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Question
If A + B = \[\frac{\pi}{3}\] and cos A + cos B = 1, then find the value of cos \[\frac{A - B}{2}\].
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Solution
Given:
A + B =\[\frac{\pi}{3}\] and cos A + cos B = 1
\[\Rightarrow 2\cos\left( \frac{A + B}{2} \right)\cos\left( \frac{A - B}{2} \right) = 1 \left[ \because \cos A + \cos B = 2\cos\left( \frac{A + B}{2} \right)\cos\left( \frac{A - B}{2} \right) \right]\]
\[ \Rightarrow 2\cos\left( \frac{\pi}{6} \right)\cos\left( \frac{A - B}{2} \right) = 1 \left[ \because A + B = \frac{\pi}{3} \right]\]
\[ \Rightarrow 2 \times \frac{\sqrt{3}}{2} \times \cos\left( \frac{A - B}{2} \right) = 1\]
\[ \Rightarrow \cos\left( \frac{A - B}{2} \right) = \frac{1}{\sqrt{3}}\]
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