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Question
Prove that:
sin (B − C) cos (A − D) + sin (C − A) cos (B − D) + sin (A − B) cos (C − D) = 0
Solution
Consider LHS:
\[\sin (B - C) \cos (A - D) + \sin (C - A) \cos (B - D) + \sin (A - B) \cos (C - D)\]
Multiplying by 2:
\[ 2\sin (B - C) \cos (A - D) + 2\sin(C - A) \cos (B - D) + 2\sin (A - B) \cos (C - D)\]
\[ = \sin \left( B - C + A - D \right) + \sin \left( B - C - A + D \right) + \sin \left( C - A + B - D \right) + \sin \left( C - A - B + D \right) + \sin \left( A - B + C - D \right) + \sin \left( A - B - C + D \right)\]
\[ = \sin\left\{ - \left( C + D - A - B \right) \right\} + \sin\left\{ - \left( A + C - B - D \right) \right\} + \sin\left\{ - \left( A + D - B - C \right) \right\} + \sin\left( C - A - B + D \right) + \sin\left( A - B + C - D \right) + \sin\left( A - B - C + D \right)\]
\[ = - \sin\left( C + D - A - B \right) - \sin\left( A + C - B - D \right) - \sin\left( A + D - B - C \right) + \sin\left( C - A - B + D \right) + \sin\left( A - B + C - D \right) + \sin\left( A - B - C + D \right)\]
\[ = 0\]
= RHS
Hence, LHS = RHS.
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