Page 1 of 44
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
Page 2 of 44
Unit 1 Answers: Chapter 9 © Macmillan Publishers Limited 2013
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
Page 3 of 44
Unit 1 Answers: Chapter 9 © Macmillan Publishers Limited 2013
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
Page 4 of 44
Unit 1 Answers: Chapter 9 © Macmillan Publishers Limited 2013
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
Page 5 of 44
Unit 1 Answers: Chapter 9 © Macmillan Publishers Limited 2013
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:
Page 6 of 44
Unit 1 Answers: Chapter 9 © Macmillan Publishers Limited 2013
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
Page 7 of 44
Unit 1 Answers: Chapter 9 © Macmillan Publishers Limited 2013
= +− −
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:
Page 8 of 44
Unit 1 Answers: Chapter 9 © Macmillan Publishers Limited 2013
−−
=−
−
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
Page 9 of 44
Unit 1 Answers: Chapter 9 © Macmillan Publishers Limited 2013
= − =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
Page 10 of 44
Unit 1 Answers: Chapter 9 © Macmillan Publishers Limited 2013
− ±∴ =
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∈
Page 11 of 44
Unit 1 Answers: Chapter 9 © Macmillan Publishers Limited 2013
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
−=
+
Page 12 of 44
Unit 1 Answers: Chapter 9 © Macmillan Publishers Limited 2013
−=
+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)
Page 13 of 44
Unit 1 Answers: Chapter 9 © Macmillan Publishers Limited 2013
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
Page 14 of 44
Unit 1 Answers: Chapter 9 © Macmillan Publishers Limited 2013
−∴ 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
−= −
Page 15 of 44
Unit 1 Answers: Chapter 9 © Macmillan Publishers Limited 2013
(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
Page 16 of 44
Unit 1 Answers: Chapter 9 © Macmillan Publishers Limited 2013
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
Page 17 of 44
Unit 1 Answers: Chapter 9 © Macmillan Publishers Limited 2013
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 )− = − °
Page 18 of 44
Unit 1 Answers: Chapter 9 © Macmillan Publishers Limited 2013
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]+
Page 19 of 44
Unit 1 Answers: Chapter 9 © Macmillan Publishers Limited 2013
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
Page 20 of 44
Unit 1 Answers: Chapter 9 © Macmillan Publishers Limited 2013
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
−= = =
− −
Page 21 of 44
Unit 1 Answers: Chapter 9 © Macmillan Publishers Limited 2013
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 = − + − −
Page 22 of 44
Unit 1 Answers: Chapter 9 © Macmillan Publishers Limited 2013
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)
Page 23 of 44
Unit 1 Answers: Chapter 9 © Macmillan Publishers Limited 2013
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
=
Page 24 of 44
Unit 1 Answers: Chapter 9 © Macmillan Publishers Limited 2013
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
=
Page 25 of 44
Unit 1 Answers: Chapter 9 © Macmillan Publishers Limited 2013
(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 )
= + °
Page 26 of 44
Unit 1 Answers: Chapter 9 © Macmillan Publishers Limited 2013
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
− ⇒ − =
Page 27 of 44
Unit 1 Answers: Chapter 9 © Macmillan Publishers Limited 2013
(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
Page 28 of 44
Unit 1 Answers: Chapter 9 © Macmillan Publishers Limited 2013
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
Page 29 of 44
Unit 1 Answers: Chapter 9 © Macmillan Publishers Limited 2013
=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
Page 30 of 44
Unit 1 Answers: Chapter 9 © Macmillan Publishers Limited 2013
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
Page 31 of 44
Unit 1 Answers: Chapter 9 © Macmillan Publishers Limited 2013
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
Page 32 of 44
Unit 1 Answers: Chapter 9 © Macmillan Publishers Limited 2013
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 +
Page 33 of 44
Unit 1 Answers: Chapter 9 © Macmillan Publishers Limited 2013
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
Page 34 of 44
Unit 1 Answers: Chapter 9 © Macmillan Publishers Limited 2013
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°
Page 35 of 44
Unit 1 Answers: Chapter 9 © Macmillan Publishers Limited 2013
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
Page 36 of 44
Unit 1 Answers: Chapter 9 © Macmillan Publishers Limited 2013
= 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
Page 37 of 44
Unit 1 Answers: Chapter 9 © Macmillan Publishers Limited 2013
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
Page 38 of 44
Unit 1 Answers: Chapter 9 © Macmillan Publishers Limited 2013
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
− − = +
Page 39 of 44
Unit 1 Answers: Chapter 9 © Macmillan Publishers Limited 2013
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
Page 40 of 44
Unit 1 Answers: Chapter 9 © Macmillan Publishers Limited 2013
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)
Page 41 of 44
Unit 1 Answers: Chapter 9 © Macmillan Publishers Limited 2013
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
= = −
Page 42 of 44
Unit 1 Answers: Chapter 9 © Macmillan Publishers Limited 2013
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
Page 43 of 44
Unit 1 Answers: Chapter 9 © Macmillan Publishers Limited 2013
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−
= =
Page 44 of 44
Unit 1 Answers: Chapter 9 © Macmillan Publishers Limited 2013
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