高一数学，求解！！！！谢谢！！！！！
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2√2(sin2A-sin2C) = (a-b)sinB
--->2√2(a2-c2)/(2R)2 = (a-b)?b/(2R)
--->a2-c2=ab-b2
--->a2+b2-c2=ab
--->cosC = (a2+b2-c2)/(2ab) = 1/2---->C=60°
SΔ = (1/2)absinC = (2R2)sinAsinBsinC
= R2[cos(A-B)-cos(A+B)]sinC
= R2[cos(A-B)-cos120°]sin60°
= √3[cos(A-B)+1/2]
≤3√3/2
--->A=B即△ABC为等边三角形时，面积的最大值=3√3/2
--->2√2(a2-c2)/(2R)2 = (a-b)?b/(2R)
--->a2-c2=ab-b2
--->a2+b2-c2=ab
--->cosC = (a2+b2-c2)/(2ab) = 1/2---->C=60°
SΔ = (1/2)absinC = (2R2)sinAsinBsinC
= R2[cos(A-B)-cos(A+B)]sinC
= R2[cos(A-B)-cos120°]sin60°
= √3[cos(A-B)+1/2]
≤3√3/2
--->A=B即△ABC为等边三角形时，面积的最大值=3√3/2
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