已知函数f(x)=cos(2x-π/3)+2sin(x-π/4)sin(x+/4),(1）求函数的最小正周期和对称轴方程

题目

已知函数f(x)=cos(2x-π/3)+2sin(x-π/4)sin(x+/4),(1）求函数的最小正周期和对称轴方程
（2）求函数在区间【-/12,π/2】的值域

答案

f(x)=cos(2x-π/3)+2sin(x-π/4)sin(x+π/4)
=cos2xcosπ/3+sin2xsinπ/3+2(sinxcosπ/4-cosxsinπ/4)(sinxcosπ/4+cosxsinπ/4)
=(1/2)cos2x+(√3/2)sin2x+2[(√2/2)sinx-(√2/2)cosx][(√2/2)sinx+(√2/2)cosx]
=(1/2)cos2x+(√3/2)sin2x+(sinx)^2-(cosx)^2
=(1/2)cos2x+(√3/2)sin2x-cos2x
=(√3/2)sin2x-(1/2)cos2x
=sin(2x-π/6)
(1)最小正周期为T=2π/2=π.
2x-π/6=kπ+π/2,则对称轴方程为x=kπ/2+π/3,k为整数.
(2)-π/12