已知△ABC的三内角分别为A B C 求证 (1)cosA=-cos(B +C ) (2)sinA[(B+C)/2]=cos(A/2)

问题描述:

已知△ABC的三内角分别为A B C 求证 (1)cosA=-cos(B +C ) (2)sinA[(B+C)/2]=cos(A/2)

1) cosA=cos(180-B-C)=cos(-B-C)=-cos(B+C)
2)sin[(B+C)/2]=sin[(180-A)/2]=sin(90-A/2)=cos(A/2)