(I)由m•n=0得 2co s 2 C 2 -2si n 2 (A+B)=0 , 即1+cosC-2(1-cos 2 C)=0;整理得2cos 2 C+cosC-1=0 解得cosC=-1(舍)或 cosC= 1 2 因为0<C<π,所以C=60° (Ⅱ)因为sin(A-B)=sinAcosB-sinBcosA 由正弦定理和余弦定理可得 sinA= a 2R ,sinB= b 2R ,cosB= a 2 + c 2 - b 2 2ac ,cosA= b 2 + c 2 - a 2 2bc 代入上式得 sin(A-B)= a 2R • a 2 + c 2 - b 2 2ac - b 2R • b 2 + c 2 - a 2 2bc = 2( a 2 - b 2 ) 4cR 又因为 a 2 - b 2 = 1 2 c 2 , 故 sin(A-B)= c 2 4cR = c 4R = 1 2 sinC=