【数学】已知α、β均为锐角,且3(sinα)^2+2(sinβ)^2=1,3sin2α-2sin2β=0,证明α+2β=π/2.

来源:学生作业帮助网 编辑:六六作业网 时间:2024/11/21 23:59:43
【数学】已知α、β均为锐角,且3(sinα)^2+2(sinβ)^2=1,3sin2α-2sin2β=0,证明α+2β=π/2.【数学】已知α、β均为锐角,且3(sinα)^2+2(sinβ)^2=1

【数学】已知α、β均为锐角,且3(sinα)^2+2(sinβ)^2=1,3sin2α-2sin2β=0,证明α+2β=π/2.
【数学】已知α、β均为锐角,且3(sinα)^2+2(sinβ)^2=1,3sin2α-2sin2β=0,证明α+2β=π/2.

【数学】已知α、β均为锐角,且3(sinα)^2+2(sinβ)^2=1,3sin2α-2sin2β=0,证明α+2β=π/2.
3(sinα)^2+2(sinβ)^2=1
3(sinα)^2=cos2β ⑴
3sin2α-2sin2β=0
2sin2β=3sin2α ⑵
⑴×⑵得
6 sin2β(sinα)^2=3sin2αcos2β=6sinαcosαcos2β
sin2β(sinα)^2-sinαcosαcos2β=0
sinαcos(α+2β)=0
α,β为锐角
sinα>0,0