Question

已知△ABC中，2根号2（sin^2A-sin^2C)=(a-b)sinB,外接圆半径为根号2.

（1）求C
（2）求△ABC面积的最大值

Answer 1

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