Chapter 7 Sums and Products of Trigonometric Functions
CHAPTER 7 Exercise 7A (p.166) 1 1. 2 sin 40° cos 20° = 2[ (sin 60° + sin 20°)] 2 = sin 60° + sin 20° 1 2. 2 cos 120° sin 80° = 2[ (sin 200° − sin 40°)] 2 = sin 200° − sin 40°
1 3. 4 cos 70° cos 50° = 4[ (cos 120° + cos 20°)] 2 = 2 cos 120° + 2 cos 20° 1 4. sin x sin 3 x = − [cos 4 x − cos( −2 x )] 2 1 = − (cos 4 x − cos 2 x ) 2 1 1 = cos 2 x − cos 4 x 2 2 1 (cos 8 A + cos 2 A) 2 1 1 = cos 8 A + cos 2 A 2 2
5. cos 5 A cos 3 A =
6. 4 cos( x + 60°)sin( x − 60°) 1 = 4[ (sin 2 x − sin 120°)] 2 = 2 sin 2 x − 2 sin 120° 7. cos 60° + cos 140° = 2 cos 100° cos( −40°) = 2 cos 100° cos 40° 8. sin 5 x − sin 3 x = 2 cos 4 x sin x 9. cos 2 y − cos 6 y = −2 sin 4 y sin( −2 y) = 2 sin 4 y sin 2 y 10. sin(3 A + B) + sin( A + 3 B) 3 A + B + A + 3B 3 A + B − A − 3B = 2 sin cos 2 2 = 2 sin 2( A + B) cos( A − B) 11. sin(
π − θ) + cos 3θ = cos θ + cos 3θ 2 = 2 cos 2θ cos( −θ) = 2 cos 2θ cos θ
12. sin A + cos B π = cos( − A) + cos B 2 1 π 1 π = 2 cos ( − A + B) cos ( − A − B) 2 2 2 2
107
13. 4 cos x cos 3 x cos 5 x 1 = 4 cos x ⋅ [cos 8 x + cos( −2 x )] 2 = 2 cos x (cos 8 x + cos 2 x ) = 2 cos x cos 8 x + 2 cos x cos 2 x 1 1 = 2 ⋅ [cos 9 x + cos( −7 x )] + 2 ⋅ [cos 3 x + cos( − x )] 2 2 = cos 9 x + cos 7 x + cos 3 x + cos x 14. cos 75° cos 45° 1 = [cos(75° + 45°) + cos(75° − 45°)] 2 1 = (cos 120° + cos 30°) 2 1 cos 75° cos 45° = (cos 120° + cos 30°) 2 2 1 1 3 cos 75°( ) = (− + ) 2 2 2 2 1 1 3 cos 75° = (− + ) 2 2 2 2 ( −1 + 3 ) = 4 1 = ( 6 − 2) 4 15.
75°+15° 75°−15° cos 75° + cos 15° 2 cos 2 cos 2 = sin 75° − sin 15° 2 cos 75°+2 15° sin 75°−2 15° 75° − 15° = cot 2 = cot 30° = 3
16. cos x + cos( x + 120°) + cos( x + 240°) x + x + 120° x − x − 120° cos = 2 cos 2 2 + cos( x + 240°) = 2 cos( x + 60°) cos( −60°) + cos( x + 240°) = 2 cos( x + 60°) cos 60° + cos( x + 240°) 1 = 2 cos( x + 60°)( ) + cos( x + 240°) 2 = cos( x + 60°) + cos( x + 240°) x + 60° + x + 240° x + 60° − x − 240° cos = 2 cos 2 2 = 2 cos( x + 150°) cos( −90°) = 2 cos( x + 150°) cos 90° =0 17. sin( A + B) =
3 7
sin A cos B + sin B cos A = sin( A − B) =
1 7
3 ..............(1) 7
108
Chapter 7 Sums and Products of Trigonometric Functions
A+ B ) 2 A+ B A+ B ) cos( ) = 2 sin( 2 2 a b = 2⋅ ⋅ 2 2 2 ± a + b ± a + b2 2 ab = 2 a + b2
1 ..............(2) 7 3 1 4 (1) + (2), 2 sin A cos B = + = 7 7 7 2 sin A cos B = 7
(b) sin( A + B) = sin 2(
sin A cos B − sin B cos A =
(1) − (2), 2 sin B cos A = sin B cos A = sin A cos B = sin B cos A tan A =2 tan B
3 1 2 − = 7 7 7
A+ B ) 2 A+ B = 2 cos 2 ( ) −1 2 2b 2 = 2 −1 a + b2 b2 − a2 = 2 a + b2
1 7
(c) cos( A + B) = cos 2(
2 7 1 7
18. sin A + sin B = a A+ B A− B 2 sin cos =a 2 2 A+ B A− B a sin cos = ..............(1) 2 2 2 A+ B A − B a2 sin 2 cos 2 ............(2) = 2 2 4 cos A + cos B = b A+ B A− B 2 cos cos =b 2 2 A+ B A− B b cos cos = ..............(3) 2 2 2 A + B A − B b2 cos 2 cos 2 ...........(4) = 2 2 4 (2) + (4), cos 2
A− B A+ B A+ B a2 b2 (sin 2 )= + cos 2 + 2 2 2 4 4
cos 2
A − B a2 + b2 = 2 4
cos
π 3π 5π 7π 9π + cos + cos + cos + cos 11 11 11 11 11 π π π π 3π 2 y sin = 2 sin cos + 2 sin cos 11 11 11 11 11 π π 5π 7π + 2 sin cos + 2 sin cos 11 11 11 11 9π π + 2 sin cos 11 11 2π 1 4π −2 π ) = sin + 2 ⋅ (sin + sin 11 2 11 11 1 6π −4 π ) + 2 ⋅ (sin + sin 2 11 11 1 8π −6 π + 2 ⋅ (sin + sin ) 2 11 11 1 10 π −8π + 2 ⋅ (sin + sin ) 2 11 11 2π 4π 2π = sin + (sin − sin ) 11 11 11 6π 4π + (sin − sin ) 11 11 6π 10 π 8π 8π + (sin − sin ) + (sin − sin ) 11 11 11 11
