chapter 9 trigonometry - macmillan caribbean · 2017-06-29 · 2 4 cot θ – 4 cot θ – 3 = 0 (2...

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Unit 1 Answers: Chapter 9 © Macmillan Publishers Limited 2013

Chapter 9 Trigonometry Try these 9.1 (a) sin 4x = 0.28 4x = sin – 1 (0. 28) 4x = 180°n + (–1)n (16.3°), n ∈ ℤ

n1x [180n ( 1) (16.3 )], n4

= + − ° ∈

(b) cos (x + 30°) = 0.6 x + 30 = cos–1 (0.6) x + 30 = 360n ± 53.13°

x 360n 23.13

nx 360n 83.13= + °

∈= − °

(c) tan (2x + 45°) = 0.7 2x + 45° = tan – 1 (0.7) 2x + 45 = 180°n + 35° 2x = 180n – 10 x = 90°n – 5° , n∈ Hence x = 90n – 5°, n∈ Exercise 9A

1 1sin 22−

θ =

n2 n , n− π θ = π + (−1) ∈ 6

n 2p 2 2p π= ⇒ θ = π −

6

p πθ = π −

12

n 2p 1 2 (2p 1) π= + ⇒ θ = + π +

6

2p +1 p2

π 13π θ = π + = π + 12 12

Hence p

12 pp

π θ = π − ∈13ππ +12

2 cos 3θ = 0

3 2n , nπ⇒ θ = π ± ∈Ζ

2

2 n , n3

πθ = π ± ∈Ζ

6

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3 tan 2 1π θ + = 3

2 nπ πθ + = π +

3 4

2 n π πθ = π + −

4 3

1 n2

π θ = π − 12

1 n , n2

πθ = π − ∈

24

4 1cos 22

π θ − = 4

2 2nπ πθ − = π ±

4 3

π πθ = π ± +

3 42 2n

72 2n , 2nπ πθ = π + π −

12 12

7n24 n

n

πθ = π + ∈π θ = π −24

5 1sin 33 2π θ − =

n3 n , n3π π

θ − = π + (−1) ∈4

n 2p] 3 2p3π π

= ⇒ θ − = π +4

3 2p3

π πθ = π + +

4

1 72p ,p3 12

π θ = π + ∈

n 2p 1] 3 (2p 1)π π= + ⇒ θ − = + π −

3 4

3 (2p 1) π πθ = + π − +

4 3

3 2p π πθ = π + π − +

4 3

132p , p12

1 θ = π + π ∈ 3

72p12

p132p12

1 π ∴θ = π + 3 ∈1 π π + 3

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6 π θ − = − 21tan 12

π πθ − = π −

2 41 n2

1 n2

π πθ = π − +

4 2

2 n , nπ θ = π + ∈ 4

7 sin 3x = 3 cos 3x

⇒ =sin 3x 3cos 3x

tan 3x = 3 3x = nπ + 1.25

1x n 0.416, n3

= π + ∈

Try these 9.2

(a) 2

2

sec cos sinRTP:sec cos cos

θ − θ θ=

θ + θ 1+ θ

Proof:

− θθ − θ θ=

θ + θ + θθ

1 cossec cos cos1sec cos cos

cos

− θθ=

21 coscos+ θ

θ

21 coscos

− θ=

+ θ

2

2

1 cos1 cos

2

2

sin1 cos

θ=

+ θ

(b) 2cosRTP: sin

sin1− θ

= θθ

Proof:

2 2cos sin

sin sin1− θ θ

=θ θ

= sin θ

(c) 2

sec tan sinRTP:cot cos cos

θ + θ θ=

θ + θ θ

Proof:

sec tancot cos

θ + θθ + θ

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1 sincos cos cos cossin

θ+

θ θ=

θ+ θ

θ

1 sincos

cos cos sinsin

+ θθ

=θ + θ θ

θ

(1 sin ) sin(cos cos sin cos

+ θ θ=

θ + θ θ) θ

+ θ

=(1 sin ) θ

θ (1+ θ)2

sincos sin

2

sincos

θ=

θ

Try these 9.3 (a) (i) 3 sin2θ = 1 + cos θ 3 (1 – cos2θ) = 1 + cos θ 3 cos2θ + cos θ – 2 = 0 (3 cos θ – 2) (cos θ + 1) = 0

2θ = θ = −1

3cos , cos

θ = 48.2°, 311.8°, 180° (ii) 4 cosec2θ – 4 cot θ – 7 = 0 4 (1 + cot2θ) – 4 cotθ – 7 = 0 4 cot2θ – 4 cot θ – 3 = 0 (2 cot θ + 1) (2 cot θ – 3) = 0

cot , cot1 3θ = − θ =

2 2

2θ = − 2, θ =

3tan tan

θ = 116.6°, 296.6°, 33.7°, 213.7° (b) 20 sec2θ – 3 tan θ – 22 = 0 20 (1 + tan2θ) – 3 tan θ – 22 = 0 20 tan2θ – 3 tan θ – 2 = 0 (4 tan θ + 1) (5 tan θ – 2) = 0

tan , tan1 2θ = − θ =

4 5

tan 180n 14 , n1θ = − ⇒ θ = − ° ∈

4

tan 180n 21.8, n2θ = ⇒ θ = + ∈

5

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Exercise 9B 1 (sin θ + cos θ)2 – 1 = sin2θ + 2 sin θ cos θ + cos2θ – 1 = 1 + 2 sin θ cos θ – 1, since sin2θ + cos2θ = 1 = 2 sin θ cos θ 2 sin x (sin x – cot x cosec x)

= − ×2 cos x 1sin x sin xsin x sin x

= −2 cos xsin xsin x

= sin2 x – cot x 3 sin4θ – cos4θ = (sin2θ – cos2θ) (sin2θ + cos2θ) = sin2θ – cos2θ 4 sin2θ (cot2θ + cosec2θ)

θ

= θ + θ θ

22

2 2

cos 1sinsin sin

θ +1

= θ θ

22

2

cossinsin

= 1 + cos2θ

5 θ −1 − − θ=

θ θ

2 2

2 2

sin (1 sin )cos cos

− θ= = −

θ

2

2

(cos ) 1cos

6 2 2 2 2sec tan sec (secsin sinθ − θ θ − θ −1)

=θ θ

1sin

=θ

= cosec θ

7 RTP: 2sin1 cos

cosθ

− = − θ1− θ

Proof:

2sin1

cosθ

−1− θ

2cos1

cos1− θ

= −1− θ

(1 cos ) (1 cos )1cos

− θ + θ= −

1− θ

= 1 – (1 + cos θ) = – cos θ

8 RTP: 2cos 1 sin

sinθ

− = θ1− θ

Proof:

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2 2cos 1 sin1 1

1 sin 1 sinθ − θ

− = −− θ − θ

(1 sin )(1 sin ) 11 sin

− θ + θ= −

− θ

= 1 + sin θ – 1 = sin θ

9 RTP: cosec cos tancos sin

θ θ− = θ

θ θ

Proof:

cosec coscos sin

θ θ−

θ θ

1 cossin cos sin

θ= −

θ θ θ

21 cos

sin cos− θ

=θ θ

2sin

sinθ

=θ

sincoscos

θ=

θθ

= tan θ

10 RTP: 2

22

1 cos cos1 sec− θ

= − θ− θ

Proof:

− θ θ=

− θ − θ

2 2

2 2

1 cos sin1 sec tan

2sin= θ2

2

cossin

θ× −

θ

= – cos2θ 11 RTP: sec4 x – sec2 x = tan4 x + tan2 x Proof: sec4x – sec2x = sec2x (sec2x – 1) = (1 + tan2 x) (tan2 x), since sec2 x – 1 = tan 2 x = tan2 x + tan4 x

