Prove that: \[ \sin(B-C)\cos(A-D) \] \[ +\sin(C-A)\cos(B-D) \] \[ +\sin(A-B)\cos(C-D)=0 \]
Solution
L.H.S.
\[ = \sin(B-C)\cos(A-D) \] \[ +\sin(C-A)\cos(B-D) \] \[ +\sin(A-B)\cos(C-D) \]Use identity:
\[ \sin X\cos Y = \frac12[\sin(X+Y)+\sin(X-Y)] \]
\[
=
\frac12[
\sin(B-C+A-D)+\sin(B-C-A+D)
]
\]
\[
+\frac12[
\sin(C-A+B-D)+\sin(C-A-B+D)
]
\]
\[
+\frac12[
\sin(A-B+C-D)+\sin(A-B-C+D)
]
\]
Rearrange terms:
\[ = \frac12[ \sin(A+B-C-D) +\sin(B+D-A-C) \] \[ +\sin(B+C-A-D) +\sin(C+D-A-B) \] \[ +\sin(A+C-B-D) +\sin(A+D-B-C) ] \]Use identity:
\[ \sin(-\theta)=-\sin\theta \]Each positive term cancels with its negative counterpart.
\[ = 0 \]Hence Proved.