已知△ABC的外接圆半径是根号2,且满足条件2倍根号2(sinA的平方—sinC的平方)=(a—b)sinB

正弦定理,a/sinA=b/sonB=c/sinC=2R,R=sqrt(2).
条件 2sqrt(2)[(sinA)^2—(sinC)^2]=(a—b)sinB,
==> 2R[(a/2R)^2—(c/2R)^2]=(a—b)b/(2R),
==> a^2-c^2+b^2=ab,
余弦定理,cosC=(a^2+b^2-c^2)/(2ab)=1/2
==> ∠C=60°.
2.△ABC面积=absinC/2=(2RsinA)(2RsinB)*sqrt(3)/4
=2sqrt(3)sinAsinB= A=B=60°.
故 当A=B=60°时,△ABC面积有最大值3sqrt(3)/2.

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+b
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