TITLE: Can $\sin n$ get arbitrarily close to $1$ for $n\in\mathbb{N}?$
QUESTION [14 upvotes]: 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?

REPLY [3 votes]: 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 $k(i-j)\mod\pi$.