Question

已知θ∈（π，3π/2）sin^2θ-（√15-√5）sinθcosθ-5√3cos^2θ=0

（1）求cosθ
（2）若f（x）=（4√15）/5sinθcos^2x-4√3cosθsinxcosx+1/2，求f（x）的最小正周期及单调递减区间

Answer 1

(1)
sin^2θ-（√15-√5）sinθcosθ-5√3cos^2θ=0
两边同时除以cos²θ
tan²θ-(√15-√5)tanθ-5√3=0
∴tanθ=√15,tanθ=-√5
∵θ∈（π，3π/2）∴tanθ>0
∴tanθ=√15
∴sinθ=√15cosθ代入sin²θ+cos²θ=1
∴ 16cos²θ=1,cos²θ=1/16
∵cosθ<0 ∴cosθ=-1/4
(2)
sinθ=-√15/4
f（x）=（4√15）/5sinθcos^2x-4√3cosθsinxcosx+1/2
=-cos²x+√3sinxcosx+1/2
=√3/2sin2x-1/2cos2x
=sin(2x-π/6)
f(x)最小正周期T=2π/2=π
由2kπ+π/2≤2x-π/6≤2kπ+3π/2,k∈Z
得 2kπ+2π/3≤2x≤2kπ+5π/3,k∈Z
∴单调递减区间是[kπ+π/3,kπ+5π/6],k∈Z