如果a+b+c=π
sin^2a+sin^2b+sin^2c-2-2cosacosbcosc
=sin^2a+sin^2b+sin^2(a+b)-2+2cosacosbcos(a+b)
=sin^2a+sin^2b+sin^2acos^2b+2sinasinbcosacosb+cos^2asin^2b-2+2cos^2acos^2b-2sinasinbcosacosb
=sin^2a+sin^2b+sin^2acos^2b+cos^2asin^2b-2+2cos^2acos^2b
=sin^2acos^2b+cos^2asin^2b-cos^2a-cos^2b+2cos^2acos^2b
=sin^2acos^2b+cos^2asin^2b-cos^2asin^2b-sin^2acos^2b
=0
即sin^2a+sin^2b+sin^2c=2+2cosacosbcosc
反之亦然
所以充要条件成立