcos A− cos B

=2 cos

A+B2

cosA−B

2

−2 sinA+B

2sin

A−B2

= −cotA+B

2cot

A−B2

≡ RHS.



sin A+ sin Bcos A+ cos B

≡ tanA+B

2


1. sin S+ sin T2. cos S+ cos T3. rearrange and simplify.

LHS =sin A+ sin Bcos A+ cos B

=2 sin

A+B2

cosA−B

2

2 cosA+B

2cos

A−B2

=sin

A+B2

cosA+B

2

= tanA+B

2

≡ RHS.


364 Trig or Treat

sin A− sin Bcos A− cos B

≡−cotA+B

2


1. sin S− sin T2. cos S− cos T3. rearrange and simplify.

LHS =sin A− sin Bcos A− cos B

=2 cos

A+B2

sinA−B

2

−2 sinA+B

2sin

A−B2

= −cos

A+B2

sinA+B

2

= −cotA+B

2

≡ RHS.



sin A+ sin 3A2 sin 2A

≡ cos A


1. sin S+ sin T2. rearrange and simplify.

LHS =sin A+ sin 3A

2 sin 2A

=2 sin

A+3A2

cosA−3A

22 sin 2A

=2 sin 2A cos(−A)

2 sin 2A

= cos(−A)

= cos A

≡ RHS.


366 Trig or Treat

sin 4A− sin 2Acos 4A+ cos 2A

≡ tan A


1. sin 4A− sin 2A suggests sin S− sin T2. cos 4A+ cos 2A suggests cos S+ cos T3. t = s/c4. rearrange and simplify.

LHS =sin 4A− sin 2Acos 4A+ cos 2A

=2 cos

4A+2A2

sin4A−2A

2

2 cos4A+2A

2cos

4A−2A2

=sin Acos A

= tan A

≡ RHS.




≡ tan A


1. cos S− cos T2. sin S+ sin T3. rearrange and simplify.

LHS =cos A− cos 3Asin 3A+ sin A

=−2 sin

A+3A2

sinA−3A

2

2 sinA+3A

2cos

A−3A2

= − sin(−A)

cos(−A)

=sin Acos A

= tan A

≡ RHS.


368 Trig or Treat

cos A+ cos 3A2 cos 2A

≡ cos A


1. cos S+ cos T2. rearrange and simplify.

LHS =cos A+ cos 3A

2 cos A

=2 cos

A+3A2

cosA−3A

22 cos 2A

=2 cos 2A cos(−A)

2 cos 2A

= cos(−A)

= cos A

≡ RHS.



sin A− sin 3A

sin2 A− cos2 A≡ 2 sin A


1. sin A− sin 3A suggests sin S− sin T formula2. s2

− c2 suggests cos 2A ≡ c2− s2


LHS =sin A− sin 3A

sin2 A− cos2 A

=2 cos

A+3A2

sinA−3A

2−cos 2A

=2 cos 2A sin(−A)

−cos 2A

= 2 sin A

≡ RHS.


370 Trig or Treat

sin(A+B)− sin(A−B)

cos(A+B)+ cos(A−B)≡ tan B


1. sin S− sin Tcos S− cos T


LHS =sin(A+B)− sin(A−B)

cos(A+B)+ cos(A−B)

=2 cos

(A+B)+(A−B)

2sin

(A+B)− (A−B)

2

2 cos(A+B)+(A−B)

2cos

(A+B)− (A−B)

2

=sin Bcos B

= tan B

≡ RHS.




≡ tan 2A


1. cos S− cos T2. sin S+ sin T3. rearrange and simplify.

LHS =cos A− cos 5Asin 5A+ sin A

=−2 sin

A+5A2

sinA−5A

2

2 sinA+5A

2cos

A−5A2

=−sin(−2A)

cos(−2A)

=sin 2Acos 2A

= tan 2A

≡ RHS.