12 RTP: cos x sin x sin x cos x1 tan x 1 cot x

+ = +− −

Proof:

+− −cos x sin x

1 tan x 1 cot x

= +− −

cos x sin xsin x cos x1 1cos x sin x

= +− −

cos x sin xcos x sin x sin x cos x

cos x sin x

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= +− −

2 2cos x sin xcos x sin x sin x cos x

−=

−

2 2cos x sin xcos x sin x

−

=(cos x sin x) +

−

(cos x sin x)cos x sin x

= cos x + sin x

13 RTP: 21 cos x (cosec x cot x)1 cos x−

= −+

Proof: Now (cosec x – cot x)2

= −

21 cos x

sin x sin x

−

=

21 cos x

sin x

−=

2

2

(1 cos x)sin x

−=

−

2

2

(1 cos x)1 cos x

−

=(1 cos x) −

−

(1 cos x)(1 cos x) +(1 cos x)

1 cos x1 cos x−

=+

14 RTP: 21 1 2 sec xsin x 1 1 sin x

+ =+ −

Proof:

1 − + ++ =

+ − + −1 1 sin x 1 sin x

sin x 1 1 sin x (1 sin x) (1 sin x)

=− 2

21 sin x

= 2

2cos x

= 2 sec2 x 15 RTP: sin4θ – sin2θ = cos4θ – cos2θ Proof: sin4θ – sin2θ = sin2θ (sin2θ – 1) = (1 – cos2θ) (– cos2θ) = – cos2θ + cos4θ

16 RTP: 2

22

cot x 1 cot xtan x 1

−= −

−

Proof:

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−−

=−

−

2

2 2

2 2

2

cos x 1cot x 1 sin xtan x 1 sin x 1

cos x

−

=2 2cos x sin x

×−

2

2 2 2

cos xsin x sin x cos x

− 1

= −2

2

cos xsin x

= – cot2 x 17 4 sec x – tan x = 6 cos x Converting to sin x and cos x:

− =4 sin x 6 cos x

cos x cos x

⇒ 4 – sin x = 6 cos2x ⇒ 4 – sin x = 6 [1 – sin2x] 4 – sin x = 6 – 6 sin2x 6 sin2 x – sin x – 2 = 0 Let y = sin x 6y2 – y – 2 = 0 (3y – 2) (2y + 1) = 0

2 1y ,3 2

= −

2 1sin x , sin x3 2

= = −

x = 41.8°, 138.2°, x = 210°, 330° ∴ x = 41.8°, 138.2°, 210°, 330° 18 3 tan2 x – sec x – 1 = 0 Replacing tan2 x = sec2 x – 1 ⇒ 3 [sec2 x – 1] – sec x – 1 = 0 3 sec2 x – sec x – 4 = 0 y = sec x 3y2 – y – 4 = 0 (3y – 4) (y + 1) = 0

= −4y , 13

= −4Now sec x = , sec x 13

3cos x cos x 14

⇒ = = −

x = 41.4°, 318.6°, x = 180° ∴ x = 41.4°, 180°, 318.6° 19 2 cot2 x + cosec x = 1 Replacing cot2 x = cosec2 x – 1 ⇒ 2 [cosec2 x – 1] + cosec x = 1 ⇒ 2 cosec2 x + cosec x – 3 = 0 y = cosec x 2y2 + y – 3 = 0 ⇒ (2y + 3) (y – 1) = 0

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= − =3y , y 12

−= =

3cosec x , cosec x 12

−= =

2sin x , sin x 13

x = 221.8°, 318.2°, x = 90° x = 90°, 221.8°, 318.2° 20 3 cos2 x = 4 sin x – 1 ⇒ 3 (1 – sin2 x) = 4 sin x – 1 ∴ 3 sin2 x + 4 sin x – 4 = 0 y = sin x 3y2 + 4y – 4 = 0 (3y – 2) (y + 2) = 0

= −2y , 23

2sin x , sin x 2 (invalid)3

∴ = = −

x = 41.8°, 138.2° 21 2 cot x = 3 sin x

⇒ =2 cos x 3 sin xsin x

⇒ 2 cos x = 3 sin2 x 2 cos x = 3 (1 – cos2 x) 3 cos2 x + 2 cos x – 3 = 0

− ±=

2 40cos x6

cos x = – 1.387, 0.72076 cox x = – 1.387 (invalid) cos x = 0.72076 ⇒ x = 43.9°, 316.1° 22 sin2 x = 3 cos2 x + 4 sin x cos2 x = 1 – sin2 x ∴ sin2 x = 3 (1 – sin2 x) + 4 sin x ⇒ sin2 x = 3 – 3 sin2 x + 4 sin x 4 sin2 x – 4 sin x – 3 = 0 (2 sin x + 1) (2 sin x – 3) = 0

= − =1 3sin x , sin x (invalid)2 2

x = 210°, 330° 23 2 cos x = tan x

sin x2 cos x =cos x

⇒ 2 cos2 x = sin x ⇒ 2 [1 – sin2 x] = sin x 2 sin2 x + sin x – 2 = 0

− ±=

1 17sin x4

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− ±∴ =

1 17sin x4

x = 51.3°, 128.7° 24 2 cot x = 1 + tan x

= +2 cos x sin x1sin x cos x

⇒ = +2 1 tan x

tan x

2 = tan x + tan2 x ∴ tan2 x + tan x – 2 = 0 (tan x – 1) (tan x + 2) = 0 tan x = 1, tan x = – 2 x = 45°, 225°, x = 116.6°, 296.6° 25 2 + 3 sin z = 2 cos2 z 2 + 3 sin z = 2 (1 – sin2 z) 2 sin 2 z + 3 sin z = 0 sin z (2 sin z + 3) = 0

3sin z 0, sin z (invalid)2−

= =

z = nπ + (– 1)n (0) z = nπ, n∈ 26 2 cot2 x + cosec x = 4 ⇒ 2 (cosec2 x – 1) + cosec x – 4 = 0 ⇒ 2 cosec2 x + cosec x – 6 = 0 ⇒ (2 cosec x – 3) (cosec x + 2) = 0

= = −3cosec x , cosec x 22

2 1sin x , sin x3 2

∴ = = −

x = nπ + (– 1)n (0.730) n∈

nx n ( 1) , n6− π = π + − ∈

27 2 sec x + 3 cos x = 7

+ =2 3 cos x 7

cos x

2 + 3 cos2 x = 7 cos x 3 cos2 x – 7 cos x + 2 = 0 (3 cos x – 1) (cos x – 2) = 0

= =1cos x , cos x 2 (invalid)3

x = 2nπ ± 1.23, n∈ (28) 5 cos x = 6 sin2 x 5 cos x = 6 (1 – cos2 x) 6 cos2 x + 5 cos x – 6 = 0 (3 cos x – 2) (2 cos x + 3) = 0

−= =

2 3cos x , cos x (invalid)3 2

x = 2nπ ± (0.841) n∈

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Try these 9.4

(a) π π π = + 12 3 47cos cos

π π π π = − cos cos sin sin

3 4 3 4

= −

1 2 3 22 2 2 2

2 64 4

= −

2 (1 3)4

= −

(b) π π π = + 5sin sin12 4 6

π π π π = + sin cos cos sin

4 6 4 6

2 3 2 12 2 2 2

= +

2 ( 3 1)4

= +

(c) π π π = + 7sin sin12 3 4

sin cos cos sin3 4 3 4π π π π

= +

= +

3 2 1 22 2 2 2

2 ( 3 1)4

= +

Exercise 9C 1 sin 75 = sin (30 + 45) = sin 30 cos 45 + cos 30 sin 45

= × + ×1 2 3 22 2 2 2

2 64 4

= +

= +2 ( 3 1)

