Question

△ABC外接圆半径R=1，且sin^A-sin^C=(根号2-a分之b）sinAsinB，求△ABC面积的最大值。

Answer 1

解：正弦定理
a/sinA=b/sinB=c/sinC=2R
a=2sinA，b=2sinB，c=2sinC
sin²A-sin²C=(√2-b/a)sinAsinB
a²/4-c²/4=(√2-b/a)ab/4
a²-c²=√2ab-b²
a²+b²-c²=√2ab
余弦定理
cosC=(a²+b²-c²)/(2ab)=(√2ab)/(2ab)=√2/2
C=45度
A+B=135度
S△ABC=1/2absinC=1/2×2sinA×2sinB×√2/2=√2sinAsinB
=√2sin(135-B)sinB=sinBcosB+sin²B
=1/2sin2B+1/2(1-cos2B)
=1/2(sin2B-cos2B)+1/2
=√2/2sin(2B-π/4)+1/2
sin(2B-π/4)=1即
2B-π/4=π/2即B=3π/8时
S有最大值=(√2+1)/2