设f(x)=sin(x+θ)+cos(x-θ)为偶函数,-π/2〈θ〈0,则θ=

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设f(x)=sin(x+θ)+cos(x-θ)为偶函数,-π/2〈θ〈0,则θ=设f(x)=sin(x+θ)+cos(x-θ)为偶函数,-π/2〈θ〈0,则θ=设f(x)=sin(x+θ)+cos(x

设f(x)=sin(x+θ)+cos(x-θ)为偶函数,-π/2〈θ〈0,则θ=
设f(x)=sin(x+θ)+cos(x-θ)为偶函数,-π/2〈θ〈0,则θ=

设f(x)=sin(x+θ)+cos(x-θ)为偶函数,-π/2〈θ〈0,则θ=
因为f(x)=sin(x+θ)+cos(x-θ)为偶函数
所以f(-x)=sin(-x+θ)+cos(-x-θ)
=sin(-x)cosθ+cos(-x)sinθ+cosxcosθ-sinxsinθ
=cosθ(cosx-sinx)+sinθ(cosx-sinx)
=(cosx-sinx)(cosθ+sinθ)
f(x)=sin(x+θ)+cos(x-θ)
=sinxcosθ+cosxsinθ+cosxcosθ+sinxsinθ
=cosθ(sinx+cosx)+sinθ(sinx+cosx)
=(sinx+cosx)(sinθ+cosθ)
因为f(-x)=f(x)
所以对于任意的x有(cosx-sinx)(cosθ+sinθ)=(sinx+cosx)(sinθ+cosθ)
所以cosθ+sinθ=0
因为-π/2〈θ〈0
所以得:θ=-π/4