Is $a_n=\sum_{\alpha=0}^n\;\text{round}(\sin\alpha)$ bounded? Is the sequence $(a_n)$ bounded if 
$$a_n=\sum_{\alpha=0}^n\;\text{round}(\sin\alpha)?$$
Edit: round$(x)$ simply means round $x$ to the nearest integer; the case where $x=\pm0.5$ may be ignored since $\pi$ is irrational.
Please see comments for our current progress.
 A: (op asked if he could post bcs it's originally a problem I found out, so I'm going to show what we've globally already found before this post was made).
Obviously round means rounding to the nearest integer. Ok, so basically, the first thing I tried to do was to find patterns. I first showed that
$$\left\lfloor\frac3\pi n-\frac12\right\rfloor=\left\{\begin{array}{ccc}2~\textrm{or}~5&\iff&\textrm{Round}(\sin n)=0\\0~\text{or}~1&\iff&\textrm{Round}(\sin n)=1\\3~\text{or}~4&\iff&\textrm{Round}(\sin n)=-1\end{array}\right.$$
This tool will be useful later. I then noticed that the sequence had a few sporadic cycling patterns which all looked the same (nothing, increment, increment, nothing, decrement, decrement, repeat). found out a way to show where those patterns broke :
$$\begin{array}{cl}
&\left(\left\lfloor\frac3\pi n-\frac12\right\rfloor-n\right)-\left(\left\lfloor\frac3\pi(n-1)-\frac12\right\rfloor-(n-1)\right)=-1\\
\iff&\left\lfloor\frac3\pi n-\frac12\right\rfloor=\left\lfloor\frac3\pi(n-1)-\frac12\right\rfloor
\end{array}$$
these are all the values of $n$ which satisfy this equation. afterwards, did some stuff to find out that the gap $G$ between two solutions is
$$\begin{array}{cl}
&\lfloor1/(1-3/\pi)\rfloor\le G\le\lceil1/(1-3/\pi)\rceil\\
\iff&22\le G\le23
\end{array}$$
Also, the sequence
$$\left(\underset{\alpha\le n,~\alpha\in\mathbb N\setminus\left\{n:\lfloor3n/\pi-1/2\rfloor=\lfloor3(n-1)/\pi-1/2\rfloor\right\}}{\sum\text{Round}\sin\alpha}\right)_{n\ge0}$$
is bounded by 0 and 2, forgot to add that.
Elie Ben Shlomo from the Facebook group "actually good math problems" found out another related question on MathOverflow.
Jack Heimrath from the same group used Birkhoff's ergodic theorem to show that the sum itself is $o(n)$
Griffin Macris, who was really helpful with this problem too, found a lot of results too, but I'd rather let him say what he found. although it's not really his major contribution, he found out that $|a_{12026980763}|=7$ and that's the highest value in the sequence found yet.
Mars Industrial found out this really good paper on arxiv which I don't really understand but sure looks related.
And a lot of other people found a lot of approaches which I unfortunately don't really get.
A: It is not an answer to your question (I believe the answer should be no), but rather to a related question, which is already nontrivial:
Is there a subsequence of $(a_n)_n$ which is bounded ?
The answer is yes.
It is a direct application of the Denjoy-Koksma inequality (see https://en.wikipedia.org/wiki/Denjoy%E2%80%93Koksma_inequality), which in this case states that, if we denote $f:[0,1[\to [0,1[, f(x)=\{x+\frac1{2\pi} \}$ and 
$$\phi(x)=round(sin(2\pi x)),$$
(which is clearly of bounded variation, namely $Var(\phi)=4$, and $\int_0^1 \phi=0$) then for any integer couple $(p,q)$ such that the following diophantine inequality holds
$$\left|\frac1{2\pi}-\frac{p}q\right|\leq \frac{1}{q^2},$$
then
$$|a_{q-1}|=\left| \sum_ {k=0}^{q-1} \phi(f^k (0)) \right|\leq Var(\phi)=4.$$
There exist infinitely many couples $p,q$ satisfying the above diophantine inequality. For example, the convergents of the continued fraction expansion of $1/2\pi$. Thus, there exists a bounded, 'explicit' subsequence of $(a_n)_n$.