372 Trig or Treat

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

≡ tan 2A


1. sin A + sin 2A + sin 3A, cos A + cos 2A + cos 3A suggest sin S + sin T andcos S+ cos T formulas using A and 3A

2. t = s/c3. rearrange and simplify.

LHS =sin A+ sin 2A+ sin 3A

cos A+ cos 2A+ cos 3A

=sin 2A+(sin A+ sin 3A)

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

=

sin 2A+

(

2 sinA+3A

2cos

A−3A2

)

cos 2A+

(

2 cosA+3A

2cos

A−3A2

)

=sin 2A+(2 sin 2A cos(−A))

cos 2A+(2 cos 2A cos(−A))

=3 sin 2A3 cos 2A

=sin 2Acos 2A

= tan 2A

≡ RHS.




≡ tan 3A



LHS =sin 4A+ sin 2Acos 4A+ cos 2A

=2 sin

4A+2A2

cos4A−2A

2

2 cos4A+2A

2cos

4A−2A2

=sin 3A cos Acos 3A cos A

= tan 3A

≡ RHS.


374 Trig or Treat

cos A− cos 3Asin 3A− sin A

≡ tan 2A


1. cos S− cos T2. sin S− sin T3. rearrange and simplify.

LHS =cos A− cos 3Asin 3A− sin A

=−2 sin

A+3A2

sinA−3A

2

2 cos3A+A

2sin

3A−A2

= − sin 2A sin(−A)

cos 2A sin A

=sin 2Acos 2A

· sin Asin A

= tan 2A

≡ RHS.




≡ tan 6A



LHS =sin 4A+ sin 8Acos 4A+ cos 8A

=2 sin

4A+8A2

cos4A−8A

2

2 cos4A+8A

2cos

4A−8A2

=sin 6Acos 6A

= tan 6A

≡ RHS.


376 Trig or Treat

sin 4A− sin 8Acos 4A− cos 8A

≡−cot 6A


1. sin S− sin T2. cos S+ cos T3. rearrange and simplify.

LHS =sin 4A− sin 8Acos 4A− cos 8A

=2 cos

4A+8A2

sin4A−8A

2

−2 sin4A+8A

2sin

4A−8A2

= −cos 6Asin 6A

= −cot 6A

≡ RHS.




≡ tan 2A tan 6A


1. cos S− cos T2. cos S+ cos T3. rearrange and simplify.

LHS =cos 4A− cos 8Acos 4A+ cos 8A

=−2 sin

4A+8A2

sin4A−8A

2

2 cos4A+8A

2cos

4A−8A2

= − sin 6Acos 6A

· sin(−2A)

cos(−2A)

=sin 6Acos 6A

· sin 2Acos 2A

= tan 2A tan 6A

≡ RHS.


378 Trig or Treat


≡ − tan 6Atan 2A


1. sin S+ sin T2. sin S− sin T3. rearrange and simplify.

LHS =sin 4A+ sin 8Asin 4A− sin 8A

=2 sin

4A+8A2

cos4A−8A

2

2 cos4A+8A

2sin

4A−8A2

=sin 6Acos 6A

· cos(−2A)

sin(−2A)

= tan 6A ·(

cos 2A−sin 2A

)

=− tan 6A

tan 2A

≡ RHS.



sin 5A+2 sin 3A+ sin A ≡ 4 sin 3A cos2 A


1. sin 5A+2 sin 3A+ sin A suggests sin S+ sin T formula using 5A and A to give3A which is present on both LHS and RHS.


LHS = sin 5A+2 sin 3A+ sin A

= (2 sin 3A)+(sin 5A+ sin A)

= (2 sin 3A)+

(

2 sin5A+A

2cos

5A−A2

)

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

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

= 2 sin 3A(1+(2 cos2 A−1))

= 2 sin 3A(2 cos2 A)

= 4 sin 3A cos2 A

≡ RHS.


