急.设f(θ)=[2cos2θ+sin2(2π-θ)+sin(π/2+θ)-3]/[2+2cos2(π+θ)+cos(-θ)]求f_数学解答

急.设f(θ)=[2cos2θ+sin2(2π-θ)+sin(π/2+θ)-3]/[2+2cos2(π+θ)+cos(-θ)]求f(π/3)
设f(θ)=[2cos2θ+sin2(2π-θ)+sin(π/2+θ)-3]/[2+2cos2(π+θ)+cos(-θ)]
求f(π/3)

人气：156 ℃　时间：2020-03-20 16:06:24

解答

f(θ)=[2cos2θ+sin2(2π-θ)+sin(π/2+θ)-3]/[2+2cos2(π+θ)+cos(-θ)]
=[2cos2θ-sin2θ+cosθ-3]/[2+2cos2θ+cosθ]
故f(π/3)=[2cos2π/3-sin2π/3+cosπ/3-3]/[2+2cos2π/3+cosπ/3]
=[-1-根号3/2+1/2-3]/[2-1+1/2]
=[-根号3-7]/3