设f(θ)=[sin2(6π+θ)+cosθ-2cos3(3π+θ)-3]/2+2cos2(θ-4π)-cos(-θ),求f(π/3)
问题描述:
设f(θ)=[sin2(6π+θ)+cosθ-2cos3(3π+θ)-3]/2+2cos2(θ-4π)-cos(-θ),求f(π/3)
答
化简f(θ)=[sin2(6π+θ)+cosθ-2cos3(3π+θ)-3]/2+2cos2(θ-4π)-cos(-θ)=[sin(12π+2θ)+cosθ-2cos(9π+3θ)-3]/2+2cos(2θ-8π)-cos(θ)=[sin2θ+cosθ-(-2cos3θ)-3]/2+2cos2θ-cos(θ)=(1/2)sin2θ+(1/...