f(x)=cos(√3x+Θ)f'(x)=√3sin(√3x+Θ)g(x)=cos(√3x+Θ)+√3sin(√3x+Θ)g(x)=g(-x)，因而cos(√3x+Θ)+√3sin(√3x+Θ)=cos(-√3x+Θ)+√3sin(-√3x+Θ)Θ=π/2
