HIGHERSCHOOLCERTIFICATEYEAR11EXTENSION1MATHEMATICS

TRIGONOMETRICFUNCTIONSFREESAMPLE

1

First published by John Kinny-Lewis in 2018 National Library of Australia Cataloguing-in-publication data ISBN:978-0-6484118-2-6 This book is copyright. Apart from any fair dealing for purposes of private study, research, criticism or review as permitted under the Copyright Act 1968, no part may be reproduced, stored in a retrieval system, or transmitted, in any form by any means, electronic, mechanical, photocopying, recording, or otherwise without prior written permission. Enquiries to be made to John Kinny-LewisCopying for educational purposes. Where copies of part or the whole of the book are made under Section 53B or Section 53D of the Copyright Act 1968, the law requires that records of such copying be kept. In such cases the copyright owner is entitled to claim payment. Typeset by John Kinny-Lewis Edited by John Kinny-Lewis

2

CONTENTS SET 1 5 SET 1 ANSWERS 17SET 1 SOLUTIONS 18SET 2 30SET 2 ANSWERS 42SET 2 SOLUTIONS 43SET 3 55SET 3 ANSWERS 67SET 3 SOLUTIONS 68SET 4 80 SET 4 ANSWERS 92SET 4 SOLUTIONS 93

3

PREFACE This book provides a thorough revision of trigonometric functions for the new Year 11 Extension 1 HSC syllabus. The book is divided into four sets. Each set contains two topic areas:

• Inverse functions

Provides questions on: examining the reflective property of the graphof a function and the graph of its inverse, writing the rule or rules forthe inverse relation by exchanging x and y in the function rules,including any restrictions, and solving for y if possible, using the

notation f −1(x) (when the inverse relation is a function) and identifying

the relationships between the domains and ranges of f (x) and f −1(x),restricting domains, when the inverse relation is not a function, so asto obtain new functions, solving problems based on the relationshipbetween a function and its inverse.

• Further Trigonometric Identities Provides questions on: converting linear and quadratic functions,

and circles from parametric form to cartesian form and vice versa,sketching linear and quadratic functions, and circles expressed inparametric form.

4

SET 1 Inverse Trigonometric Functions Question 1 Evaluate the following angles as an exact answer in terms of π.

(i) sin−1 32

⎛

⎝⎜

⎞

⎠⎟ (ii) tan−1 1

3

⎛⎝⎜

⎞⎠⎟

Question 2Evaluate the following :

(i) sin−1 sinπ3

⎛⎝⎜

⎞⎠⎟

(ii) cos cos−1 32

⎛

⎝⎜

⎞

⎠⎟

5

SET 1 Inverse Trigonometric Functions Question 3 State the domain and range of the following :

(i) y = sin−1 x2

⎛⎝⎜

⎞⎠⎟

(ii) y = 2cos−1x

Question 4Sketch the function y = 2 tan−1 x and statethe domain and range.

6

SET 1 Inverse Trigonometric Functions Question 5 Sketch the function y = 2cos−1 x −1 and statethe domain and range.

Question 6Evaluate the following :

(i) sin cos−1 35

⎛⎝⎜

⎞⎠⎟

(ii) cos sin−1 941

⎛⎝⎜

⎞⎠⎟

7

SET 1 Inverse Trigonometric Functions Question 7

Show that y = tan−1 12

⎛⎝⎜

⎞⎠⎟+ tan−1 1

3⎛⎝⎜

⎞⎠⎟= π

4, given that tan α +β( ) = tanα + tanβ

1− tanα tanβ .

Question 8

Show that 2sin−1 12

⎛⎝⎜

⎞⎠⎟= sin−1 3

2

⎛

⎝⎜

⎞

⎠⎟ .

8

SET 1 Inverse Trigonometric Functions Question 9

Find the exact value of cos sin−1 35

⎛⎝⎜

⎞⎠⎟+ cos−1 2

5⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

,

given that cos α +β( ) = cosαcosβ − sinαsinβ.

Question 10

Find the exact value of tan sin−1 35

⎛⎝⎜

⎞⎠⎟− cos−1 5

13⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

,

given that tan α −β( ) = tanα − tanβ1− tanα tanβ

9

SET 1 Inverse Trigonometric Functions Question 11 Prove that sin−1 x = cos−1 1− x2( ). Question 12

Prove that tan−1 x + tan−1 y = tan−1 x + y1− xy

⎛⎝⎜

⎞⎠⎟

, given that, tan α +β( ) = tanα + tanβ1− tanα tanβ

.

