Question

已知f(x)=sin(x+θ）+√3cos(x-θ)为偶函数，则θ=__________。

Answer 1

函数是偶函数，f(-x)=f(x)
sin(-x+θ)+√3cos(-x-θ)=sin(x+θ)+√3cos(x-θ)
-sin(x-θ)+√3cos(x+θ)=sin(x+θ)+√3cos(x-θ)
sin(x-θ)+√3cos(x-θ)=sin(x+θ)-√3cos(x+θ)
sin(x-θ+π/3)=sin(x+θ-π/3)
sin(x-θ+π/3)-sin(x+θ-π/3)=0
2cos[(x-θ+π/3+x+θ-π/3)/2]sin[(x-θ+π/3-x-θ+π/3)/2]=0
cosx sin(π/3 -θ)=0
π/3 -θ=-kπ (k∈Z)
θ=kπ +π/3 (k∈Z)

Answer 2

f(x)=sin(x+θ）+√3cos(x-θ)为偶函数
x=0时，函数取得最大值
即
f(0)=sinθ+√3cosθ
＝2（1/2sinθ+√3/2cosθ)
=2sin(θ+60°）
故
θ＝30°

Answer 3

f(x)=sin(x+θ)+根号3cos(x+θ)
=2[sinπ/6sin(x+θ)+cosπ/6*cos(x+θ)]
=2cos(x+θ-π/6)
为偶函数
θ-π/6=kπ,k∈Z
θ=π/6+kπ,k∈

Answer 4

已知f(x)=sin(x+θ）+√3cos(x-θ)为偶函数，则θ=__________。
解析：∵f(x)=sin(x+θ)+√3cos(x-θ)为偶函数
∴x=0时，函数取得最大值
即f(0)=sinθ+√3cosθ
令f(0)=sinθ+√3cosθ=1
sinθ+√3cosθ＝2(1/2sinθ+√3/2cosθ)=2sin(θ+π/3)=1
∴sin(θ+π/3)=1/2==>θ+π/3=π/6
∴θ=-π/6

验证：f(x)=sin(x-π/6)+√3cos(x+π/6)
=sinx√3/2-cosx*1/2+cosx*3/2-sinx*√3/2
=cosx
为偶函数