三角形ABC中ABC的对边为abc求证!

Question

求证a^-b^/c^=sin(A-B)/sinC

Answer 1

a^2=b^2+c^2-2bccosA
(a^2-b^2)/c^2
=1-2(b/c)cosA
=1-2sinBcosA/sinC=(sinC-2sinBcosA)/sinC
欲证(a^2-b^2)/c^2=sin(A-B)/sinC
即证
(sinC-2sinBcosA)/sinC=sin(A-B)/sinC
因为
sin(180-A-B)=sinC=sin(A+B)
=sinAcosB+cosBsinA
所以
（sinC-2sinBcosA)/sinC
=(cosBsinA-sinBcosA)/sinC
=sin(A-B)/sinC
得证

Answer 2

右=(sinAcosB-sinBcosA)/(sinAcosB+cosAsinB)=1-2sinBcosA/(sinAcosB+cosAsinB)=1-2bcosA/(acosB+bcosA)=1-((b^2+c^2-a^2)/c)/((a^2+c^2-b^2+b^2+c^2-a^2)/2c)=1-(b^2+c^2-a^2)/c^2=(a^2-b^2)/c^2=左
原式得证