10

SET 1 Further Trigonometric Identities Question 1 Prove that :

(i) cos A − B( ) = cosAcosB + sinAsinB (ii) cos A + B( ) = cosAcosB− sinAsin B

Question 2Prove that :

(i) sin A + B( ) = sinAcosB+ cosAsinB (ii) sin A − B( ) = sinAcosB− cosAsin B

11

SET 1 Further Trigonometric Identities Question 3 Prove that :

(i) tan A + B( ) = tanA + tan B1− tanA tan B

(ii) tan A − B( ) = tanA − tan B1+ tanA tan B

Question 4Prove that :

(i) sin2A = 2sinAcosA (ii) cos2A = cos2 A − sin2 A (iii) tan2A = 2 tanA1− tan2 A

= 2cos2 A −1

= 1− 2sin2 A

12

SET 1 Further Trigonometric Identities Question 5

Given that sinA = 35

find the exact value of:

(i) sin2A (ii) cos2A (iii) tan2A

Question 6Find the exact value of:

(i) 2sin150 cos150 (ii) 1− 2sin2 750 (iii) 2 tan150

1− tan2 150

13

SET 1 Further Trigonometric Identities Question 7

Given that t = tanθ2

prove that :

(i) tanθ = 2t1− t2

(ii) sinθ = 2t1+ t2

(iii) cosθ = 1− t2

1+ t2

Question 8

Given that t = tanθ2

, find an expression for each of the following in terms of t.

(i) sinθ

1+ cosθ(ii)

cosθ1− sinθ

14

SET 1 Further Trigonometric Identities Question 9 Prove that :

(i) cotθ = 12

cotθ2− tan

θ2

⎛⎝⎜

⎞⎠⎟

(ii) cotθ2= 1+ sinθ+ cosθ

1+ sinθ− cosθ

Question 10Prove that :

(i) cosAcosB = 12

cos A − B( )+ cos A + B( )( ) (ii) sinAsin B = 12

cos A − B( )− cos A + B( )( )

15

SET 1 Further Trigonometric Identities Question 11 Prove that :

(i) sinAcosB = 12

sin A + B( )+ sin A − B( )( ) (ii) cosAsin B = 12

sin A + B( )− sin A − B( )( )

Question 12Express the following products as a sum or a difference :

(i) 2sin4θcosθ (ii) 2cos350 cos150

16

SET 1 ANSWERS Inverse Trigonometric Functions Further Trigonometric

Identities 1) (i) π

3(ii) π

6

1) Proof

2) (i) π3

(ii) 32 2) Proof

3) (i) Domain − 2 ≤ x ≤ 2, Range − π2≤ y ≤ π

2(ii) Domain −1≤ x ≤1, Range 0 ≤ y ≤ 2π

3) Proof

4) See solution for graphDomain − ∞ ≤ x ≤ ∞, Range − π ≤ y ≤ π

4) Proof

5) See solution for graphDomain −1≤ x ≤1, Range −1≤ y ≤ 2π −1 5) (i) 24

25(ii) 7

25 (iii) 24

7

6) (i) 45

(ii) 4041

6) (i) 12

(ii) − 32

(iii) 13

7) Proof 7) Proof 8) Proof

8) (i) t (ii) 1+ t1− t

9) 8− 3 2125

9) Proof

10) − 3356 10) Proof

11) Proof 11) Proof 12) Proof 12) (i) sin5θ+ sin3θ (ii) cos200 + cos600

17

SET 1 SOLUTIONS Inverse Trigonometric Functions Question 1 Evaluate the following angles as an exact answer in terms of π.

