I have a couple of questions about an example from Real Analysis with Real Applications by Davidson and Donsig:
$(1)$ I was wondering if someone could elaborate what it means to say that the angles $n$ marked on a circle appear to gradually fill in a dense subset?
$(2)$ Why does it follow that there a positive integer $t$ such that $|\theta - t\psi| < \dfrac{1}{k+1}?$
Example $2.6.6$ Consider the sequence $(a_n) = (\sin n)_{n=1}^{\infty}$. As the angles $n$ radians for $n \ge 1$ are marked on a circle, they appear gradually to fill in a dense subset. If this can be demonstrated, we should be able to show that $\sin \theta$ is a limit of a subsequence of our sequence for any $\theta$ in $[0, 2\pi]$. The key is to approximate the angle $0 \mod 2\pi$ by integers. This can be done using the Pigeonhole Principle. Let $m$ be a positive integer and $\epsilon > 0$ be any positive real number. Choose an integer $N$ so large that $N\epsilon > 2\pi$. Divide the circle into $N$ arcs of length $2\pi/N$ radians each. Then consider the $N +1$ points $0, m, 2m, \cdots , Nm \mod 2\pi$ on the circle. Since there are $N + 1$ points distributed into only $N$ arcs, at least one of them contains two points, say $im$ and $jm$, where $i < j$. Then $n = jm − im$ represents an angle of at most $2\pi/N < \epsilon$ radians up to a multiple of $2\pi$. That is, $n = \psi + 2\pi s$ for some integer $s$ and real number $|\psi| < \epsilon$. In particular, $|\sin n| < \epsilon$ and $n \ge m$. Moreover, since $\pi$ is not rational, $n$ is not an exact multiple of $2\pi$
So given $\theta \in [0, 2\pi]$, construct a subsequence as follows. Let $n_{1} = 1$. Recursively we construct an increasing sequence $n_{k}$ such that $|\sin n_{k} − \sin \theta| <\dfrac{1}{k}.$
Once $n_{k}$ is defined, take $\epsilon = \dfrac{1}{k+1}$ and $m = n_{k} + 1$ in the previous paragraph. This provides an integer $n > n_{k}$ such that $n = \psi + 2\pi s$ and $|\psi| <\dfrac{1}{k+1}$. Thus there is a positive integer $t$ such that $|\theta − t\psi| < \dfrac{1}{k+1}$. Therefore
$(2.6.7)$ $|\sin(tn) − \sin(\theta)| = |\sin(t\psi) − \sin(\theta)| < |t\psi − \theta| <\dfrac{1}{k + 1}$
Set $n_{k+1} = tn$. This completes the induction. The result is a subsequence such that $\lim_{k \to \infty} \sin(n_{k}) = \sin \theta$.
(1) That the angles are dense means that, given any point/angle $\theta$ on the circle, and any positive distance $\varepsilon,$ there will be some positive integer angle $n$ on the circle whose distance to $\theta$ is at most $\varepsilon$. That is, you can approximate any point/angle on the circle as closely as you want (even though the integer angles may never actually equal $\theta$)
(2) This is actually a minor error on the author's part, since as written there might not be a positive integer $t.$ However, let's assume for the moment that $0 < \psi < \frac{1}{k+1}$ to see why it works in that case (we can fix the error later).
Now consider the values $0, \psi, 2\psi, 3\psi, \ldots$ Since $\theta > 0$ and and $\psi > 0,$ this sequence of values will eventually become larger than $\theta,$ and if we let $t\psi$ be the first element in the sequence which is greater than or equal to $\theta,$ then we have $$(t-1)\psi < \theta \leq t\psi$$ Then we subtract $t\psi$ from both sides and apply the absolute value to find $$-\psi < \theta - t\psi \leq 0 \\ |\theta-t\psi| < |\psi| < \frac{1}{k+1}$$
Fixing the minor error: As you might notice, choosing a positive $t$ this way requires $\psi > 0,$ but this isn't guaranteed by anything in the proof. If instead $\psi < 0,$ we should note that $\theta - 2\pi < 0$ corresponds to the same sine value (i.e. $\sin(\theta - 2\pi) = \sin(\theta)$) and then we can choose the smallest positive integer $t$ such that $$t\psi \leq \theta-2\pi < (t-1)\psi \implies |t\psi - (\theta-2\pi)| < \frac{1}{k+1}$$ From this point, the proof is the same until the last line, when we must write $$\left|\sin(tn) - \sin(\theta)\right| = \left|\sin(t\psi) - \sin(\theta-2\pi)\right| < \left|t\psi - (\theta-2\pi)\right| < \frac{1}{k+1}$$