If $e^{i\theta }=\cos \theta$, then $\sum _{n=0}^{\infty }\frac{\cos \left(n\theta \right)}{2^n}=$

$\sum _{n=0}^{\infty }\frac{\cos \left(n\theta \right)}{2^n}=\mathrm{Re}\left(\sum _{n=0}^{\infty }\frac{cisn\theta }{2^n}\right)$

$=\mathrm{Re}\left(\sum _{n=0}^{\infty }{\left(\frac{\mathrm{c}is\theta }{2}\right)}^n\right)$

(using De-moivre theorem)

$=\mathrm{Re}\left(\sum _{n=0}^{\infty }{\left(\frac{e^{i\theta }}{2}\right)}^n\right)$

$=\mathrm{Re}\left(1+\frac{e^{i\theta }}{2}+{\left(\frac{e^{i\theta }}{2}\right)}^2+\mathrm{....}\infty \right)$

$1+\frac{e^{i\theta }}{2}+{\left(\frac{e^{i\theta }}{2}\right)}^2+\mathrm{......}G.P$

$=\mathrm{Re}\left(\frac{1}{1-\frac{e^{i\theta }}{2}}\right)=\mathrm{Re}\left(\frac{2}{2-e^{i\theta }}\right)$

$=\mathrm{Re}\left(\frac{2}{2-\cos \theta -i\sin \theta }\right)$

$=\mathrm{Re}\left(\frac{2\left(2-\cos \theta +i\sin \theta \right)}{{\left(2-\cos \theta \right)}^2-i^2{\sin }^2\theta }\right)$

$=\mathrm{Re}\left(\frac{4-2\cos \theta }{5-4\cos \theta }+\frac{i\sin \theta }{5-4\cos \theta }\right)$

$=\frac{4-2\cos \theta }{5-4\cos \theta }$