(i) sin−1 32

⎛

⎝⎜

⎞

⎠⎟ (ii) tan−1 1

3

⎛⎝⎜

⎞⎠⎟

(i) Let y = sin−1 32

⎛

⎝⎜

⎞

⎠⎟

∴sin y = 32

y = π3

− π2≤ y ≤ π

2

∴ sin−1 32

⎛

⎝⎜

⎞

⎠⎟ =

π3

(ii) Let y = tan−1 13

⎛⎝⎜

⎞⎠⎟

∴ tan y = 13

y = π6

− π2≤ y ≤ π

2

∴ tan−1 13

⎛⎝⎜

⎞⎠⎟= π

6

Question 2Evaluate the following :

(i) sin−1 sinπ3

⎛⎝⎜

⎞⎠⎟

(ii) cos cos−1 32

⎛

⎝⎜

⎞

⎠⎟

(i) Let y = sin−1 sin π3

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

= sin−1 32

⎛

⎝⎜

⎞

⎠⎟

∴sin y = 32

y = π3

− π2≤ y ≤ π

2

∴ sin−1 sin π3

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟= π

3

(ii) Let y = cos−1 32

⎛

⎝⎜

⎞

⎠⎟

∴cos y = 32

y = π6

0 ≤ y ≤ π

cosπ6

⎛⎝⎜

⎞⎠⎟= 3

2

∴cos cos−1 32

⎛

⎝⎜

⎞

⎠⎟

⎛

⎝⎜

⎞

⎠⎟ =

32

18

SET 1 SOLUTIONS Inverse Trigonometric Functions Question 3 State the domain and range of the following :

(i) y = sin−1 x2

⎛⎝⎜

⎞⎠⎟

(ii) y = 2cos−1x

(i) For y = sin−1(x)Domain is −1≤ x ≤1

Range is − π2≤ y ≤ π

2

For y = sin−1 x2

⎛⎝⎜

⎞⎠⎟

Domain is −1≤ x2≤1

∴Domain is − 2 ≤ x ≤ 2

Range is − π2≤ y ≤ π

2

(ii) For y = cos−1(x)Domain is −1≤ x ≤1Range is 0 ≤ x ≤ πFor y = 2cos−1 xDomain is −1≤ x ≤1Range is 2× 0 ≤ y ≤ 2× π∴Range is 0 ≤ y ≤ 2π

Question 4Sketch the function y = 2 tan−1 x and statethe domain and range.

For y = 2 tan−1(x)

Domain is − ∞ ≤ x ≤ ∞

Range is − π2≤ y ≤ π

2

For y = 2 tan−1 x2

⎛⎝⎜

⎞⎠⎟

Domain is − ∞ ≤ x ≤ ∞

Range is 2× − π2≤ y ≤ 2× π

2∴Range is − π ≤ y ≤ π

19

SET 1 SOLUTIONS Inverse Trigonometric Functions Question 5 Sketch the function y = 2cos−1 x −1 and statethe domain and range.

For y = cos−1(x)Domain is −1≤ x ≤1Range is 0 ≤ y ≤ πFor y = 2cos−1 x −1Domain is −1≤ x ≤1Range is 2× 0−1≤ y ≤ 2× π −1∴Range is −1≤ y ≤ 2π −1

Question 6Evaluate the following :

(i) sin cos−1 35

⎛⎝⎜

⎞⎠⎟

(ii) cos sin−1 941

⎛⎝⎜

⎞⎠⎟

(i) sin cos−1 35

⎛⎝⎜

⎞⎠⎟

Let cos−1 35= θ

∴cosθ = 35

sinθ = 45

(3,4,5 triad)

∴ sin cos−1 35

⎛⎝⎜

⎞⎠⎟= 4

5

(ii) cos sin−1 941

⎛⎝⎜

⎞⎠⎟

Let sin−1 941

= θ

∴ sinθ = 941

cosθ = 4041

(9,40,41 triad)

∴cos sin−1 941

⎛⎝⎜

⎞⎠⎟= 40

41

20

SET 1 SOLUTIONS Inverse Trigonometric Functions Question 7

Show that y = tan−1 12

⎛⎝⎜

⎞⎠⎟+ tan−1 1

3⎛⎝⎜

⎞⎠⎟= π

4, given that tan α +β( ) = tanα + tanβ

1− tanα tanβ .

Let α = tan−1 12

⎛⎝⎜

⎞⎠⎟

and β = tan−1 13

⎛⎝⎜

⎞⎠⎟

∴ tanα = 12

and tanβ = 13

∴ tan α +β( ) =12+ 1

3

1− 12× 1

3

= 1

∴α +β = π4

∴ y = tan−1 12

⎛⎝⎜

⎞⎠⎟+ tan−1 1

3⎛⎝⎜

⎞⎠⎟= π

4

Question 8

Show that 2sin−1 12

⎛⎝⎜

⎞⎠⎟= sin−1 3

2

⎛

⎝⎜

⎞

⎠⎟ .

