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设函数f(θ)=-1/2+(sin5θ/2)/(2sinθ/2),0

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设函数f(θ)=-1/2+(sin5θ/2)/(2sinθ/2),0
设函数f(θ)=-1/2+(sin5θ/2)/(2sinθ/2),0
1.f(θ)=-1/2+(sin5θ/2)/(2sinθ/2)=-1/2+[sin(2θ)cos(θ/2)+cos(2θ)sin(θ/2)]/[2sin(θ/2)]
=-1/2+cos(2θ)/2+[2sinθcosθcos(θ/2)]/[2sin(θ/2)]
=-1/2+cos(2θ)/2+2cosθ[cos(θ/2)]^2=-1/2+[2(cosθ)^2-1]/2+cosθ(cosθ+1)
=2(cosθ)^2+cosθ-1 (0