19. Let y = cos
A − B ± a2 + b2 = 2 2
From (1), A+ B a a 2 sin = ⋅ = 2 2 2 2 2 ± a +b ± a + b2
2 y sin
= sin( π −
From (3),
= sin
A+ B b 2 b cos = ⋅ = 2 2 ± a2 + b2 ± a2 + b2 sin( A +2 B ) A+ B )= = (a) tan( 2 cos( A +2 B )
a ± a2 +b2 b ± a2 +b2
10 π π = sin 11 11
=
a b
2y = 1 1 y= 2 ∴ cos
π 11
π ) 11
π 3π 5π 7π 9π 1 + cos + cos + cos + cos = 11 11 11 11 11 2
Chapter 7 Sums and Products of Trigonometric Functions
π 2π 4π cos cos 7 7 7 π 2π π 2π 1 4π ) + cos( − )]cos = [cos( + 2 7 7 7 7 7 3π 4π 1 −π ) cos = (cos + cos 2 7 7 7 π 1 3π 4π 1 4π cos = cos + cos cos 2 7 7 2 7 7 1 1 3π + 4 π 3π − 4 π )] = [ (cos + cos 2 2 7 7 π + 4π π − 4π 1 1 )] + [ (cos + cos 2 2 7 7 1 −π 1 5π −3π = (cos π + cos ) + (cos + cos ) 4 7 4 7 7 π 1 3π 5π = (cos π + cos + cos + cos ) 4 7 7 7 1 1 = ( −1 + ) (by example 3, p.164) 4 2 1 =− 8
20. cos
sin 2 A + sin 2 B 21. cos 2 A + cos 2 B 2 sin( 2 A +2 2 B ) cos( 2 A −2 2 B ) = 2 cos( 2 A +2 2 B ) cos( 2 A −2 2 B ) =
22.
25. cos( x + y + z ) + cos( x + y − z ) + cos( x − y + z ) + cos( − x + y + z ) = 2 cos( x + y) cos z + 2 cos z cos( x − y) = 2 cos z[cos( x + y) + cos( x − y)] = 2 cos z[2 cos x cos y] = 4 cos x cos y cos z 26. tan( A + B) + tan( A − B) sin( A + B) sin( A − B) = + cos( A + B) cos( A − B) sin( A + B) cos( A − B) + cos( A + B)sin( A − B) = cos( A + B) cos( A − B) 1 (sin 2 A + sin 2 B) + 1 (sin 2 A − sin 2 B) 2 = 2 cos( A + B) cos( A − B) sin 2 A = 1 (cos A + cos 2 B) 2 2 2 sin A = cos 2 A + cos 2 B 27.
sin( 2 A +2 2 B )
cos( 2 A +2 2 B ) = tan( A + B) sin(θ + 30°) + sin(θ + 60°) cos(θ + 30°) + cos(θ + 60°) 2 sin(θ + 45°) cos 15° = 2 cos(θ + 45°) cos 15° = tan(θ + 45°) tan θ + tan 45° = 1 − tan θ tan 45° 1 + tan θ = 1 − tan θ
23. 2 sin( 45° + A)sin( 45° − B) 1 = 2 ⋅ − [cos(90° + A − B) − cos( A + B)] 2 = cos( A + B) − cos[90° − ( B − A)] = cos( A + B) − sin( B − A) = cos( A + B) + sin( A − B) 24. sin 2( x + y) − sin 2 x − sin 2 y = 2 sin( x + y) cos( x + y) − 2 sin( x + y) cos( x − y) = 2 sin( x + y)[cos( x + y) − cos( x − y)] = 2 sin( x + y)[ −2 sin x sin y] = −4 sin( x + y)sin x sin y
109
sin A + sin 3 A + sin 5 A cos A + cos 3 A + cos 5 A 2 sin 3 A cos 2 A + sin 3 A = 2 cos 3 A cos 2 A + cos 3 A sin 3 A(2 cos 2 A + 1) = cos 3 A(2 cos 2 A + 1) = tan 3 A
28. sin(3 A + B) + sin( A + 3 B) = 2 sin 2( A + B) cos( A − B) = 2[2 sin( A + B) cos( A + B)]cos( A − B) = 2 sin( A + B)[2 cos( A + B) cos( A − B)] = 2 sin( A + B)(cos 2 A + cos 2 B) 29.
sin( x − 40°) + sin( x + 80°) = 1 x − 40° + x + 80° x − 40° − x − 80° ) cos( ) =1 2 sin( 2 2 2 sin( x + 20°) cos( −60°) = 1 1 2 sin( x + 20°)( ) = 1 2 sin( x + 20°) = 1 x + 20° = 90° x = 70°
30. cos(2 x + 10°) + cos(2 x − 10°) = 0 2 x + 10° + 2 x − 10° 2 x + 10° − 2 x + 10° 2 cos( )cos( ) 2 2 =0 cos 2 x cos 10° = 0 cos 2 x = 0 (Q cos10° ≠ 0 ) 2 x = 90°, 270°, 450°, 630° x = 45°, 135°, 225°, 315°
110
Chapter 7 Sums and Products of Trigonometric Functions
31. 2 sin( x + 15°) cos( x − 15°) = 1 1 2 ⋅ (sin 2 x + sin 30°) = 1 2 1 sin 2 x + = 1 2 1 sin 2 x = 2 2 x = 30°, 150°, 390°, 510° x = 15°, 75°, 195°, 255° 32. sin 2 x + sin 3 x = 0 −x 5x =0 2 sin cos 2 2 x 5x sin cos = 0 2 2 5x x sin = 0 or cos = 0 2 2 5x x = 0°, 180°, 360°, 540°, 720° or = 90° 2 2 x = 0°, 72°, 144°, 180°, 216°, 288° 33.
