已知tan^2α=2tan^β+1 求证:sin^2β=asin^2α-1快

发布时间:2021-02-25 06:20:47

已知tan^2α=2tan^β+1 求证:sin^2β=asin^2α-1快

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题目有误已知tan^2α=2tan^2β+1 求证:sin^2 β=2sin^2α-1
tan^2α=2tan^2β+1
sin^2α/cos^2α=2sin^2β/cos^2β+1
sin^2α/cos^2α=(2sin^2β+cos^2β)/cos^2β
sin^2α/cos^2α=(sin^2β+1)/cos^2β
sin^2α/(1-sin^2α)=(sin^2β+1)/(1-sin^2β)
sin^2α-sin^2α*sin^2β=sin^2β-sin^2α*sin^2β-sin^2α+1
2sin^2α=sin^2β+1
sin^2β=2sin^2α-1
======以下答案可供参考======
供参考答案1:
1楼说的对 你的题目有误!按照1楼的题目,我的证明如下:
tan^2α+1=2(tan^2β+1)
1/cos^2α=2/cos^2β
cos^2β=2cos^2α
1-sin^2β=2(1-sin^2α)
即sin^2 β=2sin^2α-1
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