Question
If $\hat{\text{a}}$ and $\hat{\text{b}}$ are unit vectors inclined at an angle $\theta$, prove that$\cos\frac{\theta}{2}=\frac{1}{2}\big|\hat{\text{a}}+\hat{\text{b}}\big|$

Answer

Here, $\hat{\text{a}}$ and $\hat{\text{b}}$ are unit vectors, then
$\big|\hat{\text{a}}\big|=\big|\hat{\text{b}}\big|=1$
$\big|\hat{\text{a}}+\hat{\text{b}}\big|^2=\big(\hat{\text{a}}+\hat{\text{b}}\big)^2$
$=(\hat{\text{a}})^2+(\hat{\text{b}})^2+2\hat{\text{a}}.\hat{\text{b}}$
$=\big|\hat{\text{a}}\big|^2+\big|\hat{\text{b}}\big|^2+2\hat{\text{a}}.\hat{\text{b}}$
$=(1)^2+(1)^2+2\hat{\text{a}}.\hat{\text{b}}$
$\big|\hat{\text{a}}+\hat{\text{b}}\big|^2=2+2\hat{\text{a}}\times\hat{\text{b}}$
$\big|\hat{\text{a}}+\hat{\text{b}}\big|^2=2+2\times\big|\hat{\text{a}}\big|\big|\hat{\text{b}}\big|\cos\theta$ $\big[\text{since }\vec{\text{a}} .\vec{\text{b}}=\big|\hat{\text{a}}\big|\big|\hat{\text{b}}\big|\cos\theta\big]$
$\big|\hat{\text{a}}+\hat{\text{b}}\big|^2=2+2\times1\times1\times\cos\theta$
$=2+2\cos\theta$
$\big|\hat{\text{a}}+\hat{\text{b}}\big|^2=2(1+\cos\theta)$
$\big|\hat{\text{a}}+\hat{\text{b}}\big|^2=2\big(2\cos^2\frac{\theta}{2}\big)$ $\Big[\text{since}1+\cos\theta=2\cos^2\frac{\theta}{2}\Big]$
$\big|\hat{\text{a}}+\hat{\text{b}}\big|^2=4\cos^2\frac{\theta}{2}$
$\big|\hat{\text{a}}+\hat{\text{b}}\big|=\sqrt{4\cos^2\frac{\theta}{2}}$
$\big|\hat{\text{a}}+\hat{\text{b}}\big|=2\cos\frac{\theta}{2}$
$\cos\frac{\theta}{2}=\frac{1}{2}\big|\hat{\text{a}}+\hat{\text{b}}\big|$

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