L.H.S.= 2sin−1 12

⎛⎝⎜

⎞⎠⎟

= 2× π6

= π3

R.H.S.= sin−1 32

⎛

⎝⎜

⎞

⎠⎟

= π3

= L.H.S.

∴2sin−1 12

⎛⎝⎜

⎞⎠⎟= sin−1 3

2

⎛

⎝⎜

⎞

⎠⎟

21

SET 1 SOLUTIONS Inverse Trigonometric Functions Question 9

Find the exact value of cos sin−1 35

⎛⎝⎜

⎞⎠⎟+ cos−1 2

5⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

,

given that cos α +β( ) = cosαcosβ − sinαsinβ.

Let α = sin−1 35

⎛⎝⎜

⎞⎠⎟

and β = cos−1 25

⎛⎝⎜

⎞⎠⎟

∴ sinα = 35→ cosα = 4

53,4,5( )

cosβ = 25

and sinβ = 215

2, 21,5( )cos α +β( ) = cosαcosβ − sinαsinβ.

= 45× 2

5− 3

5× 21

5= 8− 3 21

25

∴cos sin−1 35

⎛⎝⎜

⎞⎠⎟+ cos−1 2

5⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟= 8− 3 21

25

Question 10

Find the exact value of tan sin−1 35

⎛⎝⎜

⎞⎠⎟− cos−1 5

13⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

,

given that tan α −β( ) = tanα − tanβ1− tanα tanβ

Let α = sin−1 35

⎛⎝⎜

⎞⎠⎟

and β = cos−1 513

⎛⎝⎜

⎞⎠⎟

∴ sinα = 35→ tanα = 3

43,4,5( )

cosβ = 513

and tanβ = 125

5,12,13( )tan α −β( ) = tanα − tanβ

1+ tanα tanβ.

=

34− 12

5

1+ 34× 12

5

=− 33

205620

= − 3356

∴ tan sin−1 35

⎛⎝⎜

⎞⎠⎟+ cos−1 5

13⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟= − 33

56

22

SET 1 SOLUTIONS Inverse Trigonometric Functions Question 11 Prove that sin−1 x = cos−1 1− x2( ). Let y = sin−1 x

∴ sin y = x1

∴cos y = 1− x2

1x, 1− x2 ,1( )

∴ y = cos−1 1− x2

∴ sin−1 x = cos−1 1− x2( )

Question 12

Prove that tan−1 x + tan−1 y = tan−1 x + y1− xy

⎛⎝⎜

⎞⎠⎟

, given that, tan α +β( ) = tanα + tanβ1− tanα tanβ

.

Let α = tan−1 x and β = tan−1 y∴ tanα = x and tanβ = y

∴ tan α +β( ) = x + y1− xy

∴α +β = tan−1 x + y1− xy

⎛⎝⎜

⎞⎠⎟

∴ tan−1 x + tan−1 y = tan−1 x + y1− xy

⎛⎝⎜

⎞⎠⎟

23

SET 1 SOLUTIONS Further Trigonometric Identities Question 1 Prove that :

(i) cos A − B( ) = cosAcosB + sinAsinB (ii) cos A + B( ) = cosAcosB− sinAsin B

(i) In ΔPOQ, PO = QO = 1∠POX = B, ∠QOX = A and ∠POQ = A − BFrom the cosine rule :∴ PQ2 = 12 +12 − 2×1×1× cos(A − B)

= 2− 2cos(A − B)From the distance formula:

PQ2 = (cosA − cosB)2 + (sinA − sin B)2

= cos2 A − 2cosAcosB+ cos2 B+sin2 A − 2sinAsin B+ sin2 B

∴ PQ2 = 2− 2(cosAcosB+ sinAsin B)∴ cos(A − B) = cosAcosB+ sinAsinB

(ii) cos A − B( ) = cosAcosB+ sinAsin B

Let B = −B∴cos(A + B) = cosAcos( − B)+ sinAsin(−B)∴cos(A + B) = cosAcosB− sinAsin B

Question 2Prove that :

