2cos²θ+sin²(2π-θ)+sin(π/2+θ)-3/2+2sin²(π/2+θ)-sin(3π/2-θ),求f（π/3）的值

2cos²θ+sin²(2π-θ)+sin(π/2+θ)-3/2+2sin²(π/2+θ)-sin(3π/2-θ),求f（π/3）的值
2cos²θ+sin²(2π-θ)+sin(π/2+θ)-3/2+2sin²(π/2+θ)-sin(3π/2-θ),求f（π/3）的值

2cos²θ+sin²(2π-θ)+sin(π/2+θ)-3/2+2sin²(π/2+θ)-sin(3π/2-θ),求f（π/3）的值
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
=3cos²θ - 1/2
先化简到这里,不知道题目中上式与函数值f(π/3)是什么关系哈?

2cos²θ=cos2θ+1；sin²(2π-θ)=（-sinθ）²=sin²θ；2sin²(π/2+θ)=2cos²θ=cos2θ+1；sin(3π/2-θ)=-cosθ。原式=1.5cos2θ+1，所以答案为0.25