f(x)=sin(2x-π/4)-2倍根号2sin方x,的最小正周期是多少,怎么求

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f(x)=sin(2x-π/4)-2倍根号2sin方x,的最小正周期是多少,怎么求
f(x)=sin(2x-π/4)-2倍根号2sin方x,的最小正周期是多少,怎么求

f(x)=sin(2x-π/4)-2倍根号2sin方x,的最小正周期是多少,怎么求
f(x)=sin(2x-π/4)-2√2sin²x
=sin2xcosπ/4-cos2xsinπ/4-√2(1-cos2x)
=√2/2 sin2x- √2/2 cos2x +√2cos2x-√2
=√2/2 sin2x+√2/2 cos2x-√2
=sin(2x+π/4)-√2
∴最小正周期是2π/2=π