证明:a²-b²/c²=sin(A-B)/sinC

证明:a²-b²/c²=sin(A-B)/sinC

应该有个条件,在三角形ABC内
sin(A-B)/sinC
=sin(A-B)sinC/sin²C
=sin(A-B)sin(A+B)/sin²C
=[(sinAcosB-cosAsinB)(sinAcosB+cosAsinB]/sin²C
=(sin²Acos²B-cos²Asin²B)/sin²C
=[sin²A(1-sin²B)-(1-sin²A)sin²B]/sin²C
=(sin²A-sin²B)/sin²C
利用正弦定理a/sinA=b/sinB=c/sinC
(sin²A-sin²B)/sin²C=(a²-b²)/c²
∴ (a²-b²)/c²=sin(A-B)/sinC