对任意三角形ABC角ABC对应边为abc求证a(sinB-sinC)+b(sinC-sinA)+c(sinA-sinC)=0
问题描述:
对任意三角形ABC角ABC对应边为abc求证a(sinB-sinC)+b(sinC-sinA)+c(sinA-sinC)=0
答
由正弦定理得:a/sina=b/sinb=c/sinc=2R
∴a(sinB-sinC)+b(sinC-sinA)+c(sinA-sinC)=a*b/(2R)-a*c/(2R)+b*c/
(2R)-b*a/(2R)+c*a/(2R)-c*b/(2R)=0
我个人觉得题目是这样的:a(sinB-sinC)+b(sinC-sinA)+c(sinA-sinB)=0