Am-sln-07(e)

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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

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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

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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