34.
3x x = cos 2 2 3x x cos − cos = 0 2 2 x −2 sin x sin = 0 2 x sin x = 0 or sin = 0 2 x x = 0°, 180° or = 0°, 180° 2 x = 0°, 180°
36. sin x + sin 2 x + sin 3 x + sin 4 x = 0 3x x 7x x 2 sin cos + 2 sin cos = 0 2 2 2 2 x 3x 7x cos (sin + sin ) = 0 2 2 2 x 3x 7x cos = 0 or sin + sin =0 2 2 2 5x x cos x = 0 = 90° or 2 sin 2 2 5x = 0 or cos x = 0 x = 180° or sin 2 5x = 0°, 180°, 360°, 540°, 720° or x = 90°, 270° 2 ∴ x = 0°, 72°, 90°, 144°, 180°, 216°, 270°, 288° 37. sin 2 A + sin 2 B + sin 2C = 2 sin( A + B) cos( A − B) + 2 sin C cos C = 2 sin C cos( A − B) + 2 sin C[ − cos( A + B)] = 2 sin C[cos( A − B) − cos( A + B)] = 4 sin A sin B sin C
cos
sin 3 x + cos 3 x = sin x + cos x sin 3 x − sin x = cos x − cos 3 x 2 cos 2 x sin x = 2 sin 2 x sin x sin x (cos 2 x − sin 2 x ) = 0 or cos 2 x − sin 2 x = 0 sin x = 0 x = 0°, 180° or tan 2 x = 1 2 x = 45°, 225°, 405°, 585° x = 22.5°, 112.5°, 202.5°, 292.5° x = 0°, 22.5°, 112.5°, 180°, 202.5°, 292.5°
35. cos x + cos 2 x + cos 3 x = 0 2 cos 2 x cos x + cos 2 x = 0 cos 2 x (2 cos x + 1) = 0 cos 2 x = 0
or 2 cos x + 1 = 0 1 2 x = 90°, 270°, 450°, 630° or cos x = − 2 x = 45°, 135°, 225°, 315° or x = 120°, 240°
∴
x = 45°, 120°, 135°, 225°, 240°, 315°
38. sin A + sin B + sin C 1 1 1 1 = 2 sin ( A + B) cos ( A − B) + 2 sin C cos C 2 2 2 2 1 1 1 1 = 2 cos C cos ( A − B) + 2 cos ( A + B) cos C 2 2 2 2 1 1 1 = 2 cos C[cos ( A − B) + cos ( A + B)] 2 2 2 A B C = 4 cos cos cos 2 2 2 39. cos 2 A + cos 2 B − cos 2C − 1
= 2 cos( A + B) cos( A − B) − 2 cos 2 C = −2 cos C cos( A − B) + 2 cos C cos( A + B) = 2 cos C[cos( A + B) − cos( A − B)] = −4 sin A sin B cos C 40. cos 2 A + cos 2 B + cos 2 C − 1 1 = (cos 2 A + cos 2 B) + cos 2 C 2 = cos( A + B) cos( A − B) + cos 2 C = − cos C cos( A − B) − cos C cos( A + B) = − cos C[cos( A − B) + cos( A + B)] = −2 cos A cos B cos C π π ) cos( x − ) 8 24 1 π π = [sin(2 x + ) + sin ] 2 12 6 1 π 1 = sin(2 x + ) + 2 12 4
41. (a) y = sin( x +
Chapter 7 Sums and Products of Trigonometric Functions
(b) y =
1 π 1 sin(2 x + ) + 2 12 4
π ) ≤1 12 π When sin(2 x + ) = 1 , y is maximum. 12 3 ∴ Maximum of y is . 4 As −1 ≤ sin(2 x +
π ) = −1 , y is minimum. 12 1 ∴ Minimum of y is − . 4 When sin(2 x +
42. (a) 1 + cos( A − B) cos( A + B) 1 = 1 + (cos 2 A + cos 2 B) 2 1 = 1 + (2 cos 2 A + 2 cos 2 B − 2) 2 = cos 2 A + cos 2 B (b) 3 + cos( P − Q) cos( P + Q) + cos(Q − R) cos(Q + R) + cos( R − P) cos( R + P) = [1 + cos( P − Q) cos( P + Q)] + [1 + cos(Q − R) cos(Q + R)] + [1 + cos( R − P) cos( R + P)] = cos 2 P + cos 2 Q + cos 2 Q + cos 2 R + cos 2 R + cos 2 P = 2(cos 2 P + cos 2 Q + cos 2 R) 43. (a) cos 3 A + cos 3 B + cos 3C = cos 3 A + cos 3 B + cos 3( π − A − B) 3 3 = 2 cos ( A + B) cos ( A − B) 2 2 − cos 3( A + B) 3 3 = 2 cos ( A + B) cos ( A − B) 2 2 2 3 ( A + B) + 1 − 2 cos 2 3 3 = 2 cos ( A + B)[cos ( A − B) 2 2 3 − cos ( A + B)] + 1 2 3 3 3 = 4 cos ( A + B)sin A sin B + 1 2 2 2 3 3 3 = 1 − 4 sin A sin B sin C 2 2 2 (b) If cos 3 A + cos 3 B + cos 3C = 1 , then by (a), 4 sin
3 3 3 A sin B sin C = 0 2 2 2
111
3 3 3 A = 0 or sin B = 0 or sin C = 0 2 2 2 3 Let sin A = 0 , 2 ∴ A = 0° or 120° Since A is an interior angle of a triangle, ∴
sin
∴ one of the angles = 120° 44. (a) Let a be the length of ladder. x = a cos β − a cos α 1 1 = a[( −2)sin (β + α )sin (β − α )] 2 2 1 1 = 2 a sin (α + β)sin (α − β) 2 2 y = a sin α − a sin β 1 1 = 2 a cos (α + β)sin (α − β) 2 2
x 1 = tan (α + β) y 2 1 x = y tan (α + β) 2 (c) From (b), 1 π 15 = 20 tan ( + β) 2 3 1 π 3 tan ( + β) = 2 3 4 1 π ( + β) = 0.643 5 2 3 β = 0.240 (corr. to 3 sig. fig.)
