Question

在△ABC中，求证a²-b²/c²=sin（A-B）/sinC

在△ABC中，求证a²-b²/c²=sin（A-B）/sinC

Answer 1

根据正弦定理：
(a+b/c)(a-b/c)
=(sinA+sinB/sinC)(sinA-sinB/sinC)
分别处理，用和化为积公粗配段式：
sinA+sinB/sinC=2sin(A+B/2)cos(A-B/2)/sin(A+B)
=2sin(A+B/2)cos(A-B/2)/2sin(A+B/2)cos(A+B/2)
=cos(A-B/2)/cos(A+B/2)
同理：a-b/c=sin(A-B/2)/sin(A+B/2)

所以原式=sin(A-B/2)cos(A-B/卖陵2)/sin(A+B/2)cos(A+B/2)
=sin(A-B)/sin(A+B)=sin(A-B)/岩誉sinC

希望对您有所帮助

Answer 2