3sinα^2+2sinβ^2=1,3sin2α-2sin2β=0,αβ均为锐角,求α+2β的值
人气:247 ℃ 时间:2020-02-06 05:13:01
解答
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
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