(b) ∴
45. (a) (i) Let P(n) be the proposition °ßsin x + sin 2 x + L + sin nx
=
sin
( n +1) x sin nx 2 2 °®. sin 2x
When n = 1 , L.H.S. = sin x R.H.S. =
sin x sin 2x sin 2x
= sin x = L.H.S.
∴ P(1) is true. Assume P(k) is true for any positive integer k. i.e. sin x + sin 2 x + L + sin kx =
sin
( k +1) x sin kx 2 2 sin 2x
112
Chapter 7 Sums and Products of Trigonometric Functions
then
= {12 [sin 12 (2 k + 1) x − sin 12 x ]
sin x + sin 2 x + L + sin kx + sin(k + 1) x sin 12 ( k + 1) x sin 12 kx = + sin(k + 1) x sin 2x =
+ 12 [sin(k + 23 ) x − sin(k + 12 ) x ]} ÷ sin 2x =
sin 12 ( k + 1) x sin 12 kx + sin(k + 1) x sin 12 x
=
sin 2x 1 1 1 = { [cos x − cos (2 k + 1) x ] 2 2 2 1 1 3 + [cos(k + ) x − cos( k + ) x ]} 2 2 2 x ÷ sin 2 1 [cos 1 x − cos 1 (2 k + 3) x ] 2 2 = 2 sin 2x = =
sin 12 ( k + 2) x sin 12 ( k + 1) x sin 2x
[( k +1) +1] x 2
sin
sin
( k +1) x 2
sin 2x Thus assuming P(k) is true for any positive integer k, P( k + 1) is also true. By the principle of mathematical induction, P(n) is true for all positive integer n. (ii) Let Q(n) be the proposition
=
1 [sin 1 (2 k 2 2
+ 3) x − sin 12 x ]
sin 2x cos 12 ( k + 2) x sin 12 ( k + 1) x sin 2x
cos
[( k +1) +1] x ( k +1) x sin 2 2 sin 2x
Thus assuming Q(k) is true for any positive integer k, Q( k + 1) is also true. By the principle of mathematical induction, Q(n) is true for all positive integer n. (b)
sin π6 + sin 2( π6 ) + sin 3( π6 ) + L + sin 11( π6 )
cos π6 + cos 2( π6 ) + cos 3( π6 ) + L + cos 11( π6 ) sin
=
(11+1)( π ) 6 2
sin
11( π ) 6 2
sin 12π (11+1)( π ) 11( π ) cos 2 6 sin 26 sin 12π
π = tan 6( ) 6 = tan π =0
°ß cos x + cos 2 x + L + cos nx cos
=
( n +1) x sin nx 2 2 °®. sin 2x
( a − 1)2 + (b + 1)2 = 1
cos x cos 2x cos 2x
= cos x = L.H.S.
