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

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  • 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