已知α,β均为锐角,且满足3sin^2α+2sin^2β=1,3sin2α-2sin2β=0 求证:α+2β=π/2
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![已知α,β均为锐角,且满足3sin^2α+2sin^2β=1,3sin2α-2sin2β=0 求证:α+2β=π/2](/uploads/image/z/8841450-66-0.jpg?t=%E5%B7%B2%E7%9F%A5%CE%B1%2C%CE%B2%E5%9D%87%E4%B8%BA%E9%94%90%E8%A7%92%2C%E4%B8%94%E6%BB%A1%E8%B6%B33sin%5E2%CE%B1%2B2sin%5E2%CE%B2%3D1%2C3sin2%CE%B1-2sin2%CE%B2%3D0+%E6%B1%82%E8%AF%81%EF%BC%9A%CE%B1%2B2%CE%B2%3D%CF%80%2F2)
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已知α,β均为锐角,且满足3sin^2α+2sin^2β=1,3sin2α-2sin2β=0 求证:α+2β=π/2
已知α,β均为锐角,且满足3sin^2α+2sin^2β=1,3sin2α-2sin2β=0 求证:α+2β=π/2
已知α,β均为锐角,且满足3sin^2α+2sin^2β=1,3sin2α-2sin2β=0 求证:α+2β=π/2
3sin²α+2sin²β=1
===> 3sin²α=1-2sin²β=cos2β…………………………(1)
3sin2α-2sin2β=0
===> 6sinαcosα=2sin2β
===> 3sinαcosα=sin2β…………………………………(2)
(1)/(2)得到:tanα=cot2β
所以,cot[(π/2)-α]=cot2β
已知α、β均为锐角
则,(π/2)-α=2β
所以,α+2β=π/2
3sin^2α+2sin^2β=1,
3sin²a=cos2B ①
3sin2α-2sin2β=0
6sinacosa=2sin2B ②
①/②
tana=cot2B
∴ tana=tan(π/2-2B)
∵ a是锐角,
B是锐角,则π/2-2B∈(-π/2,π/2)
∴ a=π/2-2B
即 a+2B=π/2
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已知α,β均为锐角,且满足3sin^2α+2sin^2β=1,3sin2α-2sin2β=0 求证α+2β=π/2尽量说详细些 谢谢 ~~
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