若а,θ为锐角,且tanθ=(sinа-cosа)/(sinа+cosа),求证:sinа-cosа=√2sinθ
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![若а,θ为锐角,且tanθ=(sinа-cosа)/(sinа+cosа),求证:sinа-cosа=√2sinθ](/uploads/image/z/8841389-5-9.jpg?t=%E8%8B%A5%D0%B0%2C%CE%B8%E4%B8%BA%E9%94%90%E8%A7%92%2C%E4%B8%94tan%CE%B8%3D%28sin%D0%B0-cos%D0%B0%29%2F%28sin%D0%B0%2Bcos%D0%B0%29%2C%E6%B1%82%E8%AF%81%3Asin%D0%B0-cos%D0%B0%3D%E2%88%9A2sin%CE%B8)
若а,θ为锐角,且tanθ=(sinа-cosа)/(sinа+cosа),求证:sinа-cosа=√2sinθ
若а,θ为锐角,且tanθ=(sinа-cosа)/(sinа+cosа),求证:sinа-cosа=√2sinθ
若а,θ为锐角,且tanθ=(sinа-cosа)/(sinа+cosа),求证:sinа-cosа=√2sinθ
【解】由结果入手,两路夹击,
sinα-cosα=(2sinβ)^1/2
平方得:
(sinα-cosα)^2=2sinβ
1-sin2α=2sinβ———(1)
sin2α=1-2sinβ———(2)
tanβ=(sinα-cosα)/(sinα+cosα)
==> (tanβ)^2=(1-sin2α)/(1+sin2α)代入(1)(2)
==> (tanβ)^2=sinβ/(1-sinβ) 切化弦
==> sinβ/(cosβ)^2= 1/(1-sinβ)得到:
==>(sinβ)^2+(cosβ)^2=1 恒成立;
得证
sinθ/cosθ=(sinа-cosа)/(sinа+cosа),设t(sinθ)=sinа-cosа,则tcosθ=sinа+cosа,于是两方程等号两边同时平方相加,有t^2=2,所以t==√2