已知函数f（x）=sin^2x+2√3sin（x+π/4）cos(x-π/4求答案.,
问题描述：
已知函数f（x）=sin^2x+2√3sin（x+π/4）cos(x-π/4求答案.,
已知函数f（x）=sin^2x+2√3sin（x+π/4）cos(x-π/4)-cos^x-√3.⑴求函数f（x）的最小正周期和单调递减区间⑵求f（x）在（-π/12,5π/12]上的值域·
答
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...