※ 引述《inch01742 (穎奇種筍子)》之銘言：
: http://i.imgur.com/ZdRX9qT.jpg
: 想請教第二小題該如何下手
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: Sent from JPTT on my Samsung SM-N9208.
(1)
f(-x) = (k + 4[sin(x)]^2)cos(-2x + θ)
要求 = -f(x)
=> cos(-2x + θ) = -cos(2x + θ)
=> θ = -3π/2
或者
cos(2x)cosθ + sin(2x)sinθ = -cos(2x)cosθ + sin(2x)sinθ
=> cosθ = 0
=> θ = -3π/2
所以f(x) = -(k + 4[sin(x)]^2)sin(2x)
f(3π/4) = 0 => 0 = (k + 2) = 0
=> k = -2
(2)
f(α/8) = 3/5
= -(-2 + 4[sin(α/8)]^2)sin(α/4)
= (2 - 2 + 2cos(α/4))sin(α/4)
= sin(α/2)
=> cos(α/2) = 4/5
=> sin(α) = 2 * 4/5 * 3/5 = 24/25
cos(α) = 7/25
=> sin(α - π/6) = 24/25 * (1/2)√3 - (7/25) * 1/2
= (12/25)√3 - 7/50
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