2cos²θ+sin²(2π-θ)+sin(π/2+θ)-3/2+2sin²(π/2+θ)-sin(3π/2-θ),求f(π/3)的值
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![2cos²θ+sin²(2π-θ)+sin(π/2+θ)-3/2+2sin²(π/2+θ)-sin(3π/2-θ),求f(π/3)的值](/uploads/image/z/2729709-45-9.jpg?t=2cos%26%23178%3B%CE%B8%2Bsin%26%23178%3B%282%CF%80-%CE%B8%29%2Bsin%28%CF%80%2F2%2B%CE%B8%29-3%2F2%2B2sin%26%23178%3B%28%CF%80%2F2%2B%CE%B8%29-sin%283%CF%80%2F2-%CE%B8%29%2C%E6%B1%82f%EF%BC%88%CF%80%2F3%EF%BC%89%E7%9A%84%E5%80%BC)
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