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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)
= – 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
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
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
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 −
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
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
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
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 )∴ + = + °
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
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
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°
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
= 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
(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
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
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