(i) sin A + B( ) = sinAcosB+ cosAsinB (ii) sin A − B( ) = sinAcosB− cosAsin B

(i) sin A + B( ) = cosπ2− A + B( )⎛

⎝⎜⎞⎠⎟

= cosπ2− A

⎛⎝⎜

⎞⎠⎟− B

⎛⎝⎜

⎞⎠⎟

= cosπ2− A

⎛⎝⎜

⎞⎠⎟

cosB+ sinπ2− A

⎛⎝⎜

⎞⎠⎟

sin B

∴ sin A + B( ) = sinAcosB+ cosAsin B

(ii) sin A + B( ) = sinAcosB+ cosAsin B

Let B = −B∴ sin(A − B) = sinAcos( − B)+ cosAsin(−B)∴ sin(A − B) = sinAcosB− cosAsin B

24

SET 1 SOLUTIONS Further Trigonometric Identities Question 3 Prove that :

(i) tan A + B( ) = tanA + tan B1− tanA tan B

(ii) tan A − B( ) = tanA − tan B1+ tanA tan B

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

= sinAcosB+ cosAsin BcosAcosB− sinAsin B

=

sinAcosBcosAcosB

+ cosAsin BcosAcosB

cosAcosBcosAcosB

− sinAsin BcosAcosB

∴ tan A + B( ) = tanA + tan B1− tanA tan B

(ii) tan A + B( ) = tanA + tan B1− tanA tan B

Let B = −B

tan A − B( ) = tanA + tan −B( )1− tanA tan −B( )

∴ tan A − B( ) = tanA − tan B1+ tanA tan B

Question 4Prove that :

(i) sin2A = 2sinAcosA (ii) cos2A = cos2 A − sin2 A (iii) tan2A = 2 tanA1− tan2 A

= 2cos2 A −1

= 1− 2sin2 A

(i) sin A + B( ) = sinAcosB+ cosAsin B

Let B = A∴sin2A = sinAcosA + cosAsinA ∴ sin2A = 2sinAcosA

(ii) cos A + B( ) = cosAcosB− sinAsin B

Let B = A∴cos2A = cosAcosA − sinAsinA

∴cos2A = cos2 A − sin2 A

= cos2A − 1− cos2 A( ) = 2cos2 A −1

= 2 1− sin2 A( )−1

= 1− 2sin2 A

(iii) tan A + B( ) = tanA + tan B1− tanA tan B

Let B = A

∴ tan2A = tanA + tanA1− tanA tanA

∴ tan2A = 2 tanA1− tan2 A

25

SET 1 SOLUTIONS Further Trigonometric Identities Question 5

Given that sinA = 35

find the exact value of:

(i) sin2A (ii) cos2A (iii) tan2A

sinA = 35

→ cosA = 45→ tanA = 3

43,4,5 triad( )

(i) sin2A = 2sinAcosA

= 2× 35× 4

5

= 2425

(ii) cos2A= 2cos2 A −1

= 2× 45

⎛⎝⎜

⎞⎠⎟

2

−1

= 725

(iii) tan2A = 2 tanA1− tan2 A

=2× 3

4

1− 34

⎛⎝⎜

⎞⎠⎟

2

= 247

Question 6Find the exact value of:

(i) 2sin150 cos150 (ii) 1− 2sin2 750 (iii) 2 tan150

1− tan2 150

(i) sin2A = 2sinAcosA

∴ sin300 = 2sin150 cos150

∴2sin150 cos150 = 12

(ii) cos2A = 1− 2sin2 A

∴cos1500 = 1− 2sin2 750

∴1− 2sin2 750 = − 32

(iii) tan2A = 2 tanA1− tan2 A

∴tan300 = 2 tan150

1− tan2 150

∴ 2 tan150

1− tan2 150 = 13

26

SET 1 SOLUTIONS Further Trigonometric Identities Question 7

Given that t = tanθ2

prove that :

(i) tanθ = 2t1− t2

(ii) sinθ = 2t1+ t2

(iii) cosθ = 1− t2

1+ t2

(i) tan2A = 2 tanA1− tan2 A

Let 2A = θ

∴ tanθ =2 tan θ

2

1− tan2 θ2

∴ tanθ = 2t1− t2

From the diagram:

AC2 = 1− t2( )2+ 2t( )2

= 1− 2t2 + t4 + 4t2

= t4 + 2t2 +1

= t2 +1( )2

∴AC = 1+ t2

(ii) ∴sinθ = 2t1+ t2

(iii) ∴cosθ = 1− t2

1+ t2

Question 8

Given that t = tanθ2

, find an expression for each of the following in terms of t.

