f(x)=sinx·cos(x-π/2)的最小正周期

问题描述:

f(x)=sinx·cos(x-π/2)的最小正周期

f(x)=sinx·cos(x-π/2)
=sinx·cos(π/2-x)
=sinx·sinx
=sin²x
=(1-cos2x)/2
所以T=2π/2=π