Per OP's request, I am making my comment into an answer.
Notice that Kronecker's theorem, as stated here, is essentially equivalent to what I suggested in the comment above.
Define $S^1 = [0,1)$, and we can think of $S^1$ as a circle of unit circumference. Let us define rotation on the circle by $\alpha\in[0,1)$ as $R_\alpha(x) = (x + \alpha)\mod 1$. Then $n$-fold application of $R_\alpha$ to $x\in[0,1]$ is
$$
R_\alpha^n(x) := \underbrace{R_\alpha\circ\cdots\circ R_\alpha}_{n-\text{ times }}(x),
$$
and has the value $(x + n\alpha)\mod 1$, or (as in the theorem of Kronecker referenced above) $(x + n\alpha) - \lfloor x + n\alpha\rfloor$. On the other hand, irrational rotation on $S^1$ is minimal, meaning that for any $x\in S^1$, if $\alpha$ is irrational, then the sequence $\{x_n\}_{n\in\mathbb{N}}$ given by $x_n = R_\alpha^n(x)$ fills $S^1$ densely.
After looking for a quick reference on the internet, I didn't find anything, so let us just prove it here quickly.
Suppose for some irrational $\alpha\in(0,1)$ and some $x\in S^1$, the sequence $\{x_n\}$, as defined above, is not dense in $S^1$. But then there exist open intervals in $S^1$ not containing $x_n$, for any $n\in\mathbb{N}$. Let $I$ be such an interval of maximal length. Obviously $R^n_\alpha(I)$ is also such an interval, of the same length as $I$, for any $n\geq 1$. If $R^n_\alpha(I)$ intersects another such interval, say $J$, and $R_\alpha(I)\neq J$, then we obtain an interval $J\cup R_\alpha^n(I)$ not containing any $x_n$, and of length greater than $I$, which by maximality of $I$ cannot happen. On the other hand, if $R_\alpha^n(I) = J$ and $J = I$, then for rational $q\in I$, we have $q = q + n\alpha \mod 1$, or $\alpha$ is rational, contradicting irrationality of $\alpha$. In other words, the sequence $\{R_\alpha^n(I)\}_{n\in\mathbb{N}}$ is a sequence of disjoint subintervals of $S^1$, each of the same positive length as $I$. But this, of course, cannot happen, since $S^1$ has finite length. We're done.
Now, say $\alpha\in (0,1)$ is irrational (say $\alpha = \pi/2$, for instance). Then the sequence $\{n\alpha\mod 1\}_{n\in\mathbb{N}}$ is dense in $[0, 1)$. In particular, there is a strictly increasing subsequence $\{n_i\}$ of natural numbers such that
$$
\lim_{i\rightarrow \infty}(n_i\alpha\mod 1) = 0.
$$
It follows that
$$
\lim_{i\rightarrow\infty}\left|n_i\alpha - n_i\right| = 0.
$$
This implies that
$$
\lim_{i\rightarrow\infty}\left|\sin(n_i) - \sin(n_i\alpha)\right| = 0.
$$
Recall that $\alpha = \pi/2$. I think you can take it from here.
