Question

在三角形ABC中 求证：(a^2-b^2)/c^2=(sin(A-B)/sinC

如题

Answer 1

(a^2-b^2)/c^2=(a+b/c)(a-b/c)
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

Answer 2

根据正弦及余弦定理可得
sin(A-B)/sinC
=(sinAcosB-cosAsinB)/sinC
=(acosB-bcosA)/c
=[(a²+c²-b²)/2c-(b²+c²-a²)/2c]/c
=(a²-b²)/c²