sin^2A+sin^2B+sin^2C-2cosAcosBcosC=3是怎么来的,明白了

Question

sin^2A+sin^2B+sin^2C-2cosAcosBcosC=
3是怎么来的,明白了

Answer 1

若A,B,C三角形中,则答案是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