在△ABC中,求证a²-b²/c²=sin（A-B）/sinC
问题描述：
在△ABC中,求证a²-b²/c²=sin（A-B）/sinC
答
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/2)/sin(A+B/2)cos(A+B/2)
=sin(A-B)/sin(A+B)=sin(A-B)/sinC
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