Evaluate Part 1

By Maclaurin series, $\displaystyle 1 + \sum_{\color{#3D99F6}n=1}^\infty \frac {(-1)^nx^{2n}}{(2n)!} = \sum_{\color{#D61F06}n=0}^\infty \frac {(-1)^nx^{2n}}{(2n)!} = \boxed{\cos x}$ .

Proof: According to Maclaurin series, we have:

$\begin{aligned} f(x) & = \sum_{n=0}^\infty \frac {\color{#3D99F6}f^{(n)}(0)}{n!}x^n \quad \quad \small \color{#3D99F6} \text{where } f^{(n)}(x) \text{ is the }n\text{th derivative of }f(x) \\ \implies \cos x & = \sum_{n=0}^\infty \frac {d^n\cos x}{dx^n} \bigg|_{x=0} \cdot \frac {x^n}{n!} \\ & = \frac {\cos 0}{0!} - \frac {\sin 0}{1!}x - \frac {\cos 0}{2!}x^2 + \frac {\sin 0}{3!}x^3 + \frac {\cos 0}{4!}x^4 - \frac {\sin 0}{5!}x^5 - \frac {\cos 0}{6!}x^6 + \frac {\sin 0}{7!}x^7 + \cdots \\ & = 1 - \frac 1{2!}x^2 + \frac 1{4!}x^4 - \frac 1{6!}x^6 + \frac 1{8!}x^8 - \frac 1{10!}x^{10} + \frac 1{12!}x^{12} - \cdots \\ & = \sum_{n=0}^\infty \frac {(-1)^nx^{2n}}{(2n)!} \end{aligned}$