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Question
In any 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 RHS of the equation
\[RHS = \frac{b - c}{a}\cos\frac{A}{2}\]
\[ = \frac{k\left( \sin B - \sin C \right)}{k\sin A}\cos\left( \frac{\pi - \left( B + C \right)}{2} \right) \left( \because A + B + C = \pi \right) \]
\[ = \frac{2\sin\left( \frac{B - C}{2} \right)\cos\left( \frac{B + C}{2} \right)}{\sin A}\]
\[ = \frac{\sin\left( \frac{B - C}{2} \right)2\cos\left( \frac{B + C}{2} \right)}{\sin A}\sin\left( \frac{B + C}{2} \right)\]
\[ = \frac{\sin\left( \frac{B - C}{2} \right)\sin\left( B + C \right)}{\sin A}\]
\[ = \frac{\sin\left( \frac{B - C}{2} \right)\sin\left( \pi - A \right)}{\sin A}\]
\[ = \frac{\sin A\sin\left( \frac{B - C}{2} \right)}{\sin A}\]
\[ = \sin\left( \frac{B - C}{2} \right) = LHS\]
\[\text{ Hence proved } .\]
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