设函数f(θ)=-1/2+(sin5θ/2)/(2sinθ/2),0

设函数f(θ)=-1/2+(sin5θ/2)/(2sinθ/2),0
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

1.f(θ)=-1/2+sin(θ/2+2θ)/(2sinθ/2)=-1/2+(sinθ/2cos2θ+cosθ/2sin2θ)/(2sinθ/2)
=-1/2+(cos2θ)/2+cosθ/2sinθcosθ/(sinθ/2)
=-1/2+(2cosθ-1)/2+2(cosθ/2)^2cosθ
=cosθ-1+2*(1+cosθ)/2*cosθ
=cosθ^2+2...

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1.f(θ)=-1/2+sin(θ/2+2θ)/(2sinθ/2)=-1/2+(sinθ/2cos2θ+cosθ/2sin2θ)/(2sinθ/2)
=-1/2+(cos2θ)/2+cosθ/2sinθcosθ/(sinθ/2)
=-1/2+(2cosθ-1)/2+2(cosθ/2)^2cosθ
=cosθ-1+2*(1+cosθ)/2*cosθ
=cosθ^2+2cosθ
2.y=acosθ+a y=f(θ)联立
cosθ^2+2cosθ-acosθ-a=0
x^2+(2-a)x-a=0在[-1,1]有根
当(2-a)/2<-1 即a>4
1+a-2-a<=0
1+2-a-a>=0 a<=3/2
当-1=<(2-a)/2<=1 即01+a-2-a>=0
1+2-a-a>=0 a<=3/2
当(2-a)/2>1 即a<0
1+a-2-a>=0
1+2-a-a<=0
综合之 0

收起

用公式cos（A+B）和cos（A-B）混合一下就行了