I am reading about hyperreal numbers defined as (to my understanding) certain equivalence classes on all sequences of real numbers. $ω$ is defined as $(1, 2, 3, ...)$, and all functions are applied element-wise. This makes sense for sequences that have an infinite limit, like $e^ω$, which is simply a bigger infinity than $ω$, but it occurs to me that there must be certain "abominable" numbers that don't fit into the standard notion of reals, omegas, and epsilons.
By the transfer principle, values like $\sin{ω} = (\sin{1},\sin{2},\sin{3},...)$ and $\frac{\sin{ω}}{ω}=(\frac{\sin{1}}{1},\frac{\sin{2}}{2},\frac{\sin{3}}{3},...)$ should exist and be finite/limited. However, for the first value I am not able to show that it is equal to a real number, or at least infinitely close to a real number (which is a necessity according to the paper I am reading), and for the second number, although it is infinitely close to $0$, its sign is ambiguous, breaking trichotomy (in the same way as $\sin{ω}$) which should hold based on the transfer principle.
So, if $\sin{ω} = r$, what is the actual value of $r$?
Your questions show that you have yet to internalize the following important fact: your source constructs the hyperreals $\:^\star\mathbb{R}$ in Section 1.3. by first fixing a non-principal ultrafilter $\mathcal{F}$ over $\mathbb{N}$.
You use $\omega$ to denote the fixed hyperreal arising from (the equivalence class of) the sequence $(0,1,2,\dots)$. But the properties of fixed hyperreals such as $\omega$ actually depend on the precise identity of the ultrafilter $\mathcal{F}$ you used to construct $\:^\star\mathbb{R}$.
Before we look at trigonometric functions, you should first think about something simpler, say the function $f: \mathbb{R} \rightarrow \mathbb{N}$ given as follows: $$f(x) = 1 \text{ if }x\text{ is an odd integer and }f(x)=0\text{ otherwise.}$$
By Section 1.6. of your source, we can get an extended function $^\star\!f: \:^\star\mathbb{R} \rightarrow \:^\star\mathbb{N}$. But what is the value of $^\star\!f(\omega)$? Since $\forall x \in \mathbb{R}. f(x) = 0 \vee f(x) = 1$, the Transfer principle says that either $^\star\!f(\omega) = 0$ or $^\star\!f(\omega) = 1$. But which?
It turns out that the answer depends on the ultrafilter $\mathcal{F}$ you used to construct $\:^\star\mathbb{R}$. You should check that if the ultrafilter contains the set of odd natural numbers, then $^\star\!f(\omega) = 1$, and if instead the ultrafilter contains the set of even natural numbers, then $^\star\!f(\omega) = 0$. Since for any set $A \subseteq \mathbb{N}$ $\mathcal{F}$ contains either $A$ or $\mathbb{N}\setminus A$, it must be one of these.
Once you understand these results, we can move on to your questions about the $\sin$ function.
Of course $\sin(\omega)$ is infinitesimally close to some real number. This follows immediately from the general result that every limited hyperreal has a shadow (Theorem 3.3. in your source). And of course $\sin(\omega)$ is limited, since we have $\forall x. |\sin(x)| < 2$.
But which real number $r$ is $\sin(\omega)$ infinitesimally close to? Well, that depends strongly on the non-principal ultrafilter $\mathcal{F}$. In fact, since the image of $\sin(\mathbb{N})$ is dense in the interval $[-1,1]$, you can choose any number $x \in [-1,1]$ and find an ultrafilter $\mathcal{F}$ such that $\sin(\omega) \approx x$ in the hyperreal field constructed using $\mathcal{F}$.
For similar reasons, the value of $\mathrm{sgn}\left(\frac{\sin(\omega)}{\omega}\right)$ is very far from ambiguous: it is either positive (so the $\mathrm{sgn}$ function takes the value $1$ or negative ($-1$), and not both. To deduce which possibility holds, you have to know more about the non-principal ultrafilter that was used to construct $\mathcal{F}$: if $\mathcal{F}$ contains the set $\left\{n \in \mathbb{N} \:|\: \frac{sin(n)}{n} > 0\right\} \subseteq \mathbb{N}$ then $\frac{\sin(\omega)}{\omega} > 0$ (exercise!) so its sign is positive. If $\mathcal{F}$ contains the complement of this set, then the sign is negative (exercise: why can't it be zero?). And $\mathcal{F}$ must contain one of these, by virtue of being an ultrafilter.