在△ABC中,求证:sin눀A+sin눀B+sin눀C=2+2cosAcosBcosC

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2025-03-10 21:50:18
推荐回答(1个)
回答1:

sina^2+sinb^2+sinc^2-2cosacosbcosc
=3-(cosa^2+cosb^2+cosc^2+2cosacosbcosc)
=3-{cosa*[cosa+2cosb*cosc]+(1/2)*[cos(2b)+cos(2c)+2]}
=3-{-cos(b+c)*[-cos(b+c)+2cosb*cosc]+(1/2)*[cos(2b)+cos(2c)]+1}
=3-{-cos(b+c)*cos(b-c)+cos(b+c)*cos(b-c)+1}
=2