Question

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

Answer 1

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<=x<=π/2，则-π/3<=2x-π/6<=5π/6。
当2x-π/6<=-π/3、即x=-π/12时，f(x)取得最小值-√3/2。
当2x-π/6<=π/2、即x=π/3时，f(x)取得最小值1。
所以，f(x)的值域为[-√3/2,1]。