Is my understanding of this theorem on the limit of a sequence correct? I want to know if I understood the theorem I need to use on this exercise correctly, so let me tell the theorem and then the problem.
$Theorem$:
If $\lim_{n\to\infty}a_n=L$ and $f$ is continuous in $L$, then $\lim_{n\to\infty}f(a_n)=f(L)$.
So the theorem states that, provided the sequence $a_n$ approaches a number $L$ when $n$ goes to infinity, then I can take any function $f$ that's continuous on $L$ and know that $\lim_{n\to\infty}f(a_n)$ will be the same than$f(L$).
$Exercise$
Let $a_n=\sin(\frac{\pi}{n})$. Find $\lim_{n\to\infty}a_n$.
Of course I could solve this by determining that $\frac{\pi}{n}\to0$ when $n\to\infty$, so that $\lim_{n\to\infty}\sin(\frac{\pi}{n})=\sin(0)=0$.
But using the theorem I could also take some function that's continuous on $\mathbb{R}$, for example $f(x)=\sin(x)$. Now evaluate $\lim_{n\to\infty}f(a_n)=f(0)=0.$
The information this would give me is that, since $\lim_{n\to\infty}f(a_n)=f(L)$, $f(L)=0$. This means $\sin(L)=0$.
Solving $\sin(L)=0$ I get $L=0$.
So after all, the ultimate "goal" of this theorem seems to be to provide us with this last equation we can solve for $L$, but it feels a bit redundant since, in order to get this equation, I have to solve the limit of the sequence first (so I already now $L$)! I get the feeling that I'm overcomplicating it. Is this a right understanding and approach to the theorem? 
 A: 
Of course I could solve this by determining that $\frac πn→0$ when $n→∞$, so that $\lim_{n→∞}\sin(\frac πn)=\sin(0)=0$

Not if you don't know the theorem in the first place, you can't!  
You are ASSUMING that $\lim_{n\to a} f(a_n)=f(\lim_{n \to a} a_n)$.  Why are you assuming that?
Consider $f(x) = \lceil x \rceil $ (the next integer equal or more than $x$) so that $f(0) = 0$ but $f(\frac 1n) = 1$ for every $n$.  
Then $\lim_{n\to \infty}  f(\frac 1n) = \lim_{n\to \infty} \lceil \frac 1n \rceil=\lim_{n\to\infty} 1 = 1$.
So your argument that $\frac 1n \to 0$ means $f(\frac 1n) \to f(0)=0$ doesn't hold.
You can NOT assume that.
The theorem is saying nothing more or less than you are allowed to assume that if $f$ is continuous.
($\lceil x \rceil$ is not continuous.)

So after all, the ultimate "goal" of this theorem seems to be to provide us with this last equation we can solve for L, but it feels a bit redundant since, in order to get this equation, I have to solve the limit of the sequence first (so I already now L)!

Yes.  You know $a_n \to L$ and you know $f(L)$.  But you don't know what $f(a_n)\to ????$ is.  The theorem is utterly essential to know that if $a_n \to L$ then we know that $f(a_n) \to f(L)$ (if $f$ is continuous).  
No redundancy.  Pure essentiality.
I think your issue is that you were taking the theorem for granted but not realizing it.
A: The purpose of the theorem is to exchange limits and continuous functions.  Under the theorem's hypotheses,
$$  \lim_{n \rightarrow \infty} f(a_n) = f \left( \lim_{n \rightarrow \infty} a_n \right)  \text{.}  $$
A: You say:

Of course I could solve this by determining that $\frac{\pi}{n}\to0$ when $n\to\infty$, so that $\lim_{n\to\infty}sin(\frac{\pi}{n})=sin(0)=0$.

In solving it this way you are using the theorem. $a_n=\frac\pi n$, and $L=\lim\limits_{n\to\infty}a_n=0$. The theorem guarantees that
$$
\begin{align}
\lim_{n\to\infty}\sin\left(\frac\pi n\right)
&=\sin(L)\\
&=\sin(0)\\[6pt]
&=0
\end{align}
$$
A: The utility of the theorem in this exercise is to allow you to use 
$\lim_{n\to\infty}\frac{\pi}{n}=0$ 
in order to infer that
$\lim_{n\to\infty}\sin(\frac{\pi}{n})=\sin(\lim_{n\to\infty}\frac{\pi}{n})=0$. 
Note that as $a_n = \sin(\frac{\pi}{n})$, L actually refers to the limit of this sequence, so by writing $\lim_{n\to\infty}f(a_n)=f(L)$ you are actually writing $\lim_{n\to\infty}f(\sin(\frac{\pi}{n}))=f(L)$,
Also note that you cannot infer L = 0 from $\sin(L) = 0$
