已知函数y=cos(2x+φ)-√3sin(2x+φ)为偶函数,则φ的值为 φ的绝对值＜派/2
问题描述：
已知函数y=cos(2x+φ)-√3sin(2x+φ)为偶函数,则φ的值为 φ的绝对值＜派/2
答
y=cos(2x+φ)-√3sin(2x+φ)
=2[1/2cos(2x+φ)-√3/2sin(2x+φ)]
=2cos(2x+φ+π/3)
2cos(2x+φ+π/3)=2cos(-2x+φ+π/3)
cos(2x+φ+π/3)-cos(-2x+φ+π/3)=0
-2sin(φ+π/3)sin(2x)=0
sin(φ+π/3)=0
φ+π/3=kπ
φ=kπ-π/3
∵|φ|2cos(2x+φ+π/3)=2cos(-2x+φ+π/3)这个为什么会等于这个啊还有cos(2x+φ+π/3)-cos(-2x+φ+π/3)=0-2sin(φ+π/3)sin(2x)=0这两步是怎么变换的 解释一下好不好2cos(2x+φ+π/3)=2cos(-2x+φ+π/3) 因为是偶函数，f(-x)=f(x) cos(2x+φ+π/3)-cos(-2x+φ+π/3)=0根据和差化积公式：cosa-cosb=-2sin[(a+b)/2]sin[(a-b)/2]