若θ,a为锐角且tanθ=(sina-cosa)/(sina+cosa),求证:sina-cosa=√2sinθ.
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![若θ,a为锐角且tanθ=(sina-cosa)/(sina+cosa),求证:sina-cosa=√2sinθ.](/uploads/image/z/2497628-20-8.jpg?t=%E8%8B%A5%CE%B8%2Ca%E4%B8%BA%E9%94%90%E8%A7%92%E4%B8%94tan%CE%B8%3D%28sina-cosa%29%2F%28sina%2Bcosa%29%2C%E6%B1%82%E8%AF%81%EF%BC%9Asina-cosa%3D%E2%88%9A2sin%CE%B8.)
若θ,a为锐角且tanθ=(sina-cosa)/(sina+cosa),求证:sina-cosa=√2sinθ.
若θ,a为锐角且tanθ=(sina-cosa)/(sina+cosa),求证:sina-cosa=√2sinθ.
若θ,a为锐角且tanθ=(sina-cosa)/(sina+cosa),求证:sina-cosa=√2sinθ.
因为tanb=(sina-cosa)/(sina+cosa),所以sinb/cosb=(sina-cosa)/(sina+cosa),
所以sinbsina+sinbcosa=sinacosb-cosacosb,所以
cosacosb+sinacosb=sinacosb-cosasinb,即sin(a-b)=cos(a-b),所以tan(a-b)=1,
所以a-b=兀/4+k兀,所以b=a-兀/4-k兀,
所以根号2sinb=(根号2)sin(a-兀/4-k兀)=(根号2)sin(a-兀/4)=
(根号2)[sinacos45-cosasin45]=sina-cosa,得证
解】由结果入手,sinα-cosα=(2sinθ)^1/2
平方得:
(sinα-cosα)^2=2sinθ
<==> 1-sin2α=2sinθ———(1)
<==> sin2α=1-2sinθ———(2)
tanθ=(sinα-cosα)/(sinα+cosα)
==> (tanθ)^2=(1-sin2α)/(1+sin2α)代入(1...
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解】由结果入手,sinα-cosα=(2sinθ)^1/2
平方得:
(sinα-cosα)^2=2sinθ
<==> 1-sin2α=2sinθ———(1)
<==> sin2α=1-2sinθ———(2)
tanθ=(sinα-cosα)/(sinα+cosα)
==> (tanθ)^2=(1-sin2α)/(1+sin2α)代入(1)(2)
==> (tanθ)^2=sinθ/(1-sinθ) 切化弦
==> sinθ/(cosθ)^2= 1/(1-sinθ)得到:
==>(sinθ)^2+(cosθ)^2=1 恒成立;
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