已知函数f(x)=cos(2x-π/3）+2sin(x-π/4）sin(x+π/4)=sin(2x-π/6) ,

题目

已知函数f(x)=cos(2x-π/3）+2sin(x-π/4）sin(x+π/4)=sin(2x-π/6) ,
已知函数f(x)=cos(2x-π/3）+2sin(x-π/4）sin(x+π/4)=sin(2x-π/6) 设g（x）=f（x+a）为偶函数,求a的最小正数

答案

f(x)=sin(2x-π/6)=cos[π/2-(2x-π/6)]=cos(2π/3-2x)=cos(2x-2π/3),故f(x)为偶函数,且其关于x=π/3对称,周期为π,因此a的最小整数为π.