在△ABC中,角A、B、C的对边分别为a、b、c.求证:(a^2-b^2)/c^2=sin(A-B)/sinC.

证明：
三角形ABC中
a/sinA=b/sinB=c/sinC=2R
左边=(a^2-b^2)/c^2
=(sin^2A-sin^2B)/sin^2C
=(sinA+sinB)(sinA-sinB)/sin^2C
=2sin(A+B)/2cos(A-B)/2*2cos(A+B)/2sin（A-B）/2/sin^2C
=2sin(A+B)/2cos(A+B)/2*2sin[(A-B)/2]cos[(A-B)/2/sin^2C
=sin(A+B)sin（A-B）/sin^2C
=sin(π-C)sin（A-B）/sin^2C
= =sinCsin（A-B）/sin^2C
=sin（A-B）/sinC
右边=sin（A-B）/sinC
左边=右边
所以(a^2-b^2)/c^2=sin(A-B)/sinC.
(中间使用的是正弦函数的和差化积)