∴ Q(1) is true. Assume Q(k) is true for any positive integer k. i.e. cos x + cos 2 x + L + cos kx =
cos
( k +1) x sin kx 2 2 sin 2x
then cos x + cos 2 x + L + cos kx + cos( k + 1) x = =
( k +1) x sin 2 x sin 2 ( k +1) x cos 2 sin
cos
1. sin θ = a − 1 , cos θ = b + 1 sin 2 θ + cos 2 θ = ( a − 1)2 + (b + 1)2
When n = 1 , L.H.S. = cos x R.H.S. =
Exercise 7B (p.170)
kx 2 kx 2
+ cos( k + 1) x + cos( k + 1) x sin 2x
sin 2x
2. x = a sec θ , y = b tan θ x y = sec θ , = tan θ a b x 2 y2 − = sec 2 θ − tan 2 θ = 1 a2 b2 b2 x 2 − a2 y2 = a2b2
3. x = cos θ + sin θ + 1, y = cos θ − sin θ + 1 cos θ + sin θ = x − 1, cos θ − sin θ = y − 1
( x − 1)2 + ( y − 1)2 = (cos θ + sin θ)2 + (cos θ − sin θ)2 = (cos 2 θ + 2 sin θ cos θ + sin 2 θ) + (cos 2 θ − 2 sin θ cos θ + sin 2 θ) =2 ∴ ( x − 1)2 + ( y − 1)2 = 2
Chapter 7 Sums and Products of Trigonometric Functions
(b) mn = (tan A + sin A)(tan A − sin A)
4. sin θ − cos θ = x , cos 2θ = y 2
= tan 2 A − sin 2 A
(sin θ − cos θ)2 = x 2
= tan 2 A − tan 2 A cos 2 A
sin 2 θ − 2 sin θ cos θ + cos 2 θ = x 2 1 − sin 2θ = x 2 sin 2θ = 1 − x 2 (1 − x 2 )2 + ( y 2 )2 = sin 2 2θ + cos 2 2θ (1 − x 2 )2 + ( y 2 )2 = 1
x 4 + y4 = 2x2 5. tan θ − sec θ = m , tan θ + sec θ = n
= tan 2 A(1 − cos 2 A) = tan 2 A sin 2 A m+n 2 m−n 2 =( ) ( ) (by (a)) 2 2 ∴ 16 mn = ( m 2 − n 2 )2 2 m − n 2 = 4 mn (as m ≥ n > 0 )
tan 2 θ − sec 2 θ = mn mn + 1 = 0
∴
6. x = a(csc θ − cot θ) , y = b(csc θ + cot θ) xy = a(csc θ − cot θ) ⋅ b(csc θ + cot θ) = ab(csc 2 θ − cot 2 θ) = ab
sin θ cos θ = x y x tan θ = y x sin θ = and 2 x + y2 y cos θ = 2 x + y2
9. (a) From (1),
(tan θ − sec θ)(tan θ + sec θ) = mn
∴
113
x2+y2
x
θ
y
(b) From (a), y2 x2 2 , cos θ = x 2 + y2 x 2 + y2 Substitute into (2), sin 2 θ =
xy = ab
7. sin θ + cos θ = m (
1 1 2 x2 y2 + )( ) ( )( )= 2 2 2 2 2 2 2 y x +y x x +y x + y2
1 sin θ cos θ = ( m 2 − 1) ....................(1) 2
(
x 2 y2 1 2 + 2) 2 = 2 2 2 y x x +y x + y2
tan θ + cot θ = n sin θ cos θ + =n cos θ sin θ sin 2 θ + cos 2 θ = n sin θ cos θ
∴
sin θ + 2 sin θ cos θ + cos θ = m 2
2
2
1 + 2 sin θ cos θ = m 2
1 .................................(2) n From (1) and (2), sin θ cos θ =
1 2 1 ( m − 1) = n 2 2 n( m − 1) = 2 8. (a) tan A + sin A = m .........(1) tan A − sin A = n .........(2) m+n 2 m−n (1) − (2), sin A = 2 (1) + (2), tan A =
x 2 y2 + =2 y2 x 2
x Let t = ( )2 = tan 2 θ , y 1 t+ =2 ∴ t t 2 − 2t + 1 = 0 ∴ t =1 x 2 ∴ ( ) =1 y
tan θ =
x = ±1 y
π , 2 tan θ = 1 π θ= 4
As 0 < θ <
∴
114
Chapter 7 Sums and Products of Trigonometric Functions
Revision Exercise 7 (p.172) 1. sin( A + C )sin( A + D) = sin[360° − ( B + D)]sin[360° − ( B + C )] = [ − sin( B + D)][− sin( B + C )] = sin( B + C )sin( B + D) 2. sin A = 2 cos B sin C sin[180° − ( B + C )] = sin( B + C ) = 2 cos B sin C sin B cos C + cos B sin C = 2 cos B sin C sin B cos C = cos B sin C tan B = tan C ∴ B=C 3. cos 2 ( x − y) − sin 2 ( x + y) 1 1 = [1 + cos 2( x − y)] − [1 − cos 2( x + y)] 2 2 1 = [cos 2( x + y) + cos 2( x − y)] 2 = cos 2 x cos 2 y 4. (a) 2 sin A cos 3 A 1 = 2 ⋅ [sin( A + 3 A) + sin( A − 3 A)] 2 = sin 4 A + sin( −2 A) = sin 4 A − sin 2 A (b) 2 sin A[cos A + cos 3 A + cos 5 A + cos 7 A + cos 9 A] = 2 sin A cos A + 2 sin A cos 3 A + 2 sin A cos 5 A + 2 sin A cos 7 A + 2 sin A cos 9 A = sin 2 A + (sin 4 A − sin 2 A) + (sin 6 A − sin 4 A) + (sin 8 A − sin 6 A) + (sin 10 A − sin 8 A) = sin 10 A ∴ cos A + cos 3 A + cos 5 A + cos 7 A + cos 9 A sin 10 A = 2 sin A 5. cos 2 A + cos 2 B − cos 2 C − 1 1 1 = (cos 2 A + 1) + (cos 2 B + 1) − cos 2 C − 1 2 2 1 = (cos 2 A + cos 2 B) − cos 2 C 2 = cos( A + B) cos( A − B) − cos 2 C = cos(180° − C ) cos( A − B) − cos C cos[180° − ( A + B)] = − cos C cos( A − B) + cos C cos( A + B) = cos C[cos( A + B) − cos( A − B)] = −2 sin A sin B cos C
