三角形ABC中,2根号2 （sin平方A-sin平方C）=（a-b）sinB,它的外接圆半径为根号21,求C的度数；2,求三角形ABC面积的最大值.
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a/sinA=b/sinB=c/sinC=2R=2√2
=>a=2RsinA,b=2RsinB,c=2RsinC
2√2（sin²A-sin²C）=（a-b）sinB
=>4R²（sin²A-sin²C）=2R（a-b）sinB
=>a²-c²=(a-b)b
=>(a²+b²-c²)/2ab=1/2=cosC
=>C=60°S△ABC=absinC/2=2RsinA*2RsinB*sinC/2
=√3(2sinAsinB)=√3[cos(A-B)-cos(A+B)]
=√3[cos(A-B)+1/2]≤3√3/2