F.4 M2 TRIGONOMETRY

26a)
i)
CD / a = sin 2θ
CD = a sin 2θ
CD = 2 a sin θ cos θ
ii)
CD / b = sin θ
CD = b sin θ
iii)
CD =
2 a sin θ cos θ = b sin θ2 a cos θ = b

cos² θ = b² / (4a²)1 - sin² θ = b² / (4a²)sin² θ = (4a² - b²) / (4a²)sin² θ = (2a - b)(2a + b) / (4a²)sin θ = √[(2a - b)(2a + b)] / (2a)
b)
i)a² : b² = 4 : 7
a : b = 2 : √7By a iii) :
sin θ
= √[(2a - b)(2a + b)] / (2a)
= √[(4 - √7)(4 + √7)] / 4
= √(16 - 7) / 4
= 3/4cos θ
= √(1 - sin² θ) for 0 < θ < π/2
= √(1 - (3/4)²)
= √7 / 4
ii)△ABC
= (1/2) sin(π - 2θ - θ)
= (1/2) sin 3θ
= (1/2) sin (θ + 2θ)
= (1/2) (sin θ cos2θ + cosθ sin2θ)
= (1/2) (sin θ (1 - 2sin²θ) + cosθ (2sinθ cosθ))
= (1/2) (sin θ - 2sin³θ + 2sinθ(1 - sin²θ) )
= (1/2) (3sin θ - 4sin³θ)
= (1/2) (3(3/4) - 4(3/4)³)
= (1/2) (36/16 - 27/16)
= 9/32


2011-07-27 00:08:48 補充：
Corrections :

△ABC
= (1/2) ab sin(π - 2θ - θ)
= (1/2) a (√7/2)a sin(π - 2θ - θ)

..........So the answer should be

= (9/32) (√7/2)a²

= (9√7) a² / 64

Sorry!