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Question
In ∆ABC, prove that \[a \left( \cos C - \cos B \right) = 2 \left( b - c \right) \cos^2 \frac{A}{2} .\]
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Solution
\[a\left( \cos C - \cos B \right)\]
\[ = k\left( \sin A\cos C - \sin A\cos B \right) \left[ \because \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k \right]\]
\[ = \frac{k}{2}\left( 2\sin A\cos C - 2\sin A\cos B \right)\]
\[ = \frac{k}{2}\left[ \sin\left( A + C \right) + \sin\left( A - C \right) - \sin\left( A + B \right) - \sin\left( A - B \right) \right]\]
\[ = \frac{k}{2}\left[ \sin\left( \pi - B \right) + \sin\left( A - C \right) - \sin\left( \pi - C \right) - \sin\left( A - B \right) \right] \left( \because A + B + C = \pi \right)\]
\[ = \frac{k}{2}\left[ \sin B - \sin C + \sin\left( A - C \right) - \sin\left( A - B \right) \right]\]
\[ = \frac{k}{2}\left[ 2\sin\left( \frac{B - C}{2} \right)\cos\left( \frac{B + C}{2} \right) + 2\sin\left( \frac{A - C - A + B}{2} \right)\cos\left( \frac{A - C + A - B}{2} \right) \right]\]
\[ = k\sin\left( \frac{B - C}{2} \right)\left[ \cos\left( \frac{\pi}{2} - \frac{A}{2} \right) + \cos\left\{ \frac{2A - \left( \pi - A \right)}{2} \right\} \right]\]
\[ = k\sin\left( \frac{B - C}{2} \right)\left( \sin\frac{A}{2} + \sin\frac{3A}{2} \right)\]
\[ = k\sin\left( \frac{B - C}{2} \right)\left[ 2\sin\left( \frac{\frac{A}{2} + \frac{3A}{2}}{2} \right)\cos\left( \frac{\frac{3A}{2} - \frac{A}{2}}{2} \right) \right]\]
\[ = 2k\sin\left( \frac{B - C}{2} \right)\sin A\cos\frac{A}{2}\]
\[ = 4k\sin\left( \frac{B - C}{2} \right)\sin\frac{A}{2} \cos^2 \frac{A}{2} . . . \left( 1 \right)\]
\[\text{ Now }, \]
\[\text{ Consider }\]
\[2\left( b - c \right) \cos^2 \frac{A}{2}\]
\[ = 2k\left( \sin B - \sin C \right) \cos^2 \frac{A}{2}\]
\[ = 2k\left[ 2\sin\left( \frac{B - C}{2} \right)\cos\left( \frac{B + C}{2} \right) \right] \cos^2 \frac{A}{2}\]
\[ = 4k\sin\left( \frac{B - C}{2} \right)\cos\left( \frac{\pi}{2} - \frac{A}{2} \right) \cos^2 \frac{A}{2}\]
\[ = 4k\sin\left( \frac{B - C}{2} \right)\sin\frac{A}{2} \cos^2 \frac{A}{2} . . . \left( 2 \right) \]
\[\text{ From } \left( 1 \right) \text{ & }\left( 2 \right), \text{ we get }\]
\[a \left( \cos C - \cos B \right) = 2 \left( b - c \right) \cos^2 \frac{A}{2}\]
\[\text{ Hence proved } .\]
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