求证(2cosθ-sin2θ)/(2cosθ+sin2θ)=tan(π/4-θ/2)
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求证(2cosθ-sin2θ)/(2cosθ+sin2θ)=tan(π/4-θ/2)
右边应该是tan(π/4-θ/2)^2
右边应该是tan(π/4-θ/2)^2
(2cosθ-sin2θ)/(2cosθ+sin2θ)=2cosθ(1-sinθ)/[2cosθ(1+sinθ)]=(1-sinθ)/(1+sinθ)
tan(π/4-θ/2)=sin(π/4-θ/2)/cos(π/4-θ/2)=(cosθ/2-sinθ/2)/(sinθ/2+cosθ/2)
=根号(cosθ/2-sinθ/2)^2/(sinθ/2+cosθ/2)^2=根号(1-sinθ)/(1+sinθ)
所以(2cosθ-sin2θ)/(2cosθ+sin2θ)=[tan(π/4-θ/2)]^2
tan(π/4-θ/2)=sin(π/4-θ/2)/cos(π/4-θ/2)=(cosθ/2-sinθ/2)/(sinθ/2+cosθ/2)
=根号(cosθ/2-sinθ/2)^2/(sinθ/2+cosθ/2)^2=根号(1-sinθ)/(1+sinθ)
所以(2cosθ-sin2θ)/(2cosθ+sin2θ)=[tan(π/4-θ/2)]^2
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