函数f(x)=sin^2(2x-派/4)的最小正周期是什么?

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π/4======以下答案可供参考======
供参考答案1:
f(x)=[sin(2x-π/4)]^2
=[1-cos(4x-π/2)]/2
=1/2-(1/2)sin4x
T=(2π)/w=2π/4=π/2
所以函数f(x)=[sin(2x-π/4)]^2的最小正周期是 π/2
供参考答案2:
f(x)=sin^2(2x-π/4)=(1/2)*[1-2cos(4x-π/2)]
=1/2-(1/2)*cos(π/2-4x)
=(1/2)-(1/2)sin4x
最小正周期=2π/4=π/2