因为(x+π/4)-(x-π/4)=π/2,所以x+π/4=π/2+(x-π/4),
所以sin(x+π/4)=sin[π/2-(x-π/4)]=cos(x-π/4)
所以sin平方(x+π/4)- sin平方(x-π/4)=cos^2(x-π/4)-sin^2(x-π/4)=cos2[(x-π/4)]=cos(2x-π/2)=sin2x
所以最小正周期2π/2=π