在三角形ABC的外接圆半径为R,且2R(sin^2A-sin^2C)=(根号2倍a-b)sinB,求三角形ABC面积的最大值.

2R[(sinA)^2-(sinC)^2]=2R[a^2/(2R)^2-c^2/(2R)^2]=[1/(2R)](a^2-c^2)
(√2a-b)sinB=[1/(2R)](√2ab-b^2)
由题意知,[1/(2R)](a^2-c^2)=[1/(2R)](√2ab-b^2)
即a^2-c^2=√2ab-b^2
cosC=(a^2+b^2-c^2)/(2ab)=√2/2,则C=π/4
c=2RsinC=√2R
√2ab=a^2+b^2-c^2>=2ab-2R^2
(2-√2)ab