sinθ+sin2θ/1+cosθ+cos2θ=
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sinθ+sin2θ/1+cosθ+cos2θ=
![sinθ+sin2θ/1+cosθ+cos2θ=](/uploads/image/z/1563904-64-4.jpg?t=sin%CE%B8%2Bsin2%CE%B8%2F1%2Bcos%CE%B8%2Bcos2%CE%B8%3D)
(sinθ+sin2θ)/(1+cosθ+cos2θ)
=(sinθ+sin2θ)/(1+cosθ+2cos²θ-1)
=(sinθ+2sinθcosθ)/(cosθ+2cos²θ)
=sinθ(1+2cosθ)/[cosθ(1+2cosθ)]
=sinθ/cosθ
=tanθ
=(sinθ+sin2θ)/(1+cosθ+2cos²θ-1)
=(sinθ+2sinθcosθ)/(cosθ+2cos²θ)
=sinθ(1+2cosθ)/[cosθ(1+2cosθ)]
=sinθ/cosθ
=tanθ
sinθ+sin2θ/1+cosθ+cos2θ=
:求证:(1-2sinθcosθ)/(cos2θ-sin2θ)=(cos2θ-sin2θ)/(1+2sinθcosθ).
已知sinθ-2cosθ=0,求sin2θ-cos2θ/1=sin2θ
sin2θ+sinθ/2cos2θ+2sin^θ+cosθ=tanθ 数学题
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