4

2 sin (A B) 5sin (A B) 13

−=

+

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−=

+sin A cos B cos A sin B 5sin A cos B cos A sin B 13

13 sin A cos B – 13 cos A sin B = 5 sin A cos B + 5 cos A sin B 18 cos A sin B = 8 sin A cos B

=sin B sin A18 8cos B cos A

9 tan B = 4 tan A

3 tan 5α =

12

cos 3

β = −5

(a) α + β = α β + α βsin ( ) sin cos cos sin

− − − = + 5 3 12 4

13 5 13 5

15 48 3365 65 65

−= − =

(b)

5 4tan tan 6312 3tan ( )

5 41 tan tan 16112 3

− − α − β α − β = = =−+ α β +

(c) α − β = α β + α βcos ( ) cos cos sin sin

12 3 5 4 1613 5 13 5 65− − − = + =

4 θ + + θ +sin ( 30) 3 cos ( 30) = θ + θ + θ − θsin cos 30 cos sin 30 3 cos cos 30 3 sin sin 30

= θ3 sin

2

+ θ + θ − θ

1 3 3cos 3 cos sin2 2 2

= θ + θ1 3cos cos2 2

= 2 cosθ. 5 sin (θ + 30) = 2 cos (θ + 60)

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sin θ cos 30 + cos θ sin 30 = 2 cos θ cos 60 – 2sin θ sin 60

3 sin cos (2) cos sin2 2

1 1 3θ + θ = θ − 2 θ

2 2

1θ + θ = θ − θ

23 sin 3 sin cos cos

2

θ = θ3 3 1sin cos

2 2

3 3 sin cosθ = θ 6 sin ( ) k sin ( )θ + α = θ − α sin cos cos sin k sin cos k cos sinθ α + θ α = θ α − θ α cos sin k cos sin k sin cos sin cosθ α + θ α = θ α − θ α cos sin (k 1) sin cos (k 1)θ α + = θ α −

sin k 1 sincos k 1 cos

α + θ= α − θ

k 1tan tank 1

+θ = α −

7 −α =

12cos13

(a) +

α =5sin

13

(b) −α =

5tan12

(c) α + = α − αcos ( 30) cos cos 30 sin sin 30

− = −

12 3 5 113 2 13 2

− −= − = +

12 3 5 1 (5 12 3)26 26 26

8 α =12sin13

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−∴ α =

5cos13

β =4sin5

−

β =3cos

5

both α and β are in the second quadrant (a) sin (α + β) = sin α cos β + cos α sin β

− − = + 12 3 5 413 5 13 5

− − −= =

36 20 5665 65 65

(b) cos (α + β) = cos α cos β – sin α sin β

− − = − 5 3 12 4

13 5 13 5

−= − =

15 48 3365 65 65

(c)

56sin ( ) 5665tan ( ) 33cos ( ) 33

65

−α + β

α + β = = =−α + β

9 α =1cos4

Since α is in the 4th quadrant sin α is negative

(a) −α =

15sin4

(b) π π π α − = α − α 6sin sin cos cos sin

6 6

− = −

15 3 1 14 2 4 2

3 5 18 8

−= −

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(c) π π π α + = α − α 3cos cos cos sin sin

3 3

− = −

1 1 15 34 2 4 2

1 3 58 8

= +

10 (a) tan (2π − θ)

tan 2 tan 0 tan1 tan 2 tan 1 0

π − θ − θ= =

+ π θ +

= −tan θ

(b) 3sin2π + θ

3 3sin cos cos sin2 2π π = θ + θ

= (−1) cosθ + (0) sinθ = −cosθ 11 tan A = y + 1 tan B = y − 1

tan A tan Btan(A B)1 tan A tan B

−− =

+

y 1 (y 1)1 (y 1) (y 1)

+ − −=

+ + −

2

21 y 1

=+ −

2

2y

=

22 cot(A B)tan(A B)

− =−

2

22y

=

= y2

12 (a) 1 tan1 tan+ θ− θ

tan tan

4

1 tan tan4

π+ θ

=π

− θ

tan4π = + θ

(b) 1 1cos sin2 2

θ + θ

cos cos sin sin4 4π π

= θ + θ

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cos4π = − θ

13 cot (θ − α) = 4

1tan( )4

⇒ θ − α =

θ − α⇒ = α = ⇒ α =

− θ αtan tan 1 1cot tan 2

1 tan tan 4 2

tan 2 11 2 tan 4

θ −⇒ =

− θ

⇒ 4 tanθ − 8 = 1 − 2 tanθ 6 tanθ = 9

9 3tan6 2

θ = =

2cot3

⇒ θ =

14 cos( ) cos( )sin( ) sin( )

α − β − α + βα + β + α − β

cos cosα β

=sin sin cos cos+ α β − α β sin sin

sin cos cos sin+ α β

α β + α β sin cos cos sin+ α β − α β

2 sin α

=sin

2 sinβ

α cos β

sincos

β=

β

= βtan 15 tan (α + β) = b

1tan2

β =

tan (α + β) = b α + β⇒ =

− α βtan tan b

1 tan tan

1tan2 b11 tan

2

α +⇒ =

− α

2 tan 1 b2 tan

α +=

− α

⇒ 2 tan α + 1 = 2b − b tan α ⇒ 2 tan α + b tan α = 2b − 1 tan α(2 + b) = 2b − 1

2b 1tanb 2−

α =+

16 2cos sin ( ) sin sin cosP VI t VI t t= φ ω − φ ω ω sin [cos sin sin cos ]VI t t t= ω φ ω − φ ω sin sin( )VI t t= ω ω − φ

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Try these 9.5 (a) 3sin θ − cos θ = r sin (θ − α) = r sin θ cos α − r cos θ sin α ⇒ r cos α = 3 [1] r sin α = 1 [2]

[2] ÷ [1] ⇒ 1tan , 18.43

α = α = °

[1]2 + [2]2⇒ r2 [cos2 α + sin2 α] = 32 + 12 r2 = 10 10r = 3 sin cos 10 sin ( 18.4 )∴ θ − θ = θ − ° 10 sin( 18.4 ) 2θ − ° =

θ − =2sin( 18.4)10

1 218.4 sin10

− θ − =

18.4 39.2 ,140.8θ − = ° ° 57.6 ,159.2θ = ° ° (b) 3 cos 2 sin 2 cos(2 )rθ − θ = θ + α = r cos 2θ cos α − r sin 2θ sin α r cos 3⇒ α = [1] r sin α = 1 [2]

1[2] [1] tan 303

÷ ⇒ α = ⇒α = °

[1]2 + [2]2] ⇒ r2 = 3 + 1 ⇒ r = 2 3 cos 2 sin 2 2 cos (2 30 )∴ θ − θ = θ + ° 2 cos(2θ + 30°) = −1

1cos(2 30 )2−

θ + ° =

2θ + 30° = 360n ± 120° 2θ = 360n + 90° 2θ = 360n − 150

180n 45

n180n 75

⇒θ = + °∈− °

(c) 2 sin x − cos x = r sin (x − α) = r sinx cos α − r cos x sin α r cos α = 2 r sin α = 1