(i) sinθ

1+ cosθ(ii)

cosθ1− sinθ

(i) sinθ

1+ cosθ=

2t1+ t2

1+ 1− t2

1+ t2

=

2t1+ t2

1+ t2 +1− t2

1+ t2

= 2t2= t

(ii) cosθ

1− sinθ=

1− t2

1+ t2

1− 2t1+ t2

=

1− t2

1+ t21+ t2 − 2t

1+ t2

=1− t( ) 1+ t( )

1− t( )2 = 1+ t1− t

27

SET 1 SOLUTIONS Further Trigonometric Identities Question 9 Prove that :

(i) cotθ = 12

cotθ2− tan

θ2

⎛⎝⎜

⎞⎠⎟

(ii) cotθ2= 1+ sinθ+ cosθ

1+ sinθ− cosθ

(i) R.H.S.= 12

cot θ2− tan θ

2⎛⎝⎜

⎞⎠⎟

= 12

1t− t

⎛⎝⎜

⎞⎠⎟

= 1− t2

2t

= 1tanθ

= cotθ= L.H.S.

∴cotθ = 12

cot θ2− tan θ

2⎛⎝⎜

⎞⎠⎟

(ii) R.H.S = 1+ sinθ+ cosθ1+ sinθ− cosθ

=1+ 2t

1+ t2+ 1− t2

1+ t2

1+ 2t1+ t2

− 1− t2

1+ t2

=

1+ t2 + 2t +1− t2

1+ t21+ t2 + 2t −1+ t2

1+ t2

=2 1+ t( )2t 1+ t( )

= 1t= cot θ

2= L.H.S.

∴cot θ2= 1+ sinθ+ cosθ

1+ sinθ− cosθ

Question 10Prove that :

(i) cosAcosB = 12

cos A − B( )+ cos A + B( )( ) (ii) sinAsin B = 12

cos A − B( )− cos A + B( )( )

(i) R.H.S.= 12

cos A − B( )+ cos A + B( )( )= 1

2cosAcosB+ sinAsin B +cosAcosB− sinAsin B

⎛⎝⎜

⎞⎠⎟

= cosAcosB= L.H.S.

∴cosAcosB = 12

cos A − B( )+ cos A + B( )( )

(ii) R.H.S.= 12

cos A − B( )− cos A + B( )( )= 1

2cosAcosB+ sinAsin B −cosAcosB+ sinAsin B

⎛⎝⎜

⎞⎠⎟

= sinsin B= L.H.S.

∴ sinAsin B = 12

cos A − B( )− cos A + B( )( )

28

SET 1 SOLUTIONS Further Trigonometric Identities Question 11 Prove that :

(i) sinAcosB = 12

sin A + B( )+ sin A − B( )( ) (ii) cosAsin B = 12

sin A + B( )− sin A − B( )( )

(i) R.H.S.= 12

sin A + B( )+ sin A − B( )( )= 1

2sinAcosB+ cosAsin B +sinAcosB− cosAsin B

⎛⎝⎜

⎞⎠⎟

= sinAcosB= L.H.S.

∴ sinAcosB = 12

sin A + B( )+ sin A − B( )( )

(ii) R.H.S.= 12

sin A + B( )− sin A − B( )( )= 1

2sinAcosB+ cosAsin B −sinAcosB+ cosAsin B

⎛⎝⎜

⎞⎠⎟

= cosAsin B= L.H.S.

∴cosAsin B = 12

sin A + B( )− sin A − B( )( )

Question 12Express the following products as a sum or a difference :

(i) 2sin4θcosθ (ii) 2cos350 cos150

(i) sinAcosB = 12

sin A + B( )+ sin A − B( )( )∴2sin4θcosθ = sin 4θ+ θ( )+ sin 4θ− θ( )

= sin5θ+ sin3θ

(ii) cosAcosB = 12

cos A − B( )+ cos A + B( )( )∴2cos350 cos150 = cos 350 −150( )+ cos 350 +150( )

= cos200 + cos600