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Question
In triangle ABC, prove the following:
\[\left( a - b \right) \cos \frac{C}{2} = c \sin \left( \frac{A - B}{2} \right)\]
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Solution
Let
\[ = k\left( \sin A - \sin B \right)\cos\frac{C}{2} \left( \text{ using }\left( 1 \right) \right) \]
\[ = k \times 2\sin\left( \frac{A - B}{2} \right)\cos\left( \frac{A + B}{2} \right)\cos\frac{C}{2}\]
\[ = 2$k\sin\left( \frac{A - B}{2} \right)\cos$\left( \frac{A + B}{2} \right)\cos\left( \frac{\pi - \left( A + B \right)}{2} \right) \left[ \because A + B + C = \pi \right]\]
\[ = 2k\sin\left( \frac{A - B}{2} \right)\cos\left( \frac{A + B}{2} \right)\sin\left( \frac{A + B}{2} \right)\]
\[ = k\sin\left( \frac{A - B}{2} \right)\sin\left( A + B \right) \left[ \because 2\cos\left( \frac{A + B}{2} \right)\sin\left( \frac{A + B}{2} \right) = \sin\left( A + B \right) \right]\]
\[ = k\sin\frac{A - B}{2}\sin\left( \pi - C \right) \left[ \because A + B + C = \pi \right]\]
\[ = k\sin C\sin\left( \frac{A - B}{2} \right) \]
\[ = C\sin\left( \frac{A - B}{2} \right) = RHS \]
Hence proved.
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