f(θ)=[2cos³θ+sin²(2π-θ)+cos(2π+θ)-3]/[2+2cos²(π+θ)+cos(-θ)]求f(π/3)

题目

f(θ)=[2cos³θ+sin²(2π-θ)+cos(2π+θ)-3]/[2+2cos²(π+θ)+cos(-θ)]求f(π/3)
f(θ)=[2cos³θ+sin²(2π-θ)+cos(2π-θ)-3]/[2+2cos²(π+θ)+cos(-θ)]求f(π/3)

答案

原式=[2cos^3 θ+sin^2 θ+cosθ-3]/[2+2cos^2 θ+cosθ]
所以,f(π/3)=[2*(1/2)^3+(√3/2)^2+(1/2)-3]/[2+2*(1/2)^2+(1/2)]
=[2*(1/8)+(3/4)+(1/2)-3]/[2+2*(1/4)+(1/2)]
=(-3/2)/3
=-1/2