Can $\sin n$ get arbitrarily close to $1$ for $n\in\mathbb{N}?$ Or put differently, does
$$\lim_{n \to \infty}\big(\max \{\sin 1, \sin 2, \ldots ,\sin n\}\big) = 1?$$
My intuition says yes, but how can one prove this?
 A: In fact, the magic element of the proof is Hurwitz's theorem : there exists a (universal) constant $c$ such that for every irrational $\zeta$ there are infinitely many rational approximants $\frac{m}{n}$ with $\left|\zeta-\dfrac{m}{n}\right| \lt \dfrac{c}{n^2}$; multiplying this through by $n$ we get that there are infinitely many $\frac{m}{n}$ with $\left|n\zeta-m\right|\lt\frac{c}{n}$.  (More accurately, this needs a slight modification of Hurwitz's theorem which says there are infinitely many such approximants with $n$ odd - but this is actually an easy consequence of the usual proof of the theorem via continued fractions.)  Now, apply this with $\zeta=\frac{\pi}{2}$; we then get that the integer $m$ is less than $\frac{c}{n}$ away from an odd multiple of $\pi/2$; if $\sin m$\lt 0 (i.e., if we have one of the 'wrong' multiples) then we can just replace $m$ with $-m$ to get a positive value for sin.  Finally, this approximation lets us bound the value of $\sin m$ from below by $1-\frac{c}{n}$.
A: Here is a proof of denseness of the sequence $k\mod\pi$ in the interval $[0,\pi]$ using the pigeon hole prinicple - possibly, the one alluded to in the comments above.
Let $\varepsilon>0$ and choose $n\in{\mathbb N}$ such that $\pi/n<\varepsilon$.  By the pigeon hole principle, at least one of the $n$ intervals in the set $\cal I$ defined by
$${\cal I}=\{[m/n,(m+1)/n]\}_{m=0}^{n-1}$$
must contain two of the first $n+1$ terms from the sequence $k\mod\pi$.  Let's call those terms $i\mod\pi$ and $j\mod\pi$.  We then have $(i-j)\mod\pi<1/n$ so that each interval in $\cal I$ contains a number of the form $k(i-j)\mod\pi$.
A: If $\alpha$ is any irrational number, then $\{a+b\alpha: a,b\in\mathbb Z\}$ is dense in $\mathbb R$.
In particular, with $\alpha=2\pi$, we can find $a+2b\pi$ which is arbitrarily close to $\frac{\pi}2$, and thus $\sin a$ can be made arbitrarily close to $1$.
The above claim, of course, requires proof, but it is true.
The key is that if you take the continued fraction expansion of $\alpha$ and write the $m$th convergent as $p_m/q_m$ then $|p_m-\alpha q_m|<\frac{1}{2q_{m-1}}$. Then given any $x\in\mathbb R$, define $$d_m = \left\lfloor \frac{x}{|p_m-\alpha q_m|}\right\rfloor$$. Then $$d_m|p_m-\alpha q_m| < x < (d_m+1)|p_m-\alpha q_m|$$ But the difference between the right and left side is less than $\frac{1}{2q_{m-1}}$, and we're done.
As Julien commented, you actually need $a>0$. That's not a hard condition to resolve. We can just pick even convergents, so that $0<p_{2m}-\alpha q_{2m}<\frac{1}{2q_{2m-1}}$. Then do the above computation without the absolute values. Ultimately, your $n=d_{2m}p_{2m}$.
