Answer
$7-\dfrac{7x^2}{2!}+\dfrac{7x^4}{4!}-\dfrac{7x^6}{6!}+..$
Work Step by Step
Since, we know that the Maclaurin Series for $\cos{x}$ is defined as:
$ \cos x=\Sigma_{n=0}^\infty \dfrac{(-1)^n x^{2n}}{(2n)!}$
Now,
$7 \cos (-x)=7 \cos x=7 \Sigma_{n=0}^\infty \dfrac{(-1)^n x^{2n}}{(2n)!}$
or, $ 7 \cos x=7-\dfrac{7x^2}{2!}+\dfrac{7x^4}{4!}-\dfrac{7x^6}{6!}+..$
