Question

Show that:
sin (B − C) cos (A − D) + sin (C − A) cos (B − D) + sin (A − B) cos (C − D) = 0

Answer 1

Consider LHS: 
\[\sin \left( B - C \right) \cos \left( A - D \right) + \sin \left( C - A \right) \cos \left( B - D \right) + \sin \left( A - B \right) \cos \left( C - D \right)\]
\[= \frac{1}{2}\left[ 2\sin \left( B - C \right) \cos \left( A - D \right) \right] + \frac{1}{2}\left[ 2\sin \left( C - A \right) \cos \left( B - D \right) \right] + \frac{1}{2}\left[ 2\sin \left( A - B \right) \cos\left( C - D \right) \right]\]
\[ = \frac{1}{2}\left[ \sin \left\{ \left( B - C \right) + \left( A - D \right) \right\} + \sin \left\{ \left( B - C \right) - \left( A - D \right) \right\} \right] + \frac{1}{2}\left[ \sin \left\{ \left( C - A \right) + \left( B - D \right) \right\} + \sin \left\{ \left( C - A \right) - \left( B - D \right) \right\} \right] + \frac{1}{2}\left[ \sin \left\{ \left( A - B \right) + \left( C - D \right) \right\} + \sin \left\{ \left( A - B \right) - \left( C - D \right) \right\} \right]\]
\[ = \frac{1}{2}\left[ \sin \left( B - C + A - D \right) + \sin \left( B - C - A + D \right) \right] + \frac{1}{2}\left[ \sin \left( C - A + B - D \right) + \sin \left( C - A - B + D \right) \right] + \frac{1}{2}\left[ \sin \left( A - B + C - D \right) + \sin \left( A - B - C + D \right) \right]\]
\[ = \frac{1}{2}\left[ \sin \left( B - C + A - D \right) + \sin \left( B - C - A + D \right) \right] + \frac{1}{2}\left[ \sin \left\{ - \left( - C + A - B + D \right) \right\} + \sin \left\{ - \left( - C + A + B - D \right) \right\} \right] + \frac{1}{2}\left[ \sin\left\{ - \left( - A + B - C + D \right) \right\} + \sin \left( A - B - C + D \right) \right]\]
\[ = \frac{1}{2}\sin\left( B - C + A - D \right) + \frac{1}{2}\sin\left( B - C - A + D \right) - \frac{1}{2}\sin\left( - C + A - B + D \right) - \frac{1}{2}\sin\left( - C + A + B - D \right) - \frac{1}{2}\sin\left( - A + B - C + D \right) + \frac{1}{2}\sin\left( A - B - C + D \right)\]
\[ = 0\]
 = RHS