已知函数f(x)＝√3sin(2x-π/6)+2sin^2(x-π/12) (1)求f(x)的最小正

题目

已知函数f(x)＝√3sin(2x-π/6)+2sin^2(x-π/12) (1)求f(x)的最小正
已知函数f(x)＝√3sin(2x-π/6)+2sin^2(x-π/12)
(1)求f(x)的最小正周期
(2)求f(x)的最值

答案

f(x)=√3sin(2x-π/6)+2sin^2(x-π/12)
=√3sin(2x-π/6)-cos(2x-π/6)+1
=2sin[(2x-π/6)-π/6]+1
=2sin(2x-π/3)+1
(1)最小正周期为T=2π/2=π.
(2)f(x)的最小值是-2+1=-1、最大值是2+1=3.