The trigonometric functions being expressed as an infinite series is something I never really understood. I understand that they can be expressed as infinite series but I never actually understood the proof. Can someone explain how we arrive to the following infinite series? I've never seen the derivation.
$${\sin}(x)=\sum_{n=0}^{\infty }\frac{(-1)^nx^{2n+1}}{(2n+1)!}$$
$${\cos}(x)=\sum_{n=0}^{\infty }\frac{(-1)^nx^{2n}}{(2n)!}$$
I think you've seen the derivation, you just failed to understand it. Naturally, it depends on the definition of the trigonometric functions. Usually, those are introduced geometrically, and with geometry, the addition theorems for $\sin$ and $\cos$ are proved, together with inequalities $$\sin x\le x\le\tan x$$ for $x\in[0,\pi/2)$. From there, you can obtain the derivatives of $\cos$ and $\sin$, and that gives the Taylor series. Known theorems show that the latter are globally convergent. So the logical structure is not all that complicated.
Of course, you can as well start from the series, or from the differential equations, and prove that those functions describe the unit circle (and I've seen textbooks doing that), but it's a bit artificial, since most students learned about trigonometric functions before power series or differential equations.
In a calculus course, you will learn about Taylor series. These are two examples of Taylor series. So wait until you study calculus; there you will find the derivations for these.
$$ \begin{align} \cos0 & = 1 \\ \cos'0 & = 0 \\ \cos''0 & = -1 \\ \cos'''0 & = 0 \\ \text{Then repeat} & \text{ this sequence of four.} \end{align} $$ The problem then is: If $\sum_{n=0}^\infty c_n x^n$ satisfies this same repeating sequence of values of its derivative at $0,$ then what is $c_n \text{?}$ You will find that $c_n$ is $n!$ times the value of the $n$th derivative at $0.$
