f(θ)=[sin^(2π-θ)+sin(π/2+θ)-3]/[2+2cos^2(π+θ)+cos(-θ)] 求f(π/3)
问题描述：
f(θ)=[sin^(2π-θ)+sin(π/2+θ)-3]/[2+2cos^2(π+θ)+cos(-θ)] 求f(π/3)
f(θ)=[sin^2(2π-θ)+sin(π/2+θ)-3]/[2+2cos^2(π+θ)+cos(-θ)] 求f(π/3) 上面是sin^2(2π-θ)
答
f(θ)=[sin^2(θ)+sin(θ)-3]/[2+2cos^2(θ)+cosθ]
=[sin^2(π/3)+sin(π/6)-3]/[2+2cos^2(π/3)+cosπ/3]
= [3/4+1/2-3]/[2+1/4+1/2]
=-7/33