△ABC中,求证(a²-b²)/(cosA+cosB) + (b²-c²)/(cosB+cosC) + (c²-a²)/(cosC+cosA)=0
问题描述:
△ABC中,求证(a²-b²)/(cosA+cosB) + (b²-c²)/(cosB+cosC) + (c²-a²)/(cosC+cosA)=0
答
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)/...