在△ABC中,a,b,c分别是角A,B,C,所对的边,且 2sin^2*A+B／2+cos2C=1 若a^2=b^2+1／2c^2,试求sin(A-B)的值
问题描述：
在△ABC中,a,b,c分别是角A,B,C,所对的边,且 2sin^2*A+B／2+cos2C=1 若a^2=b^2+1／2c^2,试求sin(A-B)的值
答
2sin^2(A+B)/2+cos2C=1
2sin^2(A+B)/2-1+cos2C=0
2sin^2(180°-C)/2-1+cos2C=0
2sin^2(90°-C/2)-1+cos2C=0
2cos^2(C/2)-1+cos2C=0
cosC+cos2C=0
cosC+2(cosC)^2-1=0
2(cosC)^2+cosC-1=0
(cosC+1)(2cosC-1)=0
cosC+1=0,2cosC-1=0
cosC=-1,cosC=1/2
C=180°(舍去),C=60°
下面用正弦定理做,自己做一下吧