6. (a) (b + c) cos A + (c + a) cos B + ( a + b) cos C = 2 R(sin B cos A + sin C cos A + sin C cos B + sin A cos B + sin A cos C + sin B cos C ) = 2 R[sin( A + B) + sin( B + C ) + sin(C + A)] = 2 R(sin A + sin B + sin C ) = a+b+c (b) (b + c) cos A + (c + a) cos B + ( a + b) cos C = a+b+c ( a + b + c) cos A + ( a + b + c) cos B + ( a + b + c) cos C = a + b + c + a cos A + b cos B + c cos C ( a + b + c)(cos A + cos B + cos C − 1) = a cos A + b cos B + c cos C A+ B A− B ) cos( ) + cos C − 1] ( a + b + c)[2 cos( 2 2 = a cos A + b cos B + c cos C A+ B A− B C ( a + b + c)[2 cos( ) cos( ) − 2 sin 2 ] 2 2 2 = a cos A + b cos B + c cos C C A− B C ( a + b + c)(2 sin )[cos( ) − sin ] 2 2 2 = a cos A + b cos B + c cos C A− B A+ B C ( a + b + c)(2 sin )[cos( ) − cos( )] 2 2 2 = a cos A + b cos B + c cos C A B C 4( a + b + c)sin sin sin 2 2 2 = a cos A + b cos B + c cos C 7. cos 2 nθ − cos 2 mθ = (cos nθ + cos mθ)(cos nθ − cos mθ) n+m n−m θ cos θ) = (2 cos 2 2 n+m n−m θ sin θ] ⋅ [( −2)sin 2 2 n+m n+m θ sin θ) = (2 cos 2 2 n−m n−m θ cos θ) ⋅ ( −2 sin 2 2 n+m n+m = (2 cos θ sin θ) 2 2 m−n m−n θ sin θ) ⋅ (2 sin 2 2 = sin(n + m)θ sin( m − n)θ Substitute n = 2 , m = 3 into the above result,
sin 5θ sin θ = cos 2 2θ − cos 2 3θ ∴ sin 5θ sin θ − sin 5θ = 0 sin 5θ(sin θ − 1) = 0 sin 5θ = 0 or sin θ = 1 π ∴ 5θ = 0, π, 2 π, 3π, 4 π, 5π or θ = 2 π 2 π π 3π 4 π θ = 0, , , , , , π 5 5 2 5 5
Chapter 7 Sums and Products of Trigonometric Functions
8. cos 2θ + cos 4θ + cos 6θ 2 θ + 4θ 2θ − 4θ = 2 cos( ) cos( ) + cos 6θ 2 2 = 2 cos 3θ cos θ + cos 6θ = 2 cos 3θ cos θ + 2 cos 2 3θ − 1 = 2 cos 3θ(cos θ + cos 3θ) − 1 θ + 3θ θ − 3θ = 2 cos 3θ[2 cos( ) cos( )] − 1 2 2 = 4 cos θ cos 2θ cos 3θ − 1
π into the above result, 12 π π π π π π cos + cos + cos = 4 cos cos cos − 1 6 3 2 12 6 4
Substitute θ =
∴
cos
3 + 1 +1 π 3 +3 = 2 2 = = 12 4 ⋅ 3 ⋅ 2 2 6 2 2
6+ 2 4
9. (a) Since α, β are the roots of the equation,
a cos α + b sin α + c = 0 .........(1)
∴
a cos β + b sin β + c = 0 ..........(2)
(1) − (2),
(b) sin A + sin B + sin C A+ B A− B cos = 2 sin + sin[π − ( A + B)] 2 2 A+ B A− B cos = 2 sin + sin( A + B) 2 2 A+ B A− B cos = 2 sin 2 2 A+ B A+ B cos + 2 sin 2 2 A+ B A− B A+ B (cos ) = 2 sin + cos 2 2 2 A+ B A B cos cos = 4 sin 2 2 2 A B C = 4 cos cos cos 2 2 2 (c) From (a),
a+b+c ⋅ sin A sin A + sin B + sin C 2m A A = ⋅ 2 sin cos 2 2 4 cos A2 cos B2 cos C2 A B C = m sin sec sec 2 2 2
a=
a(cos α − cos β) + b(sin α − sin β) = 0
(b) −2 a sin
=0 2 sin
10. (a)
α +β α −β α +β α −β sin + 2 b cos sin 2 2 2 2
cos β cos(2α + β) cos(2α + β) + cos β = cos(2α + β) 2 cos(α + β) cos α = cos(2α + β)
11. (a) 1 + k = 1 +
α−β α+β α+β (b cos − a sin )=0 2 2 2
Q
α ≠ β , and α, β lie within 0 and π,
∴
sin
∴
a sin
Q
a ≠ 0 , cos
∴
tan
α −β ≠0 2 α+β α+β = b cos 2 2 α+β ≠ 0, 2
α+β b = 2 a
a b c = = =k sin A sin B sin C a+b+c ∴ sin A + sin B + sin C k sin A + k sin B + k sin C = sin A + sin B + sin C =k a = sin A
115
(b) ( k + 1) tan(α + β) tan α 2 cos α cos(α + β) = tan(α + β) tan α cos(2α + β) 2 sin(α + β)sin α = cos(2α + β) 2( − 12 )[cos(2α + β) − cos β] = cos(2α + β) cos β − cos(2α + β) = cos(2α + β) cos β = −1 cos(2α + β) = k −1 (c) Put α = k=
