【2cos²θ+sin²(2π-θ)+sin(π/2+θ)-3】/【2+2sin²(π/2
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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.
求f(π/3)的值.
![【2cos²θ+sin²(2π-θ)+sin(π/2+θ)-3】/【2+2sin²(π/2](/uploads/image/z/18557092-28-2.jpg?t=%E3%80%902cos%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%E3%80%91%2F%E3%80%902%2B2sin%26%23178%3B%28%CF%80%2F2)
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
=(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
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