1tan 26.62

α = ⇒α = °

2 2 22 1 5r r= + ⇒ = 2 sin x cos x 5 sin (x 26.6 )− = − °

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max f (x) 5

when sin(x 26.6) 1, x 26.6 90

=

− = − =

x = 116.6˚

min f (x) 5

when sin(x 26.6) 1

= −

− = −

x − 26.6 = 270 x = 296.6˚ Try these 9.6 (a) cos 2θ − cos θ = 0 2 cos2 θ − cos θ − 1 = 0 (2 cos θ + 1) (cos θ − 1) = 0

1cos , cos 12

θ = − θ =

θ = 120°, 240°, 0°, 360° Hence θ = 0°, 120°, 240°, 360° (b) cosθ = sin 2θ cosθ − 2 sin θ cos θ = 0 cosθ (1 – 2 sinθ) = 0

cosθ = 0, 1sin2

θ =

θ = 0°, 180°, θ = 30°, 150° Hence θ = 0°, 30°, 150°, 180° (c) cos 2θ − 2cos θ = 3 2 cos2θ − 1 − 2cos θ − 3 = 0 2 cos2θ − 2cos θ − 4 = 0 cos2θ − cos θ − 2 = 0 (cos θ + 1) (cos θ − 2) = 0 cos θ = −1, cos θ = 2. θ = 180° cos θ = 2 has no solutions Hence θ = 180° Exercise 9D

1 2

sin 2x + cos xRTP: = cot x2 2 cos x + sin x−

Proof:

2

sin 2x cos x2 2cos x sin x

+− +

2

2 sin x cos x cos x2 sin x sin x

+=

+

cos x [2 sin x 1]+

=sin x [2 sin x 1]+

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cos xsin x

=

= cot x

2 RTP: 21 cos 2x tan x1 cos 2x−

=+

Proof:

1 cos2x1 cos2x−+

2=

2sin x2 2cos x

= tan2x 3 RTP: tan x − cot x = −2 cot 2x Proof:

tan x − cot x sin x cos xcos x sin x

= −

2 2 2 2sin x cos x [cos x sin x] 2 cos2x 2 cot 2x1cos x sin x sin 2xsin 2x

2

− − − −= = = = −

4 cos 2xRTP: = cos x + sin xcos x sin x−

Proof:

2 2cos2x cos x sin x

cos x sin x cos x sin x−

=− −

(cos x sin x)−

=(cos x sin x)

cos x sin x+

−

= cos x + sin x

5 RTP: 1 cos2A sin A tan Asin 2A cosA− +

=+

Proof:

1 cos2A sin Asin 2A cosA− +

+

22 sin A sin A

2 sin A cosA cos A+

=+

sin A (2 sin A 1)+

=cosA (2 sin A 1)+

sin AcosA

=

= tan A

6 RTP: 1 cos4 tan 2sin 4− θ

= θθ

Proof:

− θ=

θ1 cos4 2

sin 4

2sin θ22 θsin2 θcos2

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sin 2cos2

θ=

θ

= tan 2θ

7 1tan 2x4

=

(a) 4 4 17cos2x1717

= =

(b) cos 2x = 1 − 2 sin2x 2 sin2 x = 1 − cos 2x

2 1 4sin x 12 17

= −

1 2sin x2 17

= −

1 2 172 17

= −

8 RTP: + − −=

+ + −sin(x y) sin(x y) tan ycos(x y) cos(x y)

Proof:

sin x cos ysin(x + y) sin(x y) =

cos(x +y) + cos(x - y)− − + cos x sin y sin x cos y− + cos x sin y

cos x cos y sin x sin y− + cos x cos y + sin x sin y

=2cosx sin y2cosx

= =sin y tan ycosycosy

9 sin( ) 4sin( ) 5

θ − α=

θ + α

5 sin(θ − α) = 4 sin (θ + α) 5 sin cos 5 cos sin 4 sin cos 4 cos sin⇒ θ α − θ α = θ α + θ α 5 sin cos 4 sin cos 5 cos sin 4 cos sinθ α − θ α = θ α + θ α sin θ cos α = 9 cos θ sin α

sin sin9cos cos

θ α=

θ α

tan θ = 9 tan α

1 1tan , tan (9) 33 3

α = θ = =

2

2 tantan 21 tan

θθ =

− θ

2

2(3) 6 31 (3) 8 4

−= = =

− −

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10 5tan12

α =

5sin

13α =

12cos13

α =

cos 2α = 2cos2α − 1

212 1192 1

13 169 = − =

cos 4α = 2 cos2(2α) − 1

2119 2392 1

169 28 561− = − =

11 cos θ = p

(a) sin 2θ = 2 sin θ cos θ 22p 1 p= −

(b) 2

2 22

2

1 p 1 ptanp p

− − θ = =

(c) sin 4θ = 2 sin 2θ cos 2θ = 2 sin 2θ [2 cos2 θ − 1] 2 24p 1 p [2p 1]= − − 12 tan 2α = 1

2

2 tan 11 tan

α=

− α

2 tan α = 1 − tan2 α tan2 α + 2 tan α − 1 = 0

2 8tan2

− ±α =

2 2 22

− ±=

tan 1 2, 1 2α = − + − −

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Since 0 90 tan 2 1° < α < °⇒ α = −

13 2

2 tantan 21 tan

αα =

− α

1672

α =

2

12 tan 671 2tan 2 67 12 1 tan 672

⇒ = −

2

12 tan 672tan135 11 tan 67

2

=−

2

12 tan 6721 11 tan 67

2

− =−

⇒− + =2 1 11 tan 67 2 tan672 2

2 1 1tan 67 2 tan 67 1 02 2

⇒ − − =

1 2 8tan 672 2

±=

2 2 22

±=

1 2= ±

1tan 67 1 2, since the angle is acute2°

∴ = +

14 (a) 3tan4

θ =

tan(θ + β) = −2

tan tan 21 tan tan

θ + β= −

− θ β

3 tan4 231 tan

4

+ β= −

− β

3 3tan 2 tan4 2+ β = − + β

3 32 tan tan4 2+ = β − β

11 1 tan4 2= β

11 11tan 24 2

β = × =

(b)

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3sin

5θ =

11sin

125β =

115 5

=

11 525

=

15 cos(A B) 5cos(A B) 2

−=

+

2 cos(A − B) = 5 cos(A + B) 2 cos A cos B + 2 sin A sin B = 5 cos A cos B − 5 sin A sin B 2 sin A sin B + 5 sin A sin B = 5 cos A cos B − 2 cos A cos B ⇒ 7 sin A sin B = 3 cos A cos B

7 sin A 3 cosBcosA sin B

=

7 tan A = 3 cot B tan B = 3

1cot B3

=

3 1 1tan A7 3 7

= × =

(a) tan (A + B)

tan A tan B1 tan A tan B

+=

−

1 37

317

+=

−

22747

=

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22 114 2

= =

(b)

1 50sin A5050

= =

(c) cos 2A = 1 − 2sin2A

2

501 250

= −

2150

= −

2425

=

16 Since α, β and θ are the angles of a triangle: α + β + θ = 180° θ = 180 − (α + β) tan θ = tan (180 − (α + β))

− α + β=

− α + βtan180 tan( )

1 tan 180 tan( )

= −tan(α + β)

tan tan1 tan tan α + β

= − − α β

tan tantan tan 1

α + β=

α β −

17

(a) cos 2θ = 1 − 2 sin2 θ

211 2

4 = −

2116

= −

78

=

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(b) cos 4θ = 2 cos2(2θ) − 1

272 1

8 = −

49 132

= −

1732

=

18 3 cos x + 2 sin x = r cos(x − α) 3 cos x + 2 sin x = r [cos x cos α + sin x sin α] Comparing coefficients of cos x and sin x ⇒ r cos α = 3 [1] r sin α = 2 [2]

