问个三角函数转换的问题.

利用二倍角公式
cosx=cos2x/2=cos^2(x/2)-sin^2(x/2)=1-sin^2(x/2)-sin^2(x/2)
=1-2sin^2(x/2)
即2*sin^2(x/2)=(1-cosx)

设x/2=A
1-COS2A = 2 SIN^2 (A)
即COS2A=1- 2 SIN^2 (A) （倍角公式）

cosx=cos(x/2+x/2)=cosx/2*cosx/2-sinx/2*sinx/2=(cosx/2)^2-(sinx/2)^1=1-2(sinx/2)^2
所以1-cosx=1-[1-2(sinx/2)^2]=2(sinx/2)^2

解：1-COSx=1-COS(2*x/2)
=cos²(x/2)+sin²(x/2)-[cos²(x/2)-sin²(x/2)]
=2sin²(x/2)