∵SinC+Sin（B-A）=2Sin2A
2sinBcosA=4sinAcosA
sinB=2sinA
b=2a
c²=a²+b²-2abcosC
a=2/√3
b=4/√3
∴S△ABC=1/2absinC
=1/2×8/3×√3/2
=2√3/3
參考:
已知C=60°,則A+B=120°.得B-A=120°-2A,將其與C=60°一并代入原式：
sin60°+sin(120°-2A)=2sin2A
→sin60°+sin120°cos2A-cos120°sin2A=2sin2A
→√3/2+√3/2cos2A-(-1/2)sin2A=2sin2A
→√3/2+√3/2cos2A=3/2sin2A
→1+cos2A=√3sin2A
→1+cos2A=√3·√(1-cos²2A)
兩邊平方后解得：cos2A=1/2,或cos2A=-1
即得：A=30°,或A=90°.
相應(yīng)有：B=90°或B=30°.
故ABC為直角三角形,其60°角對(duì)邊長(zhǎng)為2,則其30°角對(duì)邊長(zhǎng)為2/√3.
得：S△ABC=½×2×2/√3=2√3/3.