y=sinx/2*sin(π/2-x/2)的最小正周期.麻烦细节.

网友回答

y=sin(x/2)*cos(x/2)
=1/2*2sin(x/2)*cos(x/2)
=1/2*sinx
所以最小正周期=2π
======以下答案可供参考======
供参考答案1:
y=sin(x/2)*sin(π/2-x/2)
=sin(x/2)cos(x/2)
=(1/2)sinx
最小正周期=2π