Question

△ABC中，求证(a²-b²)/(cosA+cosB) + (b²-c²)/(cosB+cosC) + (c²-a²)/(cos

要详细过程 谢谢
△ABC中，求证(a²-b²)/(cosA+cosB) + (b²-c²)/(cosB+cosC) + (c²-a²)/(cosC+cosA）=0

Answer 1

证明：∵ a/sinA=b/sinB=c/sinC=2R (正弦定理，其中R为△ABC的内接圆半径）
∴（a²-b²）÷(cosA+cosB)+（b²-c²）÷（cosB+cosC）+（c²-a²）÷（cosC+cosA）
=4R^2[(sinAsinA-sinBsinB)/（cosA+cosB)+(sinBsinB-sinCsinC)/(cosB+cosC)+(sinCsinC-sinAsinA)/(cosC+cosA)]
=4R^2[(1-cosAcosA-1+cosBcosB)/（cosA+cosB)+(1-cosBcosB-1+cosCcosC)/(cosB+cosC)+(1-cosCcosC-1+cosAcosA)/(cosC+cosA)]
=4R^2[(cosBcosB-cosAcosA)/（cosA+cosB)+(cosCcosC-cosBcosB)/(cosB+cosC)+(cosAcosA-cosCcosC)/(cosC+cosA)]
=4R^2[(cosB-cosA)^2/（cosA+cosB)+(cosC-cosB)^2/(cosB+cosC)+(cosA-cosC)^2/(cosC+cosA)]
=4R^2[(cosB-cosA)+(cosC-cosB)+(cosA-cosC)]
=4R^2[cosB-cosB+cosC-cosC+cosA-cosA]
=R^2*0
=0
∴ (a²-b²)/(cosA+cosB) + (b²-c²)/(cosB+cosC) + (c²-a²)/(cosC+cosA）=0