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Question
In ∆ABC, prove that if θ be any angle, then b cosθ = c cos (A − θ) + a cos (C + θ).
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Solution
Suppose
\[\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k\] ...(1)
Consider the RHS of the equation b cosθ = c cos (A − θ) + a cos (C + θ).
\[RHS = c\cos\left( A - \theta \right) + a\cos\left( C + \theta \right)\]
\[ = k\sin C\cos\left( A - \theta \right) + k\sin A\cos\left( C + \theta \right) \left( from \left( 1 \right) \right) \]
\[ = \frac{k}{2}\left[ 2\sin C\cos\left( A - \theta \right) + 2\sin A\cos\left( C + \theta \right) \right]\]
\[ = \frac{k}{2}\left[ \sin\left( A + C - \theta \right) + \sin\left( C + \theta - A \right) + \sin\left( A + C + \theta \right) + \sin\left( A - C - \theta \right) \right] \]
\[ = \frac{k}{2}\left( \sin\left( \pi - B - \theta \right) + \sin\left( C + \theta - A \right) + \sin\left( \pi - B + \theta \right) - \sin\left( C + \theta - A \right) \right) \left( \because A + B + C = \pi \right)\]
\[ = \frac{k}{2}\left( \sin\left( B + \theta \right) + \sin\left( B - \theta \right) \right)\]
\[ = \frac{k}{2}\left( \sin B\cos\theta + \sin\theta\cos B + \sin B\cos\theta - \sin\theta\cos B \right)\]
\[ = \frac{k}{2}\left( 2\sin B\cos\theta \right)\]
\[ = k\sin B\cos\theta\]
\[ = b\cos\theta = LHS\]
\[\text{ Hence proved } .\]
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