【2cos²θ+sin²(2π-θ)+sin(π/2+θ)-3】/【2+2sin²(π/2+θ)-sin(3π/2-θ)】
问题描述：
【2cos²θ+sin²(2π-θ)+sin(π/2+θ)-3】/【2+2sin²(π/2+θ)-sin(3π/2-θ)】
求f（π/3）的值.
答
f(θ)=【2cos²θ+sin²(2π-θ)+sin(π/2+θ)-3】/【2+2sin²(π/2+θ)-sin(3π/2-θ)】
=(2cos²θ+sin²θ+cosθ-3)/(2+2cos²θ+cosθ)
=(cos²θ+cosθ-2)/(2cos²θ+cosθ+2)
f(π/3)=[cos²(π/3)+cos(π/3)-2]/[2cos²(π/3)+cos(π/3)+2]
=(1/4+1/2-2)/(2*1/4+1/2+2)
=(-5/4)/3
=-5/12