已知△ABC中 2√2（sin²A-sin²C）=（a-b）sinB,△ABC外接圆半径为√2.①求角C②求△ABC面积的最大值勿复制
问题描述：
已知△ABC中 2√2（sin²A-sin²C）=（a-b）sinB,△ABC外接圆半径为√2.
①求角C
②求△ABC面积的最大值
勿复制
答
a/sinA=b/sinB=c/sinC=2R
2√2（sin²A-sin²C）=（a-b）sinB
2√2/2R^2(a²-c²)=（a-b）*b/2R
√2/(a²-c²)=（a-b）*bR
√2(a²-c²)=√2（a-b）b
a²+b²-c²=ab
cosC=(a²+b²-c²)/2ab=ab/2ab=1/2
C=60度
S=1/2absinC=√3/4*ab=√3/4*4R^2sinAsinB=2√3sinAsinB
=2√3sinAsin(120-A)
=2√3sinA[√3/2cosA+1/2sinA]
=3sinAcosA+√3sin²A
=3/2sin2A-√3/2cos2A+√3/2
=√3(√3/2sin2A-1/2cos2A)+√3/2
=√3sin(2A-60)+√3/2
0A=75时有最大值＝3√3/2