(I)由m•n=0得 2co s 2
C
2 -2si n 2 (A+B)=0 ,
即1+cosC-2(1-cos 2C)=0;整理得2cos 2C+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=
3
4
所以 sin(A-B)=
3
4 .