三角形ABC中，A、B、C成等差数列，其外接圆半径为1，且sinA-sinC+√2/2cos(A-C)=√2/2

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<A-C<2π/3
故-π/3<(A-C)/2<π/3
故(A-C)/2=π/4
A-C=π/2
解之：
A=7π/12，C=π/12。
由正弦定理：
a/sinA=c/sinC=2R
故△ABC面积=1/2acsinB
=1/2(2RsinA)(2RsinC)sinB
=2R²sinAsinCsinB
=R²[cos(A-C)-cos(A+C)]sinB
=R²[cos(π/2)-cos(2π/3)]sinπ/3
=√3/4