因A、B、C成等差数列
故A+C=2B
又A+B+C=π
故B=π/3,A+C=2π/3
sinA-sinC
=2cos(A+C)/2cos(A-C)/2
=-cos(A-C)/2
故sinA-sinC+√2/2cos(A-C)
=-cos(A-C)/2+√2/2[2cos²(A-C)/2-1]=√2/2
即2cos²(A-C)/2-√2cos(A-C)/2-2=0
解之：cos(A-C)/2=-√2/2
又-2π/3