f（x）=(sinx)^2+2√3sin（x+π/4）cos(x-π/4)-(cosx)^2-√3
=2√3[sin(x+π/4)]^2-cos2x-√3
=√3[1-cos(2x+π/2)]-cos2x-√3
=√3sin2x-cos2x
=2sin(2x-π/6),
(1)f(x)的最小正周期=π.
递减区间由（2k+1/2)π
已知函数f（x）=sin^2x+2√3sin（x+π/4）cos(x-π/4求答案.,
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