Answer
$\dfrac{1}{2}+\dfrac{1}{2}[1-\dfrac{(2x)^2}{2!}+\dfrac{(2x)^4}{4!}-\dfrac{(2x)^6}{6!}+..... +\dfrac{(-1)^n(2x)^{2n}}{(2n)!}]$
Work Step by Step
Consider the formula:$\cos^2 x=\dfrac{1+\cos 2x}{2}=\dfrac{1}{2}+\dfrac{\cos 2x}{2}$
We know that the Maclaurin series for $\cos x=\Sigma_{n=0}^{\infty} (-1)^{n} \dfrac{(x)^{2n}}{(2n)!}=1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+..... $
Replace $x$ with $2x$ in the above series.
$\cos 2 x=1-\dfrac{(2x)^2}{2!}+\dfrac{(2x)^4}{4!}-\dfrac{(2x)^6}{6!}+..... +\dfrac{(-1)^n(2x)^{2n}}{(2n)!} $
Thus, our required series is:
$f(x)=\dfrac{1}{2}+\dfrac{\cos 2x}{2}=\dfrac{1}{2}+\dfrac{1-\dfrac{(2x)^2}{2!}+\dfrac{(2x)^4}{4!}-\dfrac{(2x)^6}{6!}+..... +\dfrac{(-1)^n(2x)^{2n}}{(2n)!}}{2}\\=\dfrac{1}{2}+\dfrac{1}{2}[1-\dfrac{(2x)^2}{2!}+\dfrac{(2x)^4}{4!}-\dfrac{(2x)^6}{6!}+..... +\dfrac{(-1)^n(2x)^{2n}}{(2n)!}]$
