Question

在三角形ABC中 A B C分别对应a b c 证明(a^2-b^2)/c^2=[sin(A-B)]/sinC

Answer 1

证明：右边= [(sin(A-B)]/sinC
=（sinAcosB-cosAsinB）/sinC
=(acosB-bcosA)/c
=[(a²+c²-b²)/(2c)- (b²+c²-a²)/(2c)]/c
=(a²-b²)/c²
=左边，
∴等式得证.

Answer 2

正弦定理
a/sinA=b/sinB=c/sinC=k
a=ksinA
b=ksinB
c=ksinC
a^2-b^2/c^2=(sinA^2-sinB^2)/sinC^2

sin(A-B)/sinc
=sin(A-B)sinC/sinc^2
=sin(A-B)sin(A+B)/sinc^2
=(sinAcosB-sinBcosA)(sinAcosB+sinBcosA)/sinc^2
=(sinA^2cosB^2-sinB^2cosA^2)/sinc^2
=(sinA^2(1-sinB^2)-sinB^2(1-sinA^2))/sinc^2
=(sinA^2-sinB^2)/sinC^2

两边相等，原式得证