ΔABC的三内角对的边a、b、c满足9a^+9b^=19c^求cotC/(cotA+cotB)
热心网友
cotC/(cotA+cotB)=(cosC/sinC)/[(cosA/sinA)+(cosB/cosB)]=(cosC/sinC)/[(sinBcosA+cosBsinA)/sinAsinB]=(cosC/sinC)/[sin(B+A)/sinAsinB]=(cosC/sinC)/[sin(180-C)/sinAsinB]=(cosC/sinC)/[sinC/sinAsinB]=(cosC)×[sinAsinB/(sinC)^2],又因为sinA/sinC=a/c,sinB/sinC=b/c(正弦定理)所以原式=(cosC)×(ab/c^2),而由余弦定理有:cosC=(a^2+b^2-c^2)/2ab,代入得:原式=[(a^2+b^2-c^2)/2ab]×(ab/c^2)=(a^2+b^2-c^2)/2c^2,而条件有:a^2+b^2=19c^2/9，代入得:原式=(19c^2/9-c^2)/2c^2=(19/9-1)/2=5/9。