Answer
$1-\dfrac{5^2x^4}{2!}+\dfrac{5^4x^8}{4!}-\dfrac{5^6x^{12}}{6!}...$
Work Step by Step
Consider the Maclaurin Series for $\cos x$ as follows:
$ \cos x=1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+...$
Plug $5x^2$ instead of $x$ in the above series as follows:
$ \cos(5x^2)=1-\dfrac{(5x^2)^2}{2!}+\dfrac{(5x^2)^4}{4!}-\dfrac{(5x^2)^6}{6!}+...$
$\implies 1-\dfrac{5^2x^4}{2!}+\dfrac{5^4x^8}{4!}-\dfrac{5^6x^{12}}{6!}...$
