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