在△ABC中,角A B C所对的边a b c ,向量M=（2cos c/2,-sin(A+B)）,N=（cos c/2,2sin(A+B)）,M⊥N若a2=b2+1/2C2,试求sin(A-B)的值
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A+B+C =πA+B = π-CM⊥N=>M.N=0(2cosC/2,-sin(A+B)).(cosC/2,2sin(A+B)) =0(cosC/2)^2- (sin(A+B))^2 =0(cosC/2)^2 - (sinC)^2 =0(cosC+1)/2 - (sinC)^2 =02(cosC)^2 + cosC -1 =0(2cosC-1)(cosC+1 ) =0cosC = 1/2C...