设f( θ)=2cos^3θ + sin^2 (2π-θ)+cos(-θ)-3 / 2+2cos^2(π+θ) + cos(2π-θ),则 f(π/3) =

设f( θ)=2cos^3θ + sin^2 (2π-θ)+cos(-θ)-3 / 2+2cos^2(π+θ) + cos(2π-θ),则 f(π/3) =
设f( θ)=2cos^3θ + sin^2 (2π-θ)+cos(-θ)-3 / 2+2cos^2(π+θ) + cos(2π-θ),则 f(π/3) =

设f( θ)=2cos^3θ + sin^2 (2π-θ)+cos(-θ)-3 / 2+2cos^2(π+θ) + cos(2π-θ),则 f(π/3) =
f( θ)=2cos^3θ + sin^2 (2π-θ)+cos(-θ)-3 / 2+2cos^2(π+θ) + cos(2π-θ),
=2cos³θ +sin²θ +cosθ -3 /2+2cos²θ +cosθ
f(π/3) =[2(1/2)³+3/4+1/2-3]/[2+2(1/2)²+1/2]
=(-3/2)/(3)
=-1/2