如何根据2[sin(X/2)]^2得到2[sin(X/2)]^2=1-cosX如何化简

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如何根据2[sin(X/2)]^2得到2[sin(X/2)]^2=1-cosX如何化简
如何根据2[sin(X/2)]^2得到2[sin(X/2)]^2=1-cosX
如何化简

如何根据2[sin(X/2)]^2得到2[sin(X/2)]^2=1-cosX如何化简
二倍角公式:
cos(2α)=(cosα)^2 -(sinα)^2 =(cosα)^2 -[1 -(cosα)^2] =2(cosα)^2 -1=[1 -(sinα)^2] -(sinα)^2=1 - 2(sinα)^2
2(cosα)^2 = 1 + cos(2α)
2(sinα)^2 = 1 - cos(2α)
当 α = x/2 时就可以得到半角公式:
2[cos(x/2)]^2 = 1 + cosx
2[sin(x/2)]^2 = 1 - cosx