已知函数f(x)=sin(x+θ)+√3cos(x-θ)为偶函数,θ∈(0,π),求θ的值

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已知函数f(x)=sin(x+θ)+√3cos(x-θ)为偶函数,θ∈(0,π),求θ的值
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已知函数f(x)=sin(x+θ)+√3cos(x-θ)为偶函数,θ∈(0,π),求θ的值
已知函数f(x)=sin(x+θ)+√3cos(x-θ)为偶函数,θ∈(0,π),求θ的值

已知函数f(x)=sin(x+θ)+√3cos(x-θ)为偶函数,θ∈(0,π),求θ的值
f(x)为偶函数,则
f(x)-f(-x)≡0 (即恒等于0)
==> sin(x+θ)+√3cos(x-θ)+sin(x-θ)-√3cos(x+θ)≡0
==> sin(x+θ-π/3)+sin(x-θ+π/3)≡0
==> 2sinxcos(θ-π/3)≡0
==> cos(θ-π/3)=0
==> θ=kπ+π/2+π/3(k∈Z)
又因θ∈(0,π),所以必须k=0,从而θ=π/2+π/3=5π/6

f(x)=sinxcosθ+cosxsinθ+√3cosxcosθ+√3sinxsinθ
=sinx(cosθ+√3sinθ)+cosx(sinθ+√3cosθ)
要消去sinx,f(x)才为偶函数
∴cosθ+√3sinθ=0
2(cosθcos60°+sinθsin60°)=0
cos(θ-60°)=0
θ-60°=(180k+90)°
θ∈(0,π),
∴θ=150°