380 Trig or Treat

1− cos 2A+ cos 4A− cos 6A ≡ 4 sin A cos 2A sin 3A


1. cos 4A, cos 6A suggest cos S− cos T2. (1− cos 2A) suggests 1− cos 2A ≡ 2 sin2 A3. rearrange and simplify.

LHS = 1− cos 2A+ cos 4A− cos 6A

= (1− cos 2A)+(cos 4A− cos 6A)

= (2 sin2 A)+

(

−2 sin

(

4A+6A2

)

sin

(

4A−6A2

))

= (2 sin2 A)− (2(sin 5A) sin(−A))

= 2 sin2 A+2 sin A sin 5A

= 2 sin A(sin A+ sin 5A)

= 2 sin A

(

2 sinA+5A

2cos

A−5A2

)

= 4 sin A(sin 3A cos(−2A))

= 4 sin A sin 3A cos 2A

= 4 sin A cos 2A sin 3A

≡ RHS.




− tan 3A tan A ≡ 0


1. cos S− cos T2. cos S+ cos T3. t = s/c4. rearrange and simplify.

LHS =cos 2A− cos 4Acos 2A+ cos 4A

− tan 3A tan A

=

−2 sin

(

2A+4A2

)

sin

(

2A−4A2

)

2 cos

(

2A+4A2

)

cos

(

2A−4A2

) − tan 3A tan A

= − sin 3Acos 3A

· sin(−A)

cos(−A)− tan 3A tan A

= tan 3A tan A− tan 3A tan A

= 0

≡ RHS.


382 Trig or Treat


+tan 3Atan A

≡ 0


1. sin S+ sin T2. sin S− sin T3. t = s/c4. rearrange and simplify.

LHS =sin 2A+ sin 4Asin 2A− sin 4A

+tan 3Atan A

=

2 sin

(

2A+4A2

)

cos

(

2A−4A2

)

2 cos

(

2A+4A2

)

sin

(

2A−4A2

) +tan 3Atan A

=sin 3Acos 3A

· cos(−A)

sin(−A)+

tan 3Atan A

= tan 3A ·(

− 1tan A

)

+tan 3Atan A

= − tan 3Atan A

+tan 3Atan A

= 0

≡ RHS.



cos 2A− cos 10A ≡ tan 4A(sin 2A+ sin 10A)


1. (cos 2A− cos 10A) suggest cos S− cos T2. (sin 2A+ sin 10A) suggests sin S+ sin T3. both sides are complex; ∴ may have to simplify both sides4. tan 4A may come from (10A−2A)/25. rearrange and simplify.

LHS = cos 2A− cos 10A

= −2 sin

(

2A+102

)

sin

(

2A−10A2

)

= −2 sin 6A sin(−4A)

= 2 sin 6A sin 4A.

RHS = tan 4A(sin 2A+ sin 10A)

= tan 4A

(

2 sin

(

2A+102

)

cos

(

2A−10A2

))

= tan 4A(2 sin 6A) cos(−4A))

= tan 4A ·2 sin 6A cos 4A

=sin 4Acos 4A

·2 sin 6A cos 4A

= 2 sin 6A sin 4A

∴ LHS ≡ RHS.


384 Trig or Treat

sin 2A+ sin 4A− sin 6A ≡ 4 sin A sin 2A sin 3A


1. sin 2A− sin 6A suggests sin S− sin T formula2. sin 4A suggests “ double angle” formula to give 2A3. rearrange and simplify.

LHS = sin 2A+(sin 4A)− sin 6A

= 2 cos2A+6A

2sin

2A−6A2

+(sin 4A)

= 2 cos 4A sin(−2A)+ sin 4A

= −2 cos 4A sin 2A+2 sin 2A cos 2A

= 2 sin 2A(cos 2A− cos 4A)

= 2 sin 2A

(

−2 sin2A+4A

2sin

2A−4A2

)

= 2 sin 2A(−2 sin 3A sin(−A))

= 2 sin 2A(2 sin 3A sin A)

= 4 sin A sin 2A sin 3A

≡ RHS.



cos 3A+2 cos 5A+ cos 7A ≡ 4 cos2 A cos 5A


1. cos 3A, cos 7A suggest cos S+ cos T formula2. rearrange and simplify.

