设f(x)=【2sin^3θ+sin^2(2π-θ)+sin(π/2+θ)-3】/【2+2sin^2(π/2+θ)-sin(3π/2-θ)】,求f(π/3)的值

题目

设f(x)=【2sin^3θ+sin^2(2π-θ)+sin(π/2+θ)-3】/【2+2sin^2(π/2+θ)-sin(3π/2-θ)】,求f(π/3)的值
快啊,坐等,急!要步骤的.

答案

sin(π/2+θ)=sinπ/2cosθ+cosπ/2sinθ=cosθ
sin(3π/2-θ)=cos(π-θ)=-cosθ
sinπ/3=√3/2 sin²π/3=3/4 sin³π/3=3√3/8
cosπ/3=1/2 cos²π/3=1/4
f(θ)
=[2sin³θ+sin²(2π-θ)+sin(π/2+θ)-3]/[2+2sin²(π/2+θ)-sin(3π/2-θ)]
=[2sin³θ+sin²θ+cosθ-3]/[2+2cos²θ+cosθ]
f(π/3)
=[2*(3√3/8)+3/4+1/2-3]/[2+2*(1/4)+1/2]
=(3√3/4-7/4)/3
=√3/4-7/12