a cos A + b cos B + c cos C = 2b sin A sin C

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Solution

$By sine rule, we know that \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k (say) \Rightarrow a = k \sin A, b = k \sin B, c = k \sin C Now, LHS = a \cos A + b \cos B + c \cos C = k \sin A \cos A + k \sin B \cos B + k \sin C \cos C = \frac{k}{2} (2 \sin A \cos A + 2 \sin B \cos B + 2 \sin C \cos C) = \frac{k}{2} (\sin 2 A + \sin 2 B + 2 \sin C \cos C) = \frac{k}{2} (2 \sin \frac{2 A + 2 B}{2} \cos \frac{2 A - 2 B}{2} + 2 \sin C \cos C) = \frac{k}{2} (2 \sin (A + B) \cos (A - B) + 2 \sin C \cos C) = \frac{k}{2} (2 \sin (π - C) \cos (A - B) + 2 \sin C \cos C) (∵ A + B + C = π) = \frac{k}{2} (2 \sin C \cos (A - B) + 2 \sin C \cos C) = \frac{k}{2} \times 2 \sin C (\cos (A - B) + \cos C) = k \sin C (2 \cos (\frac{A - B + C}{2}) \cos (\frac{A - B - C}{2})) = k \sin C (2 \cos (\frac{π - B - B}{2}) \cos (\frac{B + C - A}{2})) (∵ A + B + C = π) = k \sin C (2 \cos (\frac{π - 2 B}{2}) \cos (\frac{π - 2 A}{2})) (∵ A + B + C = π) = k \sin C (2 \cos (\frac{π}{2} - B) \cos (\frac{π}{2} - A)) = 2 k \sin C (\sin B \sin A) = 2 (k \sin B) \sin A \sin C = 2 b \sin A \sin C = RHS ∴ LHS = RHS$

Hence, a cos A + b cos B + c cos C = 2b sin A sin C.

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