LHS = cos 3A+2 cos 5A+ cos 7A

= (cos 3A+ cos 7A)+2 cos 5A

=

(

2 cos3A+7A

2cos

7A−3A2

)

+2 cos 5A

= (2 cos 5A cos 2A)+2 cos 5A

= 2 cos 5A(cos 2A+1)

= 2 cos 5A(2 cos2 A−1+1)

= 4 cos2 A cos 5A

≡ RHS.


386 Trig or Treat

1+ cos 2A+ cos 4A+ cos 6A ≡ 4 cos A cos 2A cos 3A


1. cos 2A, cos 4A suggest cos S+ cos T formula2. cos 6A suggests cos 2(3A)3. rearrange and simplify.

LHS = 1+ cos 2A+ cos 4A+ cos 6A

= 1+2 cos2A+4A

2cos

2A−4A2

+ cos 6A

= 1+2 cos 3A cos(−A)+ cos 6A

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

= 1+2 cos 3A cos A+(2 cos2 3A−1)

= 2 cos 3A cos A+2 cos2 3A

= 2 cos 3A(cos A+ cos 3A)

= 2 cos 3A

(

2 cosA+3A

2· cos

A−3A2

)

= 4 cos 3A cos 2A cos(−A)

= 4 cos A cos 2A cos 3A

≡ RHS.



1− cos 2A+ cos 4A− cos 6A ≡ 4 sin A cos 2A sin 3A


1. cos 2A, cos 4A suggest cos S− cos T2. cos 6A suggests cos 2(3A)3. rearrange and simplify.

LHS = 1− cos 2A+ cos 4A− cos 6A

= 1+(cos 4A− cos 2A)− cos 6A

= 1+

(

−2 sin4A+2A

2sin

4A−2A2

)

− cos 6A

= 1−2 sin 3A sin A− cos 2(3A)

= 1−2 sin 3A sin A− (1−2 sin2 3A)

= −2 sin 3A sin A+2 sin2 3A

= 2 sin 3A(sin 3A− sin A)

= 2 sin 3A

(

2 cos3A+A

2sin

3A−A2

)

= 2 sin 3A(2 cos 2A sin A)

= 4 sin A cos 2A sin 3A

≡ RHS.


388 Trig or Treat

tan 4A(sin 2A+ sin 10A) ≡ cos 2A− cos 10A


1. t = s/c2. sin S+ sin T3. cos S− cos T4. simplify both sides before comparison.

LHS = tan 4A(sin 2A+ sin 10A)

= tan 4A

(

2 sin

(

2A+10A2

)

cos

(

2A−10A2

))

=sin 4Acos 4A

(2 sin 6A cos(−4A))

=sin 4Acos 4A

2 sin 6A cos 4A

= 2 sin 6A sin 4A

RHS = cos 2A− cos 10A

= −2 sin

(

2A+10A2

)

sin

(

2A−10A2

)

= −2 sin 6A sin(−4A)

= 2 sin 6A sin 4A

LHS ≡ RHS.



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


1. Both sides are complex; therefore it may be easier to simplify both sides beforecomparison

2. sin S+ sin T3. cos S− cos T4. rearrange and simplify.

LHS = sin A(sin 3A+ sin 5A)

= sin A

(

2 sin3A+5A

2cos

3A−5A2

)

= sin A(2 sin 4A cos(−A))

= 2 sin A cos A sin 4A

RHS = cos A(cos 3A− cos 5A)

= cos A

(

−2 sin3A+5A

2sin

3A−5A2

)

= cos A(−2 sin 4A sin(−A))


LHS ≡ RHS.


