Prove that the limit of $\sin n$ as $n \rightarrow \infty$ does not exist

Using only the delta definition of a limit, how can we prove that the sequence $\{a_n\}$, where $a_n = \sin n$, as $n$ tends to infinity does not have a limit?


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No need for $\epsilon$ actually.
If $\sin(n) \rightarrow l$, then $\sin(n+1)$ also, and $\sin(n+1)=\sin(n)\cos(1)+\sin(1)\cos(n)$.
Since both $\sin(n)$ and $\sin(n+1)$ have limit $l$ and $\sin(1) \neq 0$, $\cos(n) \rightarrow \frac{l(1-\cos(1))}{\sin(1)}$, and so $e^{in}=\cos(n)+i \sin(n)$ has a limit.
But $e^{i(n+1)}$ must then have the same limit (call it $x$), which implies $x=e^{i} x$, and since $e^{i} \neq 1$, $x$ has to be zero, which is a contradiction with the fact that $|e^{in}|=1$.

Assume $\lim \sin(n) = l$. Then so is $\lim \sin(2n) = l$. So $\lim \cos(2n) = 1 – 2l^2$, but so does the limit of $\cos(2(n + 1))$. Now apply the sum-formula to $\sin(2(n + 1) – 2n)$.

The following are true, based on standard trigonometric identities and $\sin(1) \approx 0.84147$ and $\sin(3) \approx 0.14112$:

\textrm{if } \sin(n) \le -0.4, & \textrm{ then } 0 < \sin(n+3) ; \\
\textrm{if } -0.4 \le \sin(n) \le 0.4, & \textrm{ then } \sin(n+1) < -0.4 \textrm{ or } 0.4 < \sin(n+1) ; \\
\textrm{if } 0.4 \le \sin(n),& \textrm{ then } \sin(n+3) < 0;

so there is no value $L$ where for any positive $\varepsilon < 0.2$ you have all of $\sin(n), \sin(n+1), \sin(n+3)$ and $\sin(n+4)$ within $\varepsilon$ of $L$.

Let if possible $\sin n\rightarrow x$. Then $\sin k=\sin(n+k-n)=\sin(n+k)\cos k-\cos(n+k)\sin n\rightarrow x(\cos k-\sqrt{1-x^2})$ for each positive integer k. Now as $k\rightarrow \infty$ implies that $x=x.0=0$ which shows that $\sin k=0$ for all k…a contradiction.