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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
人气:390 ℃ 时间:2020-02-05 07:05:49
解答
(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+...
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