函数f(x)=sin(x+π/6)+sin(x+π/3)在哪个区间上单调递增

题目

函数f(x)=sin(x+π/6)+sin(x+π/3)在哪个区间上单调递增
A （-π/2,π/12） B （-π/3,π/12） C（π/2,π） D （-2/π,π）

答案

∵ y=sin(x+π/6)+sin(x+π/3)
=sinxcos(π/6)+cosxsin(π/6)+sinxcos(π/3)+cosxsin(π/3)
=(1/2+√3/2)sinx+(1/2+√3/2)cosx
=(1/2+√3/2)*√2[sinx*(√2/2)+cosx*(√2/2)]
=(1/2+√3/2)*√2[sinx*cos(π/4)+cosx*sin(π/4)]
=(1/2+√3/2)*√2sin(x+π/4)
∵ y=sinx在（-π/2,π/2）上递增
∴ y=sin(x+π/4)在(-3π/4,π/4)上递增
参考选项,选 B