Question

在△ABC中，求证（a²-b²）/c²=sin（A-B）/sinC

Answer 1

由正弦定理:

a/sinA=c/sinC

a/c=sinA/sinC,两边同时乘以2cosB,左边分子分母同乘以c.得：

2ac*cosB/c²=2sinAcosB/sinC.

由余弦定理a²+c²-b²=2ac*cosB得：

(a²+c²-b²)/c²=2sinAcosB/sinC

两边同时减去1,可得:

(a²-b²)/c²=(2sinAcosB-sinC)/sinC

且有2sinAcosB-sinC=2sinAcosB-sin(A+B)
=2sinAcosB-(sinAcosB+cosAsinB)
=sinAcosB-cosAsinB
=sin(A-B)

则原式得证.