π π , β= 4 6 π cos 6
cos(2 ⋅ π4 + π6 )
=− 3
116
Chapter 7 Sums and Products of Trigonometric Functions
From (b), k − 1 = ( k + 1) tan(α + β) tan α π π π − 3 − 1 = ( − 3 + 1) tan( + ) tan 4 6 4 5π − 3 − 1 = tan 12 − 3 + 1 3 +1 = 3 −1 =2+ 3
Enrichment 7 (p.173) 1. (a) (i) a − b = k sin A − k sin B = k (sin A − sin B) cos C2 (ii) cot C = 2 sin C2 = =
∴
cos
cos( π2 − sin( π2 −
A+ B ) 2 A+ B ) 2
sin( A +2 B )
cos( A +2 B )
A+ B C A+ B cot = sin 2 2 2
(b) (i) ( a − b) cot
C 2
C 2 A+ B A+ B A − B sin 2 = 2 k cos sin ⋅ 2 2 cos A +2 B A+ B A− B sin = 2 k sin 2 2 = − k (cos A − cos B) = k (sin A − sin B) ⋅ cot
(ii) Similarly,
A = − k (cos B − cos C ) 2 B (c − a) cot = − k (cos C − cos A) 2 C A ∴ ( a − b) cot + (b − c) cot 2 2 B + (c − a) cot 2 = − k (cos A − cos B) −k (cos B − cos C ) − k (cos C − cos A) =0 (b − c) cot
2. (a) cos A + cos C = 2(1 − cos B) A C ∴ (1 − 2 sin 2 ) + (1 − 2 sin 2 ) 2 2 2 B = 2(1 − 1 + 2 sin ) 2
A C B − 2 sin 2 = 4 sin 2 2 2 2 2 A 2 C 2 B 1 − sin − sin = 2 sin 2 2 2 2 A 2 B 2 C ∴ sin + sin + sin 2 2 2 2 B 2 B = 1 − sin = cos 2 2 (b) cos A + cos C = 2(1 − cos B) A+C A−C B 2 cos cos = 2(1 − 1 + 2 sin 2 ) 2 2 2 2 B = 4 sin 2 A+C 2 π ) = 4 sin ( − 2 2 A+C ) = 4 cos 2 ( 2 A−C A+C ∴ cos = 2 cos 2 2 A C A C A C cos cos + sin sin = 2 cos cos 2 2 2 2 2 2 A C − 2 sin sin 2 2 A C A C 3 sin sin = cos cos 2 2 2 2 A C 1 ∴ tan tan = 2 2 3 C A A C tan 2 + tan 2 (c) (i) cot + cot = 2 2 tan A2 tan C2 2 − 2 sin 2
=
tan
A 2
= 3(tan
(ii) 2 cot
+ tan C2 1 3
A C + tan ) 2 2
B A+C = 2 tan 2 2 2(tan A2 + tan C2 ) = 1 − tan A2 tan C2 =
2(tan A2 + tan C2 )
1 − 13 A C = 3(tan + tan ) 2 2 C A tan 2 + tan 2 = tan A2 tan C2 A C = cot + cot 2 2
3. (a) cos(θ − φ) = cos θ cos φ + sin θ sin φ ............(1) a 2 = (cos θ + cos φ)2 = cos 2 θ + 2 cos θ cos φ + cos 2 φ ............(2)
Chapter 7 Sums and Products of Trigonometric Functions
b 2 = (sin θ + sin φ)2 = sin 2 θ + 2 sin θ sin φ + sin 2 φ ...............(3)
− (sin C + cos C )2 − (sin 2 B + 2 sin B cos B + cos 2 B)
a2 + b2 = cos 2 θ + sin 2 θ + 2(cos θ cos φ + sin θ sin φ) + cos 2 φ + sin 2 φ = 2 + 2(cos θ cos φ + sin θ sin φ) ..................(4) Substitute (1) into (4), a + b = 2 + 2 cos(θ − φ) 1 ∴ cos(θ − φ) = ( a 2 + b 2 − 2) 2 b sin θ + sin φ = a cos θ + cos φ 2
2
θ+φ θ−φ cos 2 2 θ+φ θ−φ 2 cos 2 cos 2
2 sin
= tan
4. (a) (sin A + cos A)2 − (sin B + cos B)2 = (sin 2 A + 2 sin A cos A + cos 2 A)
(2) + (3),
=
117
θ+φ 2
cos θ + cos φ = 2 ..........................(1) (b) sin θ + sin φ = 2 ...........................(2) ∴ a=b= 2 θ+φ From (a), cos(θ − φ) = 1 and tan =1 2 θ + φ π 5π ∴ θ − φ = 0 and = , 2 4 4 π 5π θ = φ and θ + φ = , 2 2 5π When θ + φ = , 2 5π 2θ = 2 5π θ= 4 5π Substitute θ = into (1), 4 5π 2 cos =− 2 ≠ 2 4 5π ∴ is not a solution. 4 π When θ + φ = , 2 π θ= 4 π Substitute θ = into (1) and (2), 4 π π 2 cos = 2 , 2 sin = 2 4 4 π ∴ θ=φ= 4
− (sin 2 C + 2 sin C cos C + cos 2 C ) = (1 + sin 2 A) − (1 + sin 2 B) − (1 + sin 2C ) = (sin 2 A − sin 2 B − sin 2C ) − 1 (b) From (a), (sin A + cos A)2 − (sin B + cos B)2 − (sin C + cos C )2 = (sin 2 A − sin 2 B − sin 2C ) − 1 = 2 sin A cos A − [2 sin( B + C ) cos( B − C )] − 1 = 2 sin A cos A − [2 sin(180° − A) cos( B − C )] − 1 = 2 sin A[cos A − cos( B − C )] − 1 A+ B−C ) = 2 sin A[ −2 sin( 2 A− B+C sin( )] − 1 2 180° − 2C 180° − 2 B = −4 sin A[sin( )sin( )] − 1 2 2 = −4 sin A[sin(90° − C )sin(90° − B)] − 1 = −4 sin A cos B cos C − 1 From (a) and (b), we have (sin 2 A − sin 2 B − sin 2C ) − 1 = −4 sin A cos B cos C − 1 sin 2 A − sin 2 B − sin 2C = −4 sin A cos B cos C .............................(1) Substitute A = 90° into (1),
