Show that: \( \sin A \sin(B-C)+\sin B \sin(C-A)+\sin C \sin(A-B)=0 \)
Solution:
\[
\sin A \sin(B-C)+\sin B \sin(C-A)+\sin C \sin(A-B)
\]
Using identity,
\[
\sin(x-y)=\sin x \cos y-\cos x \sin y
\]
\[
=\sin A(\sin B\cos C-\cos B\sin C)
\]
\[
+\sin B(\sin C\cos A-\cos C\sin A)
\]
\[
+\sin C(\sin A\cos B-\cos A\sin B)
\]
\[
=\sin A\sin B\cos C-\sin A\cos B\sin C
\]
\[
+\sin B\sin C\cos A-\sin B\cos C\sin A
\]
\[
+\sin C\sin A\cos B-\sin C\cos A\sin B
\]
Cancelling like terms,
\[
=0
\]
\[
\boxed{\sin A \sin(B-C)+\sin B \sin(C-A)+\sin C \sin(A-B)=0}
\]