Show that $\cos(2^n)$ diverges Problem:
I have sequence defined as follows:
$$
a_n = \cos(2^n)
$$
I need to show that it diverges.
My progress:
I tried common method here supposing that this sequence has limit $\lim_{n\to\infty}\cos(2^n) = a$ and then trying to get to contradiction with $\lim_{n\to\infty}(\cos(2^{n+1})-\cos(2^n))=0$. But it led me nowhere.
How can I show that this sequence diverges?
 A: $\textbf{Claim:}$ If $a_{n}$ converges then it converges to either $1$ or $-\frac{1}{2}$.
Proof: Suppose it converges to some $c$ then for each $\epsilon > 0$ $\exists N_{\epsilon} \in \mathbb{N}$ so that if $n \geq N_{\epsilon}$ we have
$$|a_{n+1}-a_{n}| < \epsilon$$
$$|2a_{n}^2-a_{n}-1|<\epsilon$$
This means that $c$ lies an arbitrarily small neighbourhood of one of the roots of the polynomial $2x^2-x-1$ which are $1$ and $-\frac{1}{2}$ respectively as we can take $\epsilon \rightarrow 0$.
$\textbf{Claim:}$ $a_{n}$ does not converge to $-\frac{1}{2}$.
Suppose it converges to $-\frac{1}{2}$ then
$$a_{n+1}+\frac{1}{2} = 2a_{n}^2-1+\frac{1}{2} = 2a_{n}^2-\frac{1}{2} = 2(a_{n}+\frac{1}{2})(a_{n}-\frac{1}{2})$$
Let us choose $M \in \mathbb{N}$ so that for all $n \geq M$ we have $|a_{n}-\frac{1}{2}| \geq \frac{3}{4}$. For $n \geq M$ one has
$$  |a_{n+1}+\frac{1}{2}| \geq \frac{3}{2}|a_{n}+\frac{1}{2}| $$
One must clearly have $a_{M} = \frac{1}{2}$ because otherwise
$\liminf_{k \rightarrow\infty}|a_{k}+\frac{1}{2}| \geq (\frac{3}{2})^{k-M}|a_{M}+\frac{1}{2}| = \infty.$
But if $a_{M} = \frac{1}{2}$ then $2^{m} \equiv \frac{2\pi}{3} \mod 2\pi$ or $2^{m} \equiv \frac{4\pi}{3} \mod 2\pi$; this would imply that $\pi$ is rational which is impossible. This is a contradiction.
$\textbf{Claim:}$ $a_{n}$ does not converge to $1$.
Suppose it converges to $1$ then
$$a_{n+1}-1 = 2a_{n}^2-2 = 2(a_{n}-1)(a_{n}+1)$$
Let us choose $K \in \mathbb{N}$ so that for $n \geq K$ we have $|a_{n}+1| \geq \frac{3}{2}$.
For $n \geq K$ one has
$$|a_{n+1}-1| \geq 3|a_{n}-1|$$
One must have $a_{K} = 1$ because otherwise we have
$$\liminf_{k \rightarrow \infty}|a_{k}-1| \geq 3^{k-K}|a_{K}-1| = \infty$$
But if $a_{K} = 1$ then $2^{K} \equiv 0 \mod 2\pi$; this implies that $\pi$ is rational. This is a contradiction.