B + C = 90° 2 B + 2C = 180° sin 2 B + sin 2C = sin(180° − 2C ) + sin 2C = 2 sin 2C π Q 0
5. (a) sin 4 x = 2 sin 2 x cos 2 x = 2(2 sin x cos x ) cos 2 x = 4 sin x cos x cos 2 x
118
Chapter 7 Sums and Products of Trigonometric Functions
(b) Put x = 36° , then
sin 4(36°) = 4 sin 36° cos 36° cos 72°
A+ B A− B sin 2 2 1 A+ B A− B ) = − [cos( + 2 2 2 A+ B A− B − cos( − )] 2 2 1 = − (cos A − cos B) 2 1 = (cos B − cos A) 2
(d) sin
sin 144° 4 sin 36° 1 = 4
cos 36° cos 72° =
(c) From (b), 1 (cos 108° + cos 36°) 2 1 = 4 cos(180° − 72°) + cos 36° = cos 36° − cos 72° 1 = 2
cos 36° cos 72° =
2 A + 3A 2 A − 3A ) cos( ) 2 2 −A 5A = 2 sin cos 2 2 A 5A = 2 sin cos 2 2
2. (a) sin 2 A + sin 3 A = 2 sin(
(d) (cos 36° + cos 72°)2 = (cos 36° − cos 72°)2 + 4 cos 36° cos 72° 1 1 = ( ) 2 + 4( ) 2 4 5 = 4 5 (Q It is positive.) cos 36° + cos 72° = 2 (e) Solve (c) and (d), cos 36° =
5 +1 4
cos 72° =
5 −1 4
Classwork 1 (p.165) A+ B A− B cos 2 2 A+ B A− B A+ B A− B 1 ) + sin( )] = [sin( + − 2 2 2 2 2 1 = (sin A + sin B) 2
1. (a) sin
A+ B A− B sin 2 2 1 A+ B A− B ) = [sin( + 2 2 2 A+ B A− B − sin( − )] 2 2 1 = (sin A − sin B) 2 A+ B A− B (c) cos cos 2 2 1 A+ B A− B ) = [cos( + 2 2 2 A+ B A− B + cos( − )] 2 2 1 = (cos A + cos B) 2 (b) cos
(b) cos(2 A + B) + cos(2 A − B) 2A + B + 2A − B ) = 2 cos( 2 2A + B − 2A + B ) ⋅ cos( 2 = 2 cos 2 A cos B (c)
1 (sin 6 A − sin 2 A) 2 1 6A + 2A 6A − 2A )sin( )] = [2 cos( 2 2 2 = cos 4 A sin 2 A
(d) cos( A − B) − cos( A + B) A− B+ A+ B A− B− A− B = −2 sin( )sin( ) 2 2 = −2 sin A sin( − B) = 2 sin A sin B
Classwork 2 (p.166) 1. (a) 2 sin( 45° + A)sin( 45° − A) 1 = 2{− [cos( 45° + A + 45° − A) 2 − cos( 45° + A − 45° + A)]} = −(cos 90° − cos 2 A) = cos 2 A (b) cos 2 ( A − B) − cos 2 ( A + B) = [cos( A − B) + cos( A + B)] ⋅ [cos( A − B) − cos( A + B)] A− B+ A+ B A− B− A− B = (2 cos cos ) 2 2 A− B+ A+ B A− B− A− B ⋅ ( −2 sin sin ) 2 2 = [2 cos A cos( − B)][−2 sin A sin( − B)] = 4 cos A cos B sin A sin B = sin 2 A sin 2 B
Chapter 7 Sums and Products of Trigonometric Functions
(c) 2 sin 3 A cos A + 2 cos 4 A sin 2 A 1 1 = 2[ (sin 4 A + sin 2 A)] + 2[ (sin 6 A − sin 2 A)] 2 2 = sin 4 A + sin 2 A + sin 6 A − sin 2 A = sin 4 A + sin 6 A = 2 sin 5 A cos( − A) = 2 sin 5 A cos A 2. cos 2 x + cos 3 x = 0 , 0° ≤ x < 360° 5x −x 2 cos cos =0 2 2 5x x 2 cos cos = 0 2 2 5x x cos cos = 0 2 2 5x cos =0 2 5x = 90°, 270°, 450°, 630°, 810° 2 x = 36°, 108°, 180°, 252°, 324° or cos
∴
x = 0 ( 0° ≤ x < 180° ) 2 x = 90° 2 x = 180°
x = 36°, 108°, 180°, 252°, 324°
Classwork 3 (p.170) 1. sin θ + cos θ = m (sin θ + cos θ)2 = m 2 sin 2 θ + 2 sin θ cos θ + cos 2 θ = m 2
1 + sin 2θ = m 2 sin 2θ = m 2 − 1 ........(1) cos 2θ = n ...............(2) (1)2 + (2)2 ,
sin 2 2θ + cos 2 2θ = ( m 2 − 1)2 + n 2 1 = ( m 2 − 1)2 + n 2 n2 = 2m2 − m 4
2. x = cot θ + cos θ ......................(1) y = cot θ − cos θ ......................(2) x+y (1) + (2), = cot θ 2 ( x + y)2 = cot 2 θ 4 4 = tan 2 θ ....................(3) 2 ( x + y)
x−y = cos θ 2 ( x − y)2 = cos 2 θ 4 4 = sec 2 θ ......(4) 2 ( x − y) By (3) and (4), (1) − (2),
sec 2 θ = tan 2 θ + 1 4 4 = +1 2 ( x − y) ( x + y)2 4 ( x + y ) 2 = 4( x − y ) 2 + ( x 2 − y 2 ) 2 16 xy = ( x 2 − y 2 )2
119