Let $P_n$ be the point on the unit circle with coordinates $(\cos n, \sin n)$. The points $P_n$ are all distinct. For if $P_a=P_b$, where $a\ne b$ then $a$ and $b$ differ by a multiple of $2\pi$. This means that $a-b=2k\pi$ for some integer $k$. It follows that $\pi=\frac{a-b}{2k}$, which is impossible, since $\pi$ is irrational.
Now let $\epsilon$ be a (very small) positive real, and let $N\gt \frac{2\pi}{\epsilon}$. Consider the $N$ points $P_1, P_2,\dots,P_N$. They are all distinct, so two of them, say $P_a$ and $P_{a+t}$, must be distance $\lt \epsilon$ from each other, along the unit circle. Let this distance be $\delta$.
Let $Q$ be any point on the unit circle. Consider the sequence of points $P_a,P_{a+t}, P_{a+2t},\dots$. These form a sequence of points that travel clockwise or counterclockwise around the circle, separated by $\delta$. So one of them will be within $\delta$ of $Q$.
We have shown that for any $\epsilon \gt 0$, there are infinitely many points in our sequence $P_1,P_2,P_3,\dots$ that are within distance $\epsilon$ of $Q$. So every point on the unit circle is a limit point of our sequence $(P_n)$.
As a consequence, the set of subsequential limits of the sequences $(\cos n)$ and $(\sin n)$ are each equal to the full interval $[-1,1]$.
Remark: From the wording of the question, it is not clear to me whether you are expected to prove that the sequence $(P_n)$ is dense in the unit circle. Perhaps you are just expected to guess that it is, and draw the appropriate conclusion about subsequential limits of $(\sin n)$.