Why is $\sin{x} = x -\frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$ for all $x$? I'm pretty convinced that the Taylor Series (or better: Maclaurin Series):
$$\sin{x} = x -\frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$$
Is exactly equal the sine function at $x=0$
I'm also pretty sure that this function converges for all $x$
What I'm not sure is why this series is exactly equal to the sine function for all $x$. 
I know exactly how to derive this expression, but in the process, it's not clear that this will be equal the sine function everywhere. Convergence does not mean this will be equal, it just mean that it will have a defined value for all $x$.
Also, I would want to know: is this valid for values greater than $\frac{\pi}{2}$? I mean, don't know how I can proof that this Works for values greater than the natural definition of sine.

First, let's take the Taylor's polynomial $\displaystyle T_n(x) = \sum\limits_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(x-a)^k$ at a given point $a$. We can now say:
$\displaystyle R_n(x) = f(x) - T_n(x)$, where $R_n$ can be called the remainder function.
If we can prove that $\lim\limits_{n\rightarrow \infty}R_n(x) = 0$, then $f(x) = \lim\limits_{n\rightarrow\infty}T_n(x)$, that is, Taylors series is exactly equal to the function. Now we can use the Taylor's theorem:
$\mathbf{Theorem\,\, 1}$ If a function is $n+1$ times differentiable in an interval $I$ that contains the point $x=a$, then for $x \in I$ there exists $z$ that is strictly between $a$ and $x$, such that:
$\displaystyle R_n = \frac{f^{(n+1)}(z)}{(n+1)!}(x-a)^{(n+1)}$.  (This is known as the Lagrange remainder.)$\hspace{1cm}\blacksquare$
Ok, now for every Taylor's Series we want, we must prove that $\lim\limits_{n\rightarrow \infty}R_n(x) = 0$ in order for the series to be exactly equal to the function.


Example: Prove that for $f(x) = \sin(x)$, Maclaurin's series (Taylor's series with $a=0$) is exactly equal to the function for all $x$.
First, we know that $\displaystyle\left|\, f^{(n+1)}(z)\right| \leq 1$, because $\left|\sin(x)\right|\leq 1$, and $|\cos(x)|\leq 1$. So we have:
$\displaystyle 0\leq \left|R_n\right| = \frac{\left|f^{(n+1)}(z)\right|}{(n+1)!}\left|x\right|^{n+1} \leq \frac{\left|x\right|^{n+1}}{(n+1)!}$
It's easy to prove that $\displaystyle\lim\limits_{n\rightarrow\infty}\frac{\left|x\right|^n}{n!} = 0$ for all $x\in \mathbb{R}$ (for example with D'Alambert ratio criterion for series convergence) Therefore, according to the squeeze theorem (In my country we call it the Two Policemen and the Burglar theorem :D), it follows that $\left|R_n\right| \rightarrow 0$ when $n\rightarrow\infty$. This in turn is equivalent to: 
$\displaystyle R_n\rightarrow 0$ as $n\rightarrow\infty$ (because $\left|R_n - 0\right| = \left|\left|R_n\right| - 0\right|$).
Therefore, according to the Theorem 1, it holds that $f(x) = \lim\limits_{n\rightarrow\infty}T_n$, for all $x$ in the radius of convergence.
**
Now, let's find the radius of convergence:
$\mathbf{Theorem\,\, 2}$ For a Maclaurin Series of a function: 
$f(x) = \displaystyle\sum\limits_{k=0}^{+\infty}c_k x^k = c_0 + c_1x + c_2x^2 + \dots$
The equality holds for $x$ in the radius of convergence:
$\displaystyle\frac{1}{R} = \limsup\limits_{k \rightarrow \infty} \left|c_k\right|^{\large{\frac{1}{k}}}$   (Cauchy-Hadamard formula)
where $R$ is the radius of convergence of the given function, or more concretely, the power series converges to the function for all $x$ that satisfies:  $\displaystyle\left|\,x\, \right| < R$, where $R\in [0,+\infty]$. $\hspace{1cm}\blacksquare$
Let's look at our given example for $\sin(x)$. We have:
$\displaystyle\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots = \sum\limits_{k=0}^{+\infty}c_k x^{2k+1}$
where $\displaystyle c_k = \frac{(-1)^k}{(2k+1)!}$.
Substituting for $R$ we get: 
$\displaystyle \frac{1}{R} = \limsup\limits_{k\rightarrow\infty}\frac{1}{\sqrt[k]{(2k+1)!}}=0,$ (this can be proven with Stirling's approximation) and
$\displaystyle\implies \boxed{R = +\infty}$
So, our series converges for all $\displaystyle|x| < +\infty \, \Longleftrightarrow\, \boxed{ -\infty < x < +\infty}$, which we expected.
Finally, we have that $f(x) = \lim\limits_{n\rightarrow\infty}T_n$, for all $x\in\mathbb{R}$, which is what we wanted to prove. $\blacksquare$
