Question

△ABC中,求证(a²-b²)/c²=sin(A-B)/sinC

Answer 1

证明：由sin(A-B)=sinAcosB-cosAsinB
得sin(A-B)/sinC==(sinAcosB-cosAsinB)/sinC
又由正弦定理有a/sinA=b/sinB=c/sinC
所以有sin(A-B)/sinC=acosB/c-bcosA/c
又余弦定理cosB=(a²+c²-b²)/2ac
cosA=(b²+c²-a²)/2bc
得sin(A-B)/sinC=a(a²+c²-b²)/2ac²-b(b²+c²-a²)/2bc²
=(a²+c²-b²)/2c²-(b²+c²-a²)/2c²
=(a²-b²)/c²
得证。

Answer 2

sin(A-B)=sinAcosB-cosAsinB
sin(A-B)/sinC=(sinAcosB-cosAsinB)/sinC=acosB/c-bcosA/c
cosB=(a^2+c^2-b^2)/2ac cosA=(b^2+c^2-a^2)/2bc
sin(A-B)=(a^2+c^2-b^2)/2c^2- (b^2+c^2-a^2)/2c^2=(a^2+c^2-b^2-b^2-c^2+a^2)/2c^2
=(2a^2-2b^2)/2c^2=(a^2-b^2)/c^2