3sinα^2+2sinβ^2=1,3sin2α-2sin2β=0,αβ均为锐角,求α+2β的值
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3sinα^2+2sinβ^2=1,3sin2α-2sin2β=0,αβ均为锐角,求α+2β的值
cos(α+2β)
=cosαcos2β-sinαsin2β
=cosα(1-2(sinβ)^2)-sinαsin2β
=cosα(3(sinα)^2)-sinα(3*sin(2α)/2)
=3sin(2α)sinα/2-3sin(2α)*sinα/2
=0
又α,β为锐角
所以0
∵3sin²α+2sin²β-2sinα=0,
∴2sin²β=2sinα-3sin²α=sinα(2-3sinα)≥0
∴0≤sinα≤2/3
∴cos²α+cos²β=cos²α+(1-sin²β)
=cos²α+[1-1/2...
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∵3sin²α+2sin²β-2sinα=0,
∴2sin²β=2sinα-3sin²α=sinα(2-3sinα)≥0
∴0≤sinα≤2/3
∴cos²α+cos²β=cos²α+(1-sin²β)
=cos²α+[1-1/2(2sinα-3sin²α)]
=1/2sin²α-sinα+2
=1/2(sinα-1)²+7/4
∴sin²α+sin²β=2-1/2(sinα-1)²-7/4
=1/4-1/2(sinα-1)²
所以当sinα=2/3时,
sin2α+sin2β取最大值 4/9
故答案为:
4/9
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