Even integer approximations to multiples of pi I admit that I'm probably out of my depth with this question, but I can't help but feel curious. 
I wanted to show that, in the sequence $\{\sin(n)\}$, there is never a largest term (the sequence never attains its limit superior). My reasoning was that, given $$ \left| \sin(x) - \sin(y) \right| \leq \left| x - y \right|,$$ if we can furnish an even $M$ with $M = N\pi \pm \epsilon$, then either $\frac{3}{2}M$ or $\frac{1}{2}M$ will nearly equal an odd multiple $K$ of $\frac{\pi}{2}$ such that $\sin(K\frac{\pi}{2}) = 1$.
By the inequality, the difference between $\sin(\frac{3}{2}M)$ and $1$ -- or $\sin(\frac{1}{2}M)$ and $1$, whichever -- would be at most $\frac{3}{2} \epsilon$. Thus, the problem reduces to showing that we can get an arbitrarily small $\epsilon$. 
(I recognize that the inequality above is pretty watered down: the mean value theorem shows that the inequality is as stark as $\cos(\xi) \leq 1$ for $\xi \in (x,y)$, which, if both points $x$ and $y$ are close to $(2k+ \frac{1}{2}) \pi$, is really much stronger than what I've got. This seems like a hard way to prove the claim, so if anybody has a better one, I'd also like to hear about that.)
But my main question, which I came to because of the above, is about approximating multiples of $\pi$ by integers. If $\pi = \frac{p}{q} + \epsilon$, then $q\pi - q\epsilon = p$; thus, the size of $q$ becomes important to the accuracy, since the $\epsilon$ we were considering above is $q\epsilon$ in these terms. 
Spivak's Calculus has a little discussion about this when he proves $e$ is transcendental. He notes that the proof of $e$'s irrationality shows that $\sum_{k=1}^n \frac{n!}{k!} = n!e - R_n,$ with $|R_n| < \frac{3}{n+1}$. The sum on the left can be controlled for parity, since choosing $n$ odd leaves $(\cdots + n + 1)$ at the tail of this sum, and all other terms multiplied by $(n-1)$. So there must exist, given $\epsilon > 0$, an $N$ and an even $M$ such that $M = Ne \pm \epsilon$.
(In other words, if $e$ were $\pi$, I'd be home already!) 
Spivak mentioned that this property - good approximations existing with small denominators - is somehow characteristic of transcendental numbers. 

"The number $e$ is by no means unique in this respect: generally speaking, the better a number can be approximated by rational numbers, the worse it is."

So I wonder:


*

*Can we furnish an approximation $\frac{p}{q}$ to $\pi$ with $q\epsilon$ arbitrarily small? (For my purposes, can we do better and furnish one with an even $p$?)

*More generally (and I am out of my depth here, but would enjoy references), what can we prove about the "goodness" of rational approximations to transcendental numbers, in the sense of small denominators?

 A: The following general result of Dirichlet in particular answers your question about approximations to $\pi$. Let $\alpha$ be irrational. Then there exist infinitely many (reduced) fractions $\frac{p}{q}$ such that
$$\left|\alpha-\frac{p}{q}\right|\lt \frac{1}{q^2}.$$
For discussion of the general topic, and further links, please see this Wikipedia entry.
A: In the spirit of experimental mathematics, I did a calculation of the quantity $|q\epsilon|$ for the first $100$ rational approximations for $\pi$ up to $10^{-n}$ accuracy, and it looks like it in fact decreases exponentially:

So if this trend continues, you can indeed approximate $\pi$ by large enough integers. Of course, this isn't a proof or full solution to your question, just a suggestion in the right direction.
A: Here's an alternative approach to your original problem
that doesn't use Diophantine approximation.
A topological result (cf. Kelley's Topology, page 58) says that an
 additive subgroup $G$ of the real line is either dense, or of the form $r\mathbb{Z}$ for
 some $r\in\mathbb{R}$.   If we consider the group $G$ generated by $1$ and $2\pi$, we see
 that the purported $r$ would be  both rational and irrational. Such $r$ doesn't exist so $G$ must be dense.
Letting $(m_n)$ and $(\ell_n)$ be integer sequences with $m_n+\ell_n\,2\pi\to\pi/2$, 
we see that $\sin(m_n)=\sin(m_n+\ell_n\,2\pi)\to\sin(\pi/2)=1$. 
