Limit of $\sin 2^n$ I am trying to show that
$$\lim_{n\to \infty}\sin 2^n$$
diverges for
$n \in \mathbb N$
I could show that assuming the limit converges, say to $L$ then
$$L=\lim_{n\to \infty}\sin 2^{n+1}$$
$$=2\lim_{n\to \infty}\sin 2^n\lim_{n\to \infty}\cos 2^n$$
$$=2L\lim_{n\to \infty}\cos 2^n$$
It cannot be
$$\lim_{n\to \infty}\cos 2^n=\frac{1}{2}$$
since it implies
$$\frac{1}{2}=\lim_{n\to \infty}\cos 2^{n+1}$$
$$=2(\lim_{n\to \infty}\cos 2^n)^2-1$$
$$=-\frac{1}{2}$$
So either $$\lim_{n\to \infty}\sin 2^n=0$$
or it diverges.
For the sequence to converge, necessarily it must be that $2^n$ gets  arbitrarily close to $m\pi$ for some integer $m(n)$ as $n$ goes to infinity, which seems counterintuitive. But I couldn't prove it, so anyone has some good idea?
 A: One can see with the double angle formula that
$$\sin2^{n+1}=2\sin2^n\cos2^n=\pm2\sin2^n\sqrt{1-\sin^22^n}\approx\pm2\sin2^n$$
So if $\sin2^n$ gets close to $0$, $\sin2^{n+1}$ will double and get farther from $0$.  Indeed, for if the limit to exist, the double angle theorem says it must be $\sqrt3/2$, which is a contradiction to your statement.
$$L=2L\sqrt{1-L^2}$$

edit:
To be more specific, if $0<|\sin2^n|<\frac12$, then it follows from the above that
$$\sqrt3|\sin2^n|<|\sin2^{n+1}|<2|\sin2^n|$$
A: If, for some $a$, $$
\lim\limits_{n \to \infty} \sin 2^n =\sin a
$$ Introduce $$
b := \pi/2 -a
$$ Suppose, with both $0 <|\epsilon_1| <|\epsilon$ and $0 <|\epsilon_2| < |\epsilon|$, $$
2^n =\pm a \pm 2b +\epsilon_1 \quad (\mathrm{mod}\; 2\pi) \\
2^{n+1} =\pm a \pm 2b +\epsilon_2 \quad (\mathrm{mod}\; 2\pi)
$$ (They may be different numbers, and not necessarily same sign, so there are 4 signs.) 
Then by your construction, if we subtract them, either: $$
2^n =\pm 2a \pm 2b +2|\epsilon_3|\quad (\mathrm{mod}\; 2\pi) \\
$$ where $0 <|\epsilon_3| <|\epsilon$. 
Make $\epsilon$ so small that this is absurd, unless $a =0$ or $a =\pi/2$. 
For $a=\pi/2$, we know $2^n$ is $2k\pi \pm \pi/2 +\epsilon_4$. But then $2^{n+1} =4k\pi +\pi +2\epsilon_4$ (or minor case, which is similar). Make $\epsilon$ so small that this is absurd.  
It remains to deal with $a=0$. Here, similarly,  $2^n =k\pi +\epsilon_5$. If $\epsilon_5 \neq 0$, then times 2 so many times, that $\epsilon_5$ exceeds $\epsilon$.
That $2^n =k\pi$, for some $k$, is outrageous.
(A bit hard to write very clearly, but draw a graph and you will get it. If someone finds a better way to put it, edit by all means.)
A: $\pi\not\in \mathbb Q.$   Let $2^n=k_n\pi +d_n$ where $k_n\in \mathbb Z$ and  $0<|d_n|< \pi /2.$
Suppose $\lim_{n\to \infty}\sin 2^n=0.$ Then $\lim_{n\to \infty} |d_n|=0.$
If $n\geq n_0\implies |d_n|<\pi /8,$ consider $j\in \mathbb N$ such that $2^j|d_{n_0}|<\pi /2<2^{j+1}|d_{n_0}|.$ Then $$2^{j-1}|d_{n_0}|=(1/2)\cdot 2^j|d_{n_0}|<\pi /4$$ so we have$$|d_{n_0+j-1}|=2^{j-1}|d_{n_0}|.$$
Then  $$\pi /8<(1/4)\cdot 2^{j+1}|d_{n_0}|=2^{j-1}|d_{n_0}|=|d_{n_0+j-1}|<\pi/8$$  a contradiction.
A: Hint:
$$2^n=2\pi\frac{2^n}{2\pi}$$ so that the quadrant of the angle $2^n$ is given by the first two bits of the fractional part of $\dfrac{2^n}{2\pi}$, which are nothing but the $n^{th}$ and $n+1^{th}$ bits of the fractional part of $\dfrac1{2\pi}$, a transcendental number.
If you can show that all bit pairs occur infinitely many times, you are done.
