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Question
In triangle ABC, prove the following:
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Solution
Let
\[\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k\]
Then,
Consider the LHS of the equation \[b \cos B + c \cos C = a \cos \left( B - C \right)\]
\[LHS = b\cos B + c\cos C\]
\[ = k\left( \sin B\cos B + \sin C\cos C \right) \]
\[ = \frac{k}{2}\left( 2sinBcosB + 2sinCcosC \right)\]
\[ = \frac{k}{2}\left( \sin2B + \sin2C \right) . . . \left( 1 \right)\]
\[ RHS = a\cos\left( B - C \right)\]
\[ = k\sin A\cos\left( B - C \right) \]
\[ = \frac{k}{2}\left[ 2\sin A\cos\left( B - C \right) \right]\]
\[ = \frac{k}{2}\left[ \sin\left( A + B - C \right) + \sin\left( A - B + C \right) \right] \left[ \because 2\sin A\cos B = \sin\left( A + B \right) + \sin\left( A - B \right) \right]\]
\[ = \frac{k}{2}\left[ \sin\left( \pi - C - C \right) + \sin\left( \pi - B - B \right) \right] \left[ \because \sin\left( \pi - A \right) = \sin A, A + B + C = \pi \right]\]
\[ = \frac{k}{2}\left( \sin2C + \sin2B \right)\]
\[ = \frac{k}{2}\left( \sin2B + \sin2C \right) = LHS \left[ from\left( 1 \right) \right]\]
\[\text{ Hence proved } .\]
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