Sine limit proof Prove that $\lim_{n\to\infty} \sin (n)$ doesn't exist. 
Use $\sin^2 n + \cos^2 n = 1$ and $\sin(2n) = 2\sin (n)\cos(n)$ and $\sin(n+1)=\sin(n)\cos(1) + \cos(n)\sin(1)$.
 A: Suppose that the limit exists, and call it $L$. We have that $L=\lim_{n\to\infty}\sin(n+1)$ and also $L=\lim_{n\to\infty}\sin(2n)$. (Really, any subsequence of $\sin(1),\sin(2),\dots$ will have limit $L$ as well.)
The third of your equations tells us that $\displaystyle \cos(n)=\frac{\sin(n+1)-\sin(n)\cos(1)}{\sin(1)}$, so $\displaystyle \lim_{n\to\infty}\cos(n)=\frac{L(1-\cos(1))}{\sin(1)}$.
The second of your equations then gives us that $L=2L^2(1-\cos(1))/\sin(1)$, which means that either $L=0$, or $\displaystyle L=\frac{\sin(1)}{2(1-\cos(1))}$. 
The first of your equations gives us that $\displaystyle L^2+\frac{L^2(1-\cos(1))^2}{\sin^2(1)}=1$. This immediately implies that $L\ne 0$. So we must have 
 $$\frac{\sin^2(1)}{4(1-\cos(1))^2}\left(1+\frac{1-2\cos(1)+\cos^2(1)}{\sin^2(1)}\right)=1, $$
which (using that $\sin^2(1)+\cos^2(1)=1$), reduces to $\displaystyle \frac1{2(1-\cos(1))}=1$, which means that $\cos(1)=1/2$. But this is false, and we have reached a contradiction.

For different approaches, see here, here, and here.
