已知函数f(x)=cos^2(x-π/6)-1/2cos2x,求函数f(x)的最小正周期

(1) f(x)=cos^2(x-π/6)-1/2cos2x
=[cos(2x-π/3)+1]/2-1/2cos2x
=1/2[1/2cos2x-√3/2sin2x]-1/2cos2x
=-1/2(1/2cos2x+√3/2sin2x)
=-1/2sin(2x+π/6)
所以,最小正周期为：T=π；
因为sin2x的对称轴为：x=π/4+kπ/2
所以f(x)图像对称轴的方程为：x=(π/4-π/6)+kπ/2=π/12+kπ/2
(2) 因为 -π/12≤x≦π/12
所以-π/6≤2x≤π/6
所以0≦2x+π/6≦π/3
由正弦函数图像可知,f(x)的值域为：[0,√3/2]