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Question
If \[\hat{a} \text{ and } \hat{b}\] are unit vectors inclined at an angle θ, prove that \[\cos\frac{\theta}{2} = \frac{1}{2}\left| \hat{a} + \hat{b} \right|\]
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Solution
\[\text{ Given that } \hat{ a }\ \text{ and } \hat{b}\ \text{ are unit vectors }.\]
\[So,\left| \hat{a} \right|=1,\left| \hat{b} \right|=1\]
\[\text{We have}\]
\[ \left| \hat{a} + \hat{b} \right|^2 = \left| \hat{a} \right|^2 + \left| \hat{b} \right|^2 + 2 \hat{a} . \hat{b} \]
\[ = 1 + 1 + 2 \left| \hat{a} \right| \left| \hat{b} \right| \cos \theta\]
\[ = 2 + 2\cos \theta\]
\[ \Rightarrow \cos\theta = \frac{\left| \hat{a} + \hat{b} \right|^2 - 2}{2} .....................\left( 1 \right)\]
\[ \left| \hat{a} - b \right|^2 = \left| \hat{a} \right|^2 + \left| \hat{b} \right|^2 - 2 \hat{a} .\hat {b} \]
\[ = 1 + 1 - 2 \left| \hat{a} \right| \left| \hat{b} \right| \cos \theta\]
\[ = 2 - 2\cos \theta\]
\[ \Rightarrow \cos\theta = \frac{2 - \left| \hat{a} - \hat{b} \right|^2}{2}...................... \left( 2 \right)\]
\[ \text{ Now },\]
\[\cos \frac{\theta}{2} = \sqrt{\frac{1 + \cos \theta}{2}}\]
\[ = \sqrt{\frac{1 + \frac{\left| \hat{a} + \hat{b} \right|^2 - 2}{2}}{2}} ...............\left[\text{ From }\left( 1 \right) \right]\]
\[ = \sqrt{\frac{2 + \left| \hat{a} + \hat{b} \right|^2 - 2}{4}}\]
\[ = \sqrt{\frac{\left| \hat{a} + \hat{b} \right|^2}{4}}\]
\[ = \frac{1}{2}\left| \hat{a} + \hat{b} \right|\]
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