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.
3个回答

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)

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