# $\cos(n\vartheta)=\frac{a_n}{3^n}$

I want to show,

if i know that $$\cos(\vartheta)=\frac{1}{3}$$ than $$\cos(n\vartheta)=\frac{a_n}{3^n}$$ for $$n\in \mathbb{N}$$, where $$a_n \in \mathbb{Z}$$,$$3 \nmid a_n$$

My approach was to do it by induction. For n=1 it is clear. Than i used $$\cos(\alpha+\beta)=\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)$$ with $$\alpha=n\vartheta$$ and $$\beta=\pm \vartheta$$. I added the results and got $$\cos((n+1)\vartheta)=2\cos(\vartheta)\cos(n\vartheta)-\cos((n-1)\vartheta)$$.

I get then

$$\cos((n+1)\vartheta)=2\frac{1}{3}\frac{a_n}{3^n}-\cos((n-1)\vartheta) = \frac{2a_n-\cos((n-1)\vartheta)3^{n+1}}{3^{n+1}}$$

I don't know how to argue that $$2a_n-\cos((n-1)\vartheta)3^{n+1} \in \mathbb{Z}$$ and that $$3 \nmid 2a_n-\cos((n-1)\vartheta)3^{n+1}$$.

If $$a_n = 3^n \cos(n \vartheta)$$, your equation $$\cos((n+1) \vartheta) = 2 \cos(\vartheta) