f（x）=1/2*sin2x*(sin2x+cos2x)=1/2*sin²2x+1/2*sin2x *cos2x=1/2*[(1-cos4x)/2]+1/4sin4x=1/4*(sin4x-cos4x+1)=1/4*[√2 sin(4x-π/4) +1]=√2/4* sin(4x-π/4) +1/4,所以函数的最小正周期是2π/4=π/2....