设f(θ)=(2cos^3θ+sin^2(2π-θ)+sin(π/2+θ)-3)/(2-2cos^2(π+θ)+cos(-θ)) 求f(π/3)的值
问题描述:
设f(θ)=(2cos^3θ+sin^2(2π-θ)+sin(π/2+θ)-3)/(2-2cos^2(π+θ)+cos(-θ)) 求f(π/3)的值
答
f(θ)=(2cos^3 θ+sin^2 θ+cosθ-3)/(2-2cos^2 θ+cosθ)
f(π/3)=[2*(1/2)^3+(√3/2)^2+1/2-3]/[2-2*(1/2)^2+1/2]
=[1/4+3/4+1/2-3]/[2-1/2+1/2]
=-3/4.