Show that: \( \sin(B-C)\cos(A-D)+\sin(C-A)\cos(B-D)+\sin(A-B)\cos(C-D)=0 \)
Solution:
\[
\sin(B-C)\cos(A-D)+\sin(C-A)\cos(B-D)
\]
\[
+\sin(A-B)\cos(C-D)
\]
Using identity,
\[
\sin x \cos y=\frac12[\sin(x+y)+\sin(x-y)]
\]
\[
=\frac12[\sin(B-C+A-D)+\sin(B-C-A+D)]
\]
\[
+\frac12[\sin(C-A+B-D)+\sin(C-A-B+D)]
\]
\[
+\frac12[\sin(A-B+C-D)+\sin(A-B-C+D)]
\]
\[
=\frac12[\sin(A+B-C-D)+\sin(B+D-A-C)
\]
\[
+\sin(B+C-A-D)+\sin(C+D-A-B)
\]
\[
+\sin(A+C-B-D)+\sin(A+D-B-C)]
\]
Using,
\[
\sin(-\theta)=-\sin\theta
\]
all terms cancel out.
\[
=0
\]
\[
\boxed{\sin(B-C)\cos(A-D)+\sin(C-A)\cos(B-D)+\sin(A-B)\cos(C-D)=0}
\]