Question

三角函数问题

求具体过程

Answer 1

解：y=cos(2x+Φ)向右平移π/2得到：y=cos(2x-π+Φ）
y=cos(2x-π+Φ）=cos(π-2x-Φ)=cos[π/2+π/2-2x-Φ)
=cos[π/2-(2x+Φ-π/2)]=sin(2x+Φ-π/2)=sin(2x+π/3)
所以：Φ-π/2=π/3
所以：Φ=5π/6

追问

谢谢

Answer 2

y=cos(2x+φ)向右平移π/2后得到：y=cos[2(x-π/2)+φ]
=cos(2x+φ-π)
=cos[(2x+φ-π/2)-(π/2)]
=sin[2x+φ-(π/2)]
它与y=sin[2x+(π/3)]重合
所以，φ-(π/2)=π/3
则，φ=5π/6