已知函数f(x)=sin(x+θ)+根号3cos(x-θ) (θ∈[-π/2,π、2])是偶函数,则θ的值为

题目

已知函数f(x)=sin(x+θ)+根号3cos(x-θ) (θ∈[-π/2,π、2])是偶函数,则θ的值为

答案

f(x)是偶函数.
则 f(π/2)=f(-π/2)
即 sin(π/2+θ)+根号3cos(π/2-θ) = sin(-π/2+θ)+根号3cos(-π/2-θ)
即 cosθ+√3sinθ=-cosθ-√3sinθ
∴ cosθ+√3sinθ=0
即 2sin(θ+π/6)=0
∴ θ+π/6=kπ
即 θ=kπ-π/6
根据条件,则θ=-π/6
此时f(x)=sin(x-π/6)+√3cos(x+π/6)
=sinxcos(π/6)-cosx*sin(π/6)+√3cosxcos(π/6)-√3sinxsin(π/6)
=(√3/2)sinx-(1/2)cosx+(3/2)cosx-(√3/2)sinx
=cosx
是偶函数
∴ θ=-π/6