Answer
$1+\Sigma_{n=1}^\infty (-1)^n\dfrac{ 2^{(2n-1)}x^{(2n)}}{(2n)!}$
Work Step by Step
Consider the Maclaurin Series for $\cos x$ is defined as:
$ \cos x=1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+...$
Use formula such as: $1+\cos 2x=2 \cos^2 x$
Then, we have $\cos^2 x=\dfrac{1}{2}+\dfrac{\cos 2x}{2}=\dfrac{1}{2}+1-\dfrac{(2x)^2}{2!}+\dfrac{(2x)^4}{4!}-\dfrac{(2x)^6}{6!}+...$
$\implies 1+\Sigma_{n=1}^\infty (-1)^n\dfrac{ 2^{(2n-1)}x^{(2n)}}{(2n)!}$
