题目

已知锐角三角形ABC中,2根号2(sin^2(A)-siin^2(C)=(a-b)sinB,外接圆半径根号2,求三角形ABC面积的取值范围

提问时间：2020-10-31

答案

根据正弦定理：a/sinA=b/sinB=c/sinC=2R
∴sinA=a/2R,sinB=b/2R,sinC=c/2R
∴2√2[sin^2(A)-siin^2(C)]=(a-b)sinB
2√2[a²/4R²-c²/4R²]=(a-b)b/2R
2√2[a²-c²]=2Rb(a-b) .//左右同乘以4R²
2√2[a²-c²]=2√2×b(a-b)
a²-c²=ab-b²
a²+b²-c²=ab
cosC=(a²+b²-c²)/2ab=1/2
∴C=60°
△ABC面积=(1/2)absinC
=(√3/4)ab
=(√3/4)2RsinA×2RsinB
=2√3×sinAsinB
=2√3×sinAsin(180°-60°-A)
=2√3×[sinAsin120°cosA-sinAcos120°sinA]
=2√3×[√3/2×sinAcosA+1/2×sin²A]
=√3×[√3sinAcosA+sin²A]
=√3×[√3/2×sin2A-1/2×cos2A+1/2]
=√3×sin(2A-30°)+√3/2
∵A+B=120°,A