Answer
$\cos \ 4 x =\Sigma_{n=0}^{\infty} (-1)^{n} \dfrac{4^{2n}(x)^{2n}}{(2n)!}$
Work Step by Step
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 $4 x$ in the above series.
$\cos 4 x=1-\dfrac{(4x)^2}{2!}+\dfrac{(4x)^4}{4!}-\dfrac{(4x)^6}{6!}+..... =1-2x^2+\dfrac{2x^4}{3}-.......$
Thus, our required series is:
$\cos \ 4 x =\Sigma_{n=0}^{\infty} (-1)^{n} \dfrac{4^{2n}(x)^{2n}}{(2n)!}$
