已知△ABC的外接圆半径为R,且满足2R[(sinA)^2-(sinC)^2]=(√2a-b)sinB,求△ABC面积的最大值

2R[(sinA)^2-(sinC)^2]=(√2a-b)sinB;两边同乘以2R得：[(2RsinA)^2-(2RsinC)^2]=(√2a-b)2RsinB
a^2-c^2=b(√2a-b); a^-c^2=√2ab-b^2; a^2+b^2-c^2=√2ab;
由余弦定理：cosC=(a^2+b^2-c^2)/(2ab)=√2ab/(2ab)=√2/2; 0