sin^2A+sin^2B+sin^2C-2cosAcosBcosC=2sina^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)+co

Question

sin^2A+sin^2B+sin^2C-2cosAcosBcosC=2
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
除了这种方法之外还有没有别的方法?

Answer 1

有将sin2B+sin2C移到另一侧和2联立用三角函数的基本关系化成角B、C的余弦,进而再根据A=π-B-C将cosA化为角B、C的关系即可证．
证明：（1）要证sin^2A+sin^2B+sin^2C-2cosAcosBcosC=2成立
即证sin2A=2-sin2B-sin2C+2cosAcosBcosC成立
又因为2-sin2B-sin2C+2cosAcosBcosC=cos2B+cos2C+2cos（π-B-C）cosBcosC
=cos2B+cos2C-2cos（B+C）cosBcosC=cos2B+cos2C-2（cosBcosC-sinBsinC）cosBcosC
=cos2B+cos2C-2cos2Bcos2C+2sinBsinCcosBcosC
=（cos2B-cos2Bcos2C）+（cos2C-cos2Bcos2C）+2sinBsinCcosBcosC
=cos2Bsin2C+cos2Csin2C+2sinBsinCcosBcosC
=（cosBsinC+cosCsinC）2
=sin2（B+C）=sin2（π-A）=sin2A
即证,有点乱,仔细看看- -
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