已知3sin^2 α+2sin^2 β=1,3sin2α-2sin2β=0 求证cos(α+2β)=0
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已知3sin^2 α+2sin^2 β=1,3sin2α-2sin2β=0 求证cos(α+2β)=0
![已知3sin^2 α+2sin^2 β=1,3sin2α-2sin2β=0 求证cos(α+2β)=0](/uploads/image/z/7450770-66-0.jpg?t=%E5%B7%B2%E7%9F%A53sin%5E2+%CE%B1%2B2sin%5E2+%CE%B2%3D1%2C3sin2%CE%B1-2sin2%CE%B2%3D0+%E6%B1%82%E8%AF%81cos%28%CE%B1%2B2%CE%B2%29%3D0)
3sin^2 α+2sin^2 β=1①
3sin2α-2sin2β=0 ②
①==>3sin^2 α=1-2sin^2β=cos2β
②==>sin2β=3/2sin2α=3sinαcosα
∴cos(α+2β)
=cosαcos2β-sinαsin2β
=cosα*3sin^2α-sinα*3sinαcosα
=0
3sin2α-2sin2β=0 ②
①==>3sin^2 α=1-2sin^2β=cos2β
②==>sin2β=3/2sin2α=3sinαcosα
∴cos(α+2β)
=cosαcos2β-sinαsin2β
=cosα*3sin^2α-sinα*3sinαcosα
=0
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