390 Trig or Treat

sin A(sin A+ sin 3A) ≡ cos A(cos A− cos 3A)


1. Both sides appear to be complex and therefore there may be a need to simplifyboth sides before comparison is made

2. sin S+ sin T3. cos S− cos T4. rearrange and simplify.

LHS = sin A(sin A+ sin 3A)

= sin A

(

2 sinA+3A

2cos

A−3A2

)

= sin A(2 sin 2A cos(−A))


RHS = cos A(cos A− cos 3A)

= cos A

(

−2 sinA+3A

2sin

A−3A2

)

= cos A(−2 sin 2A sin(−A))


LHS ≡ RHS.



tan A+ tan(A+120◦)+ tan(A+240◦) ≡ 3 tan 3A∗


1. tan(A+120◦), tan(A+240◦), tan 3A suggest expansion of “compound angles”2. tan 120◦ = tan(180◦−60◦) =− tan 60◦ = −

√

33. tan 240◦ = tan(180◦+60◦) = tan 60◦ =

√

34. both LHS and RHS are complex; therefore may have to expand both sides for

ease of comparison5. tan 3A suggest tan(2A+A) and tan 2A expansions6. rearrange and simplify.

LHS = tan A+ tan(A+120◦)+ tan(A+240◦)

= tan A+

(

tan A+ tan 120◦

1− tan A tan 120◦

)

+

(

tan A+ tan 240◦

1− tan A tan 240◦

)

= tan A+

(

tan A−√

3

1+√

3 tan A

)

+

(

tan A+√

3

1−√

3 tan A

)

= tan A+(tan A−

√3)(1−

√3 tan A)+(tan A+

√3)(1+

√3 tan A)

(1+√

3 tan A)(1−√

3 tan A)

= tan A+[(tan A−√

3 tan2 A−√

3+3 tan A)

+(tan A+√

3 tan2 A+√

3+3 tan A)]/(1−3 tan2 A)

= tan A+8 tan A

1−3 tan2 A

=tan A(1−3 tan2 A)+8 tan A

1−3 tan2 A

=9 tan A−3 tan3 A

1−3 tan2 A

=3(3 tan A− tan3 A)

1−3 tan2 A


392 Trig or Treat

RHS = 3 tan 3A

= 3 tan(A+2A)

= 3

(

tan A+ tan 2A1− tan A tan 2A

)

= 3

tan A+

(

2 tan A1− tan2 A

)

1− tan A

(

2 tan A1− tan2 A

)

= 3

tan A(1− tan2 A)+2 tan A1− tan2 A

(1− tan2 A)−2 tan2 A1− tan2 A

= 3

(

tan A− tan3 A+2 tan A1− tan2 A−2 tan2 A

)

=3(3 tan A− tan3 A)

1−3 tan2 A

∴ LHS ≡ RHS.

An alternative proof is to write the second and third terms of the LHS in terms ofsin/cos; then add them using common denominator, followed by “compound angleformula” for the numerator, and “cos S+cos T formula” for the denominator. Thissecond proof is shorter and more elegant. Try it out for yourself.∗Note that there is greater beauty if the identity is written as:

tan A+ tan(A+120◦)+ tan(A+240◦) ≡ 3 tan(A+360◦).

Of course, tan(A+360◦) = tan A.

September 17, 2007 21:40 Book:- Trig or Treat (9in x 6in) 12˙addenda

Addenda


394 Trig or Treat

Proof for sin(AA + BB)

O

T1

cos

A

A

sin

A

P

Fig. 1

RO S

B

B

BTQ

1

cos

A

A

sin

A

P

Fig. 2


Addenda 395

Proof for sin(AA + BB)

Let us draw a right-angled triangle with angle A as in Fig. 1 (opposite page)

then : sin A = O/H = sin A/1

cos A = A/H = cos A/1

Let’s add a second triangle with angle B to the first triangle. Figure 2(opposite page)

now : sin(A+B) =OH

=PR1

= PQ+QR (from Fig. 2)From geometry, we know that ]QTO = B (alt ]

