If $\cos \mathrm{\theta }=\frac{1}{2}\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)$ then $\frac{1}{2}\left({\mathrm{x}}^2+\frac{1}{{\mathrm{x}}^2}\right)$ is equal to

Given that, $\cos \mathrm{\theta }=\frac{1}{2}\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)$

$\Rightarrow \mathrm{x}+\frac{1}{\mathrm{x}}=2\cos \mathrm{\theta }----\left(\mathrm{i}\right)$

We know that,

$\begin{array}{l}{\mathrm{x}}^2+\frac{1}{{\mathrm{x}}^2}={\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)}^2-2\\ =(2\cos \mathrm{\theta }{)}^2-2=4{\cos }^2\mathrm{\theta }-2\\ =2\cos 2\mathrm{\theta }\\ \therefore \frac{1}{2}\left({\mathrm{x}}^2+\frac{1}{{\mathrm{x}}^2}\right)=\frac{1}{2}\times 2\cos 2\mathrm{\theta }=\cos 2\mathrm{\theta }\end{array}$