Question

急急！！！已知函数f（x）=sin^2x+2√3sin（x+π/4）cos(x-π/4求答案。，过程

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·

Answer 1

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<2x-π/6<(2k+3/2)π,k∈Z确定，
各加π/6，(2k+2/3)π<2x<(2k+5/3)π,
各除以2，（k+1/3)π<x<(k+5/6)π.
(2)x∈(-π/12,5π/12],
∴u=2x-π/6的值域是（-π/3,2π/3],
∴f(x)=2sinu的值域是(-√3，2].