′s between // lines)Similarly, ]QPT = B (both are complementary ]s of ]PTQ)

For triangle PQT ,

cos B =PQPT

=PQ

sin A

∴ PQ = sin A cos B

For triangle TOS,

sin B =TSOT

=QR

cos A

∴ QR = cos A sin B

sin(A+B) = PQ+QR

= sin A cos B+ cos A sin B

∴ sin(A+B) ≡ sin A cos B+ cos A sin B


396 Trig or Treat

Proof for cos(AA + BB)

RO S

B

B

BTQ

1

cos

A

A

sin

A

P


Addenda 397

Proof for cos(AA + BB)

From the Fig (opposite page)

cos(A+B) =OROP

=OR1

= OS−RS

]QTO = B (alt ]′s between // lines)

]QPT = ]QTO (both are complementary angles to ]PTQ)

= B

For triangle PQT :

sin B =QTPT

=QT

sin A

∴ QT = sin A sin B

For triangle TOS;

cos B =OSOT

=OS

cos A

∴ OS = cos A cos B

cos(A+B) = OS−RS

= OS−QT

= cos A cos B− sin A sin B

∴ cos(A+B) ≡ cos A cos B− sin A sin B

(since RS=QT)


398 Trig or Treat

Proof for tan(AA + BB)

tan(A+B) =sin(A+B)

cos(A+B)

=sin A cos B+ cos A sin Bcos A cos B− sin A sin B

=tan A+ tan B

1− tan A tan Bdivide allfour terms bycos A cos B

Substitute B by (−B)

tan(A+(−B)) =tan A+ tan(−B)

1− tan A tan(−B)

∴ tan(A−B) =tan A− tan B

1+ tan A tan B

Since tan(−B) = − tan B.


Addenda 399

Appetisers for Higher Trigonometry

sin x = x−x3

3!+

x5

5!−

x7

7!+

x9

9!−·· ·

cos x = 1−x2

2!+

x4

4!−

x6

6!+

x8

8!−·· ·

tan x = x+x3

3+

2x5

15+ · · ·

arcsin x = x−12

x3

3+

1 ·32 ·4

x5

5−

1 ·3 ·52 ·4 ·6

·x7

7+ · · ·

arctan x = x−x3

3+

x5

5−

x7

7+ · · ·

ex = 1+x1!

+x2

2!+

x3

3!+

x4

4!+ · · ·

eix = 1+(ix)1!

+(ix)2

2!+

(ix)3

3!+

(ix)4

4!+ · · ·

=

1−x2

2!+

x4

4!−

x6

6!+

x8

8!−·· ·

+ix1!

−ix3

3!+

ix5

5!−·· ·

eix = cos x+ i sin x

eiπ = cos π + i sin π

= (−1)+ i(0)

eiπ = −1

September 12, 2007 19:19 Book:- Trig or Treat (9in x 6in) 13˙about-author

About the Author

Dr Y E O Adrian graduated from the University of Singapore with firstclass Honours in Chemistry in 1966, and followed up with a Master ofScience degree in 1968.

He received his Master of Arts and his Doctor of Philosophy degreesfrom Cambridge University in 1970, and did post-doctoral research atStanford University, California.

For his research, he was elected Fellow of Christ’s College, Cambridgeand appointed Research Associate at Stanford University in 1970.

His career spans fundamental and applied research and development,academia, and top appointments in politics and industry. His public ser-vice includes philanthropy and sports administration. Among his numer-ous awards are the Charles Darwin Memorial Prize, the Republic ofSingapore’s Distinguished Service Order, the International OlympicCommittee Centenary Medal, and the Honorary Fellowship of Christ’sCollege, Cambridge University.

adrian yeo trig or treat- an encyclopedia of trigonometric identity proofs with intellectually...

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