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Question
a cos A + b cos B + c cos C = 2b sin A sin C
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Solution
\[\text{ By sine rule, we know that }\]
\[\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k \left( say \right)\]
\[ \Rightarrow a = k \sin A, b = k \sin B, c = k \sin C\]
\[\text{ Now }, \]
\[LHS = a \cos A + b \cos B + c \cos C\]
\[ = k \sin A \cos A + k \sin B \cos B + k \sin C \cos C\]
\[ = \frac{k}{2} \left( 2 \sin A \cos A + 2 \sin B \cos B + 2 \sin C \cos C \right)\]
\[ = \frac{k}{2} \left( \sin 2A + \sin 2B + 2 \sin C \cos C \right)\]
\[ = \frac{k}{2} \left( 2 \sin \frac{2A + 2B}{2}\cos\frac{2A - 2B}{2} + 2 \sin C \cos C \right)\]
\[ = \frac{k}{2} \left( 2 \sin \left( A + B \right) \cos \left( A - B \right) + 2 \sin C \cos C \right)\]
\[ = \frac{k}{2} \left( 2 \sin \left( \pi - C \right) \cos \left( A - B \right) + 2 \sin C \cos C \right) \left( \because A + B + C = \pi \right)\]
\[ = \frac{k}{2} \left( 2 \sin C \cos \left( A - B \right) + 2 \sin C \cos C \right)\]
\[ = \frac{k}{2} \times 2 \sin C\left( \cos \left( A - B \right) + \cos C \right)\]
\[ = k \sin C\left( 2 \cos \left( \frac{A - B + C}{2} \right)\cos \left( \frac{A - B - C}{2} \right) \right)\]
\[ = k \sin C\left( 2 \cos \left( \frac{\pi - B - B}{2} \right)\cos \left( \frac{B + C - A}{2} \right) \right) \left( \because A + B + C = \pi \right)\]
\[ = k \sin C\left( 2 \cos \left( \frac{\pi - 2B}{2} \right)\cos \left( \frac{\pi - 2A}{2} \right) \right) \left( \because A + B + C = \pi \right)\]
\[ = k \sin C\left( 2 \cos \left( \frac{\pi}{2} - B \right)\cos \left( \frac{\pi}{2} - A \right) \right)\]
\[ = 2k \sin C\left( \sin B \sin A \right)\]
\[ = 2 \left( k \sin B \right) \sin A \sin C\]
\[ = 2b \sin A \sin C\]
\[ = RHS\]
\[ \therefore LHS = RHS\]
Hence, a cos A + b cos B + c cos C = 2b sin A sin C.
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