2[2] [1] tan3

÷ ⇒ α =

α = 33.7° [1]2 + [2]2⇒ r2 = 32 + 22

r 13=

3 cos x 2 sin x 13 cos(x 33.7 )∴ + = − ° (a) Max values = 13 When cos (x − 33.7°) = 1 ⇒ x − 33.7° = 0 x = 33.7° (b) 3 cos x + 2 sin x = 2 13 cos(x 33.7 ) 2⇒ − ° =

2cos(x 33.7 )13

− ° =

1 2x 33.7 cos 13

− − ° =

x − 33.7° = 360° n ± 56.3°

x 360 n 90

nx 360 n 22.6= ° + °

∈= ° − °

19 (a) 2 sin x + 4 cos x = r sin (x + α) 2 sin x + 4 cos x = r sin x cos α + r cos x sin α Equating coefficients of sin x and cos x ⇒ r cos α = 2 [1] r sin α = 4 [2]

4[2] [1] tan 2 63.42

÷ ⇒ α = = ⇒α = °

[1]2 + [2]2 ⇒ r2 = 22 + 42 20r = 2 sin x 4 cos x 20 sin (x 63.4 )∴ + = + °

(b) 2Max2 sin x + 4 cos x

2Max20 sin(x 63.4 )

= + °

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2 2020

=

4 5) 520 5

= =

20 4 cos x − 3 sin x = r cos (x + α) 4 cos x − 3 sin x = r cos x cos α − r sin x sin α (a) Equating coefficients of cos x and sin x ⇒ r cos α = 4 [1] r sin α = 3 [2]

3[2] [1] tan 36.74

÷ ⇒ α = ⇒α = °

[1]2 + [2]2 ⇒ r2 = 32 + 42 r 25 5= = ∴ 4 cos x − 3 sin x = 5 cos (x + 36.7°) (b) 5 cos (x + 36.9°) = 2

2cos (x 36.9 )5

+ ° =

x + 36.9° = 66.4°, 293.6° x = 29.5°, 256.7° (c) max (4 − 5 cos (x + 36.9°)) = 4 + 5 = 9

cos(x + 36.9) = 1 x + 36.9 = 0°, 360 x = −36.9, 323.1 Try these 9.7 (a) sin (C + D) = sin C cos D + cos C sin D [1] sin (C − D) = sin C cos D − cos C sin D [2] [1] − [2] ⇒ sin(C + D) − sin(C − D) = 2 cos C sin D

A +B A BLet C = , D =2 2

−

A + B A B A + B A - B A + B A Bsin + sin = 2 cos sin2 2 2 2 2 2

− − ⇒ − −

A + B A Bsin A sin B 2 cos sin2 2

− ⇒ − =

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(b) cos(C + D) = cos C cos D − sin D sin C [1] cos(C − D) = cos C cos D + sin C sin D [2] [1] + [2] ⇒ cos(C + D) + cos (C − D) = 2 cos C cos D

A B A BLet C , D2 2+ −

= =

A B A B A B A B A B A Bcos cos 2 cos cos2 2 2 2 2 2

+ − + − + − ⇒ + + − =

+ − ⇒ + = A B A BcosA cosB 2 cos cos

2 2

(c) cos(C + D) = cos C cos D − sin C sin D [1] cos(C − D) = cos C cos D + sin C sin D [2] [1] − [2] ⇒ cos (C + D) − cos(C − D) = −2 sin C sin D

A B A BLet C , D2 2+ −

= =

A B A B A B A B A B A Bcos cos 2 sin sin2 2 2 2 2 2

+ − + − + − ⇒ + − − = −

A B A BcosA cosB 2 sin sin2 2+ − ⇒ − = −

Exercise 9E 1 sin 4 x − sin x

4x x 4x x2 cos sin2 2+ − =

5 32 cos x sin x2 2

=

2 cos 3x + cos 2x

3x 2x 3x 2x2 cos cos2 2+ − =

5 12 cos x cos x2 2

=

3 5A A 5A Acos5A cosA 2sin sin2 2+ − − = −

= −2 sin 3A sin 2A

4 4A 4B 4A 4Bsin 4A sin 4B 2 sin cos2 2+ − + =

= 2 sin 2(A + B) cos 2(A − B)

5 6A 4A 6A 4Acos6A cos4A 2 sin sin2 2+ − − = −

= − 2 sin 5A sin A

6 2A 8A 2A 8Acos2A cos8A 2 sin sin2 2+ − − = −

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

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7 7x 3x 7x 3xsin 7x sin3x 2 sin cos2 2+ − + =

= 2 sin 5x cos 2x

8 5x 3x 5x 3xcos5x cos3x 2 cos cos2 2+ − + =

= 2 cos 4x cos x

9 6x 2x 6x 2xsin 6x sin 2x 2 cos sin2 2+ − − =

= 2 cos 4x sin 2x

10 7x 5x 7x 5xsin 7x sin5x 2 sin cos2 2+ − + =

= 2 sin 6x cos x

11 (a) 5cos cos12 12π π+

5 512 12 12 122 cos cos

2 2

π π π π + − =

2 cos cos4 6π π =

2 322 2

=

62

=

(b) 5cos cos12 12π π−

2 sin sin4 6π π = −

2 122 2

= −

22

−=

(c) 5sin sin12 12π π −

2 cos sin4 6π π =

2 122 2

=

22

=

12 cos cossin sin

α + βα + β

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=2 α + β α − β

cos cos

2 2

2 α + β α − β

sin cos2 2

cos

2 cot2sin

2

α + β α + β = = α + β

13 sin 40 + cos 70 = cos 50 + cos 70

50 70 50 702 cos cos2 2+ −

=

= 2 cos 60 cos (−10) = 2 cos 60 cos 10 14 cos 5x + cos x = 0 ⇒ 2 cos 3x cos 2x = 0 ∴ cos 3x = 0, cos 2x = 0

3x 2n OR 2x 2n2 2π π

= π ± = π ±

2π πx = nπ ± or x = nπ ± , n3 6 4

∈

15 sin 6x + sin 2x = 0 2 sin 4x cos 2x = 0 sin 4x = 0, cos 2x = 0

4x = nπ, 2x 2n2π

= π ±

nπ πx = or x = nπ ± , n4 4

∈

16 cos 6x − cos 4x = 0 ⇒ −2 sin 5x sin x = 0 sin 5x = 0, sin x = 0 5x = nπ + (−1)n or x = nπ + (−1)n (0)

nπx = or x = nπ, n5

∈

17 sin 3x = sin x ⇒ sin 3x − sin x = 0 2 cos 2x sin x = 0 cos 2x = 0, sin x = 0

π= π ± = π2x 2n , x n

2

πx = nπ +

n4x = nπ

∴ ∈

18 cos 6x + cos 2x = cos 4x ⇒ 2 cos 4x cos 2x = cos 4x ⇒ 2 cos 4x cos 2x − cos 4x = 0 ⇒ cos 4x (2 cos 2x − 1) = 0

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cos 4x = 0, 2 cos 2x − 1 = 0

1cos2x2

=

π π4x = 2nπ ± , 2x = 2nπ ±2 3

2nπ π 2nπ πx = ± , x = ±4 8 2 6

nπ πx = +2 8 n

πnπ ±6

∈

19 sin 7x + sin x = sin 4x ⇒ sin 7x + sin x − sin 4x = 0 ⇒ 2 sin 4x cos 3x − sin 4x = 0 sin 4x [2 cos 3x − 1] = 0

sin 4x = 0, 1cos32

x =

π= π = π ±4x n , 3x 2n

3

nπ 2 πx = , x = nπ ± , n4 3 9

∈

20 cos 5x − sin 3x − cos x = 0 ⇒ cos 5x − cos x − sin 3x = 0 ⇒ −2 sin 3x sin 2x − sin 3x = 0 ⇒ −sin 3x [2sin 2x + 1] = 0

1sin3x 0, sin 2x2

= = −

n3x n , 2x n ( 1)6−π = π = π + −

nnπ nπ πx = , x = + ( 1) ,n3 2 12

− − ∈

21 sin 3x + sin 4x + sin 5x = 0 sin 5x + sin 3x + sin 4x = 0 2 sin 4x cos x + sin 4x = 0 sin 4x (2 cos x + 1) = 0

1sin 4x 0, cos x2

= = −

24x = nπ, x = 2n3π

π ±

nπ 2πx = or x = 2nπ ± , n4 3

∈

22 sin x + 2 sin 2x + sin 3x = 0 sin 3x + sin x + 2 sin 2x = 0 2 sin 2x cos x + 2 sin 2x = 0 2 sin 2x (cos x + 1) = 0 sin 2x = 0, cos x + 1 = 0 2x = nπ, cos x = −1 x = 2nπ ± π

of 44


nπx = or x = 2nπ ± π, n2

∈

23 cos 3x + cos x + 2 cos 2x = 0 2 cos 2x cos x + 2 cos 2x = 0 2 cos 2x (cos x + 1) = 0 cos 2x = 0, cos x = −1

2x 2n , x 2n2π

= π ± = π ± π

πx = n ± , x = 2nπ ± π, n4

π ∈

24 sin 4 sincos4 cos

θ + θθ + θ

2

=

5 3sin cos2 2θ θ

2 5 3cos cos2 2θ θ

5tan2θ =

25 sin 6 sin 2cos6 cos 2

θ − θθ + θ

2 cos4θ

=sin 2

2 cos4θ

θ cos2θ

= tan 2θ

26 sin8 sin 4cos8 cos 4

θ + θθ − θ

2 sin 6θ

=cos2

2 sin 6θ

− θ sin 2θ

= − cot 2θ

27 cos7 cossin 7 sin

θ + θθ + θ

2=

cos4 cos32

θ θsin 4 cos3θ θ

= cot 4θ

28 sin x + 2 sin 3x + sin 5xsin 3x + 2 sin 5x + sin 7x

sin5x sin x 2sin3xsin 7x sin3x 2 sin5x

+ +=

+ +

2 sin3x cos2x 2 sin3x2 sin5x cos2x 2 sin5x

+=

+

2

=sin 3x( cos 2x +1)

2 sin 5x (cos 2x +1)sin 3x=sin 5x

of 44


29 sin x sin 2xcos x cos2x

+−

3x2 sin2

=

xcos2

3x2 sin2

− xsin

2−

xcos2xsin2

=

xcot2

=

30

5x x2 sin cossin3x sin 2x 5x x2 2 tan cot

5x xsin3x sin 2x 2 22 cos sin2 2

+ = = −

31 cos3x cos x 2sin3x sin x

+=

+cos 2x cos x

2 sin 2x cos x

= cot 2x

32 sin 7 sin 2cos7 cos

θ + θ=

θ + θsin 4 cos3θ θ

2 cos4 cos3θ θ

= tan 4θ

33 θ + θ + θθ − θ + θ

cos 2 cos2 cos3cos 2 cos2 cos3

cos3 cos 2 cos2cos3 cos 2 cos2

θ + θ + θ=

θ + θ − θ

2 cos2 cos 2 cos22 cos2 cos 2 cos2

θ θ + θ=

θ θ − θ

2 cos2θ

=(cos 1)

2 cos2θ +

θ (cos 1)θ −

cos 1cos 1

θ +=

θ −

2

2

2

2 cos2 cot

22 sin2

θθ = = − θ −

34 sin 4 sin 6 sin5cos4 cos6 cos5

θ + θ + θθ + θ + θ

sin 6 sin 4 sin5cos6 cos4 cos5

θ + θ + θ=

θ + θ + θ

2 sin5 cos sin52 cos5 cos cos5

θ θ + θ=

θ θ + θ

sin5 [2 cos 1]θ θ +

=cos5 [2 cos 1]θ θ +

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sin5cos5

θ=

θ

= tan 5θ

35 sin 6 sin 7 sin sin 2cos2 cos cos6 cos7

θ + θ + θ + θθ + θ + θ + θ

sin 7 sin 2 sin 6 sincos7 cos2 cos6 cos

θ + θ + θ + θ=

θ + θ + θ + θ

9 5 7 52 sin cos 2 sin cos2 2 2 2

9 5 7 52 cos cos 2 cos cos2 2 2 2

θ θ θ θ + =

θ θ θ θ +

52 cos2θ

=

9 7sin sin2 2

52 cos2

θ θ + θ

9 7cos cos2 2

θ θ +

9 7sin sin2 2

9 7cos cos2 2

θ θ + =θ θ +

2 sin 4 cos

2

2 cos4 cos2

θ θ =θ θ

sin 4cos4

θ=

θ

= tan 4θ

36 sin sin3 cos5 cos7sin 4 cos8 cos4θ + θ + θ + θ

θ + θ + θ

2 sin 2 cos 2cos6 cos2 sin 2 cos2 2cos6 cos2

θ θ + θ θ=

θ θ + θ θ

2

=cos [sin 2 cos6 ]θ θ + θ

2 cos2 [sin 2 cos6 ]θ θ + θ

2

cos2 cos 1

θ=

θ −

2

2

2

cosseccos

1 2 sec2cos

θθθ= =

− θ−θ

37 cos 2 cos3 cos7cos 2cos3 cos7

θ − θ + θθ + θ + θ

2 cos4 cos3 2 cos32 cos4 cos3 2 cos3

θ θ − θ=

θ θ + θ

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2 cos3θ

=[cos4 1]

2 cos3θ −

θ [cos4 1]θ +

cos4 1cos4 1

θ −=

θ +

2

2

2 sin 22 cos 2− θ

=θ

= −tan2(2θ)

38 cos5 2 cos7 cos9cos5 2 cos7 cos9

θ + θ + θθ − θ + θ

2 cos7 cos 2 2 cos72 cos7 cos 2 2 cos7

θ θ + θ=

θ θ − θ

2 cos7θ

=[cos2 1]

2 cos7θ +

θ

22

2

cos2 1 2 cos cotcos2 1 2 sin[cos2 1]

θ + θ= = = − θ

θ − − θθ −

39 2 sin 6θ cos θ = sin 7θ + sin 5θ 40 −2 sin 8θ cos 4θ = − [2 sin 8θ cos 4θ] = −[sin 12θ + sin 4θ] 41 2 cos 6θ cos 2θ = cos 8θ + cos 4θ 42 2 sin 7θ sin θ = −[−2 sin 7θ sin θ] = −[cos 8θ − cos 6θ] = cos 6θ − cos 8θ 43 2 cos 7θ cos 3θ = cos 10θ + cos 4θ 44 −2 sin 7θ cos 3θ = −[2 sin 7θ cos 3θ] = −[sin 10θ + sin 4θ] = −sin 10θ − sin 4θ 45 RTP: 2 cos x (sin 3x − sin x) = sin 4x Proof: 2 cos x (sin 3x − sin x) = 2 cos x [2 cos 2x sin x] = 2 cos 2x [2 sin x cos x] = 2 cos 2x sin 2x = sin 4x

46 5x x 5x xsin5x sin x 2 sin cos2 2+ − + =

= 2 sin 3x cos 2x. sin 5x + sin x + cos 2x = 0 ⇒ 2 sin 3x cos 2x + cos 2x = 0 cos 2x [2 sin 3x + 1] = 0

1cos2x 0, sin3x2

= = −

n2x 2n , 3x n ( 1)2 6π −π = π ± = π + −

n

πx = nπ ±4 n

nπ πx= + ( 1)3 18

∈− −

47 (a) RTP: sin 2P + sin 2Q + sin 2R = 4 sin P sin Q sin R where P + Q +R = 180°

of 44


Proof: sin 2P + sin 2Q + sin 2R = sin 2P + 2 sin (Q + R) cos (Q − R) = sin 2P + 2 sin (180 − P) cos (Q − R), [Q + R = 180 − P] = 2 sin P cos P + 2 sin P cos (Q − R), [sin (180 − P) = sin P] = 2 sin P [cos P + cos (Q − R)]

P Q R P Q R2 sin P 2 cos cos2 2

+ − − + =

= 2 sin P [2 cos (90 − R) cos (90 − Q)] [P + Q − R = 180 − 2R

P Q R 90 R2

+ −= −

P + R − Q = 180 − 2Q

P + R Q = 90 Q2−

− ]

= 4 sin P sin Q sin R [cos(90 − R) = sin R cos (90 − Q) = sin Q] (b) sin 2P + sin 2Q − sin 2R = 2 sin P cos P + 2 cos (Q + R) sin (Q − R) = 2 sin P cos P + 2 cos(180 − P) sin (Q − R) = 2 sin P cos P − 2 cos P sin (Q − R). = 2 cos P [sin P − sin (Q − R)]

P Q R P R Q2 cosP 2 cos sin2 2

+ − + − =

= 2 cos P [2 cos(90 − R) sin (90 − Q)] = 4 cos P cos Q sin R 48 (a) Proof: P + Q + R = 180° sin(Q + R) = sin(180 − P) = sin P (b) cos(Q + R) = cos(180 − P) = −cos P 49 (a) α + β + γ = 180°. sin β cos γ + cos β sin γ = sin (β + γ) = sin (180 − α) = sin α (b) cos γ + cos β cos α = cos (180 − (β + α)) + cos β cos α = − cos(β + α) + cos β cos α cos cos= − β α sin sin cos cos+ β α + β α = sin β sin α (c) sin α − cos β sin γ sin(180α) cosβsin γ= − − sin cos cos sin= β γ + β γ cos sin− β γ = sin β cos γ 50 1 + cos 2θ + cos 4θ + cos 6θ = 1 + cos 2θ + 2 cos 5θ cos θ

of 44


= 2 cos2θ + 2 cos 5θ cos θ = 2 cos θ [cos θ + cos 5θ] = 2 cos θ [2 cos 3θ cos 2θ] = 4 cos θ cos 2θ cos 3θ Now 1 + cos 2θ + cos 4θ + cos 6θ = 0 ⇒ 4 cos θ cos 2θ cos 3θ = 0. ⇒ cos θ = 0, cos 2θ = 0, cos 3θ = 0

π π πθ = 2nπ ± , 2θ = 2nπ ± , 3θ = 2nπ ±2 2 2

πθ = 2nπ ±2

πθ = nπ ± n4

2nπ πθ = ±3 6

∈

51 1 − cos 2θ + cos 4θ − cos 6θ = 2 sin2θ + [−2 sin(−θ) sin 5θ] = 2 sin2 θ + 2 sin θ sin 5θ = 2 sin θ [sin θ + sin 5θ] = 2 sin θ [2 cos 2θ sin 3θ] = 4 sin θ cos 2θ sin 3θ Now 1 − cos 2θ + cos 4θ − cos 6θ = 0 ⇒ 4 sin θ cos 2θ sin 3θ = 0 ⇒ sin θ = 0, cos 2θ = 0, sin 3θ = 0

n , 2 2n , 3 n2π

θ = π θ = π ± θ = π

θ = nππnπ ± n4

nπ3

∈

Review Exercise 9

1 (a) cosRTP: sec tan1 sin

θ= θ + θ

− θ

Proof:

cos cos 1 sin1 sin 1 sin 1 sin

θ θ + θ= ×

− θ − θ + θ

2

cos (1 sin )1 sinθ + θ

=− θ

2

cos (1 sin )cosθ + θ

=θ

1 sincos+ θ

=θ

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1 sincos cos

θ= +

θ θ

= sec θ + tan θ

(b) 1 1 sec x tan xsec x tan x sec x tan x sec x tan x

+= ×

− − +

2 2

sec x tan xsec x tan x

+=

−

= sec x + tan x [since sec2x − tan2x = 1] 2 (a) 3 cos2x = 1 + sin x ⇒ 3(1 − sin2x) = 1 + sin x ⇒ 3 − 3 sin2x = 1 + sin x ⇒ 3 sin2x + sin x − 2 = 0. y = sin x 3y2 + y − 2 = 0 (3y − 2) (y + 1) = 0

2y , 13

= −

2sin x , sin x 13

= = −

n

n

x = nπ + ( 1) (0.730)nπx = nπ + ( 1) ,

2

− ∈− −

(b) 3 cos x = 2 sin2x 3 cos x = 2(1 − cos2x) 2 cos2 x + 3 cos x − 2 = 0 y = cos x 2y2 + 3y − 2 = 0 (2y − 1) (y + 2) = 0

1y , 22

= −

1cos x , cos x 2 No solutions2

= = − ⇒

x 2n3π

= π ±

3 + α + α=

9 9 cos6 9(1 cos6 )2 2

29 cos (3 )= α = 3 cos 3α

4 cos3 sin3cos sin

α α+

α α

cos3 sin sin3 coscos sin

α α + α α=

α α

sin( 3 )1 sin 22

α + α=

α

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sin 41 sin 22

α=

α

α=

2 sin2 α

α

cos21 sin22

= 4 cos 2α 5 RTP: sin (α + β) sin(α − β) = sin2 α − sin2 β Proof: sin (α + β) sin (α − β) = (sin α cos β + cos α sin β) (sin α cos β − cos α sin β) = sin2α cos2β − cos2α sin2β 2 2 2 2sin (1 sin ) sin (1 sin )= α − β − β − α

2 2 2sin sin sin= α − α β 2 2 2sin sin sin− β + β α

2 2sin sin= α − β

6 5sin13

θ =

12cos

13θ =

5tan12

θ =

3sin

5−

α =

4cos5−

α =

3tan4

α =

(a) sin (θ + α) = sin θ cos α + cos θ sin α

5 4 12 313 5 13 5

− − = +

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20 3665 65−

= −

5665−

=

(b) cos(θ + α) = cosθ cosα − sin θ sin α

12 4 5 313 5 13 5

− = − −

48 15 3365 65 65− −

= + =

7 (a) RTP:

2cos 2cos3 cos7 tan (2 )cos 2cos3 cos7

θ − θ + θ= − θ

θ + θ + θ

Proof:

cos 2cos3 cos7cos 2cos3 cos7

θ − θ + θθ + θ + θ

2 cos4 cos3 2 cos32 cos4 cos3 2 cos3

θ θ − θ=

θ θ + θ

2cos3θ=

(cos4 1)2 cos3

θ −θ (cos4 1)θ +

cos 4 1cos 4 1

θ −=

θ +

1

=22sin 2 1− θ −

22 cos 2 1θ − 1+

2−=

2sin 22

θ2cos 2θ

2

2

sin 2cos 2− θ

=θ

= − tan2(2θ) (b) RTP:

2cos5 2 cos7 cos9 cotcos5 2 cos7 cos9

θ + θ + θ= − θ

θ − θ + θ

Proof:

cos5 2 cos7 cos9cos5 2 cos7 cos9

θ + θ + θθ − θ + θ

2 cos 7 cos2 2 cos72 cos 7 cos2 2 cos7

θ θ + θ=

θ θ − θ

2 cos7θ

=[cos2 1]

2 cos7θ +

θ [cos2 1]θ −

22 cos 1θ −

=1+

1 22 sin 1− θ −

2=

2cos2

θ− 2sin θ

= −cot2θ

of 44


8 cos 15° = cos(60° − 45°) = cos 60 cos 45 − sin 60 sin 45

2 2 34 2 2

= −

2 [1 3]4

= −

9 sin( ) sin cos cos sinsin sin sin sin

θ − α θ α − θ α=

θ α θ α

θ=

sin αθ

cossin

θ α−

αcos sin

sin θ αsin sin

cos cossin sin

α θ= −

α θ

= cot α − cot θ

10 sin x cos x4 4π π + +

1 sin 2x2 2

π = +

1 cos2x2

=

Since sin (90 ) cos+ α = α 11 f(θ) = 3 sin θ + 4 cos θ 3 sin θ + 4 cos θ = r sin (θ + α) = r sin θ cos α + r cos θ sin α Comparing coefficients of sin θ and cos θ ⇒ r cos α = 3 [1] r sin α = 4 [2]

1r sin 4 4 4[2] [1] tan , tan 53.1r cos 3 3 3

−α ÷ ⇒ = ⇒ α = α = = ° α

2 2 2 2 2 2 2 2[1] [2] sin r cos 4 3r+ ⇒ α + α = + r2 = 25 r = 5 ∴ f(θ) = 5 sin (θ + 53.1°) max f(θ) = 5

1 1 1min10 f ( ) 10 5 15

= = + θ +

12 1f (x) sin 4x2 2

π = +

(a) 1 1Range : f (x)2 2−

≤ ≤

(b) Period :2π

(c)

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13 (a) 2 cos(2x) + 3 sin 2x = r cos (2x − θ) = r cos 2x cos θ + r sin 2x sin θ Equating coefficients of cos 2x and sin 2x ⇒ r cos θ = 2 [1] r sin θ = 3 [2]

r sin θ 3[2] [1]r cos θ 2

÷ ⇒ =

13 3tan tan 56.32 2

− θ = ⇒ θ = = °

2 2 2 2 2 2 2 2[1] [2] r cos r sin 2 3+ ⇒ θ + θ = + r2 = 13 r 13= ∴ 3 cos 2x + 3 sin 2x = − °13 cos (2x 56.3 ) (b) 13 cos(2 56.3 ) 2x − ° =

2cos(2x 56.3 )13

− ° =

1 22x 56.3 cos13

− − =

⇒ 2x − 56.3 = 360n ± 56.3° 2x = 360n + 112.6°, 360°n x = 180°n + 56.3°, 180°n, n ∈ ℤ (c) Maximum value 13= cos(2x − 56.3) = 1 2x − 56.3 = 0 x = 28.2° 14 2 sin 6θ cos θ = sin 7θ + sin 5θ 15 −2 sin 8θ cos 4θ = −[sin 12θ + sin 4θ] = −sin 12θ − sin 4θ 16 2 cos 6θ cos 2θ = cos 8θ + cos 4θ 17 (a) 2 tan x− 1 = 3 cot x

32 tan x 1tan x

− =

⇒ 2 tan2 x − tan x − 3 = 0 (2 tan x −3) (tan x + 1) = 0

3tan x , tan x 12

= = −

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x = nπ + 0.983

nπx = nπ4

∈

−

(b) 6 sec2 z = tan z + 8 6 (1 + tan2 z) = tan z + 8 6 tan2 z − tan z − 2 = 0 (2 tan z + 1) (3 tan z − 2) = 0

1 2tan z tan z2 3−

= =

z = nπ 0.464

nz = nπ + 0.588

− ∈

18 (a) RTP: sin x sin x 2 cos x cot x1 sec x 1 sec x

+ = −− +

Proof:

sin x sin x1 sec x 1 sec x

+− +

sin x(1 sec x) sin x(1 sec x)(1 sec x) (1 sec x)+ + −

=− +

sin x sin x sec x+

=sin x sin x sec x+ −21 sec x−

2 sin x1

=1− 2tan x−

2

2

2

2 sin x 2 cos xsin x sin x

cos x

−= =−

= −2 cot x cos x

(b) RTP: −=

+1 sin x cosx

cosx 1 sin x

Proof:

1 sin x 1 sin x 1 sin xcos x cos x 1 sin x− − +

= ×+

21 sin x

cos x(1 sin x)−

=+

2cos

=x

cos x (1 sin x)+

cos x1 sin x

=+

OR:

+ −−

= −

2 2

2 2

x x x xcos sin 2 sin cos1 sin x 2 2 2 2x xcosx cos sin2 2

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2x xcos sin2 2

− =

x xcos sin2 2−

x xcos sin2 2

+

x xcos sin2 2x xcos sin2 2

−=

+

x x x xcos sin cos sin2 2 2 2x x x xcos sin cos sin2 2 2 2

− += ×

+ +

−

=+ +

2 2

2 2

x xcos sin2 2

x x x xcos 2sin cos sin2 2 2 2

cos x1 + sin x

=

(c) cot θ + tan θ

cos sinsin cos

θ θ= +

θ θ

2 2cos sin

sin cosθ + θ

=θ θ

1 1sin cos

= ×θ θ

= sec θ cosec θ 19 (a) cos 5x − sin 3x − cos x = 0 ⇒ cos 5x − cos x − sin 3x = 0 ⇒ −2sin 3x sin 2x − sin 3x = 0 ⇒ −sin 3x (2 sin 2x + 1) = 0

1sin3x 0, sin 2x2−

= =

n3x n x3π

= π⇒ =

n2x n ( 1)6−π = π + −

nnπ πx = + ( 1)2 12 n

nπx =3

− − ∈

(b) sin 3x + sin 4 x + sin 5x = 0 sin 4x + sin 3x + sin 5x = 0 sin 4x + 2 sin 4x cos x = 0 sin 4x (1 + 2 cos x) = 0

1sin 4x 0, cos x2−

= =

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24x n , x 2n3π

= π = π ±

nπx =4 n

πx = 2nπ ±3

∈

20 (a) sin 7 sincos7 cos

θ − θθ + θ

2 cos4θ

=sin3

2 cos4θ

θ cos3θ

sin3cos3

θ=

θ

= tan 3θ

(b) cos 2 cos2 cos3cos 2 cos2 cos3

θ + θ + θθ − θ + θ

2 cos2 cos 2 cos22 cos2 cos 2 cos2

θ θ + θ=

θ θ − θ

2 cos2θ

=(cos 1)

2 cos2θ +

θ (cos 1)θ −

=2 θ

−

2cos2

2 θ

2sin2

θ = − 2cot

2

chapter 9 trigonometry - macmillan caribbean · 2017-06-29 · 2 4 cot θ – 4 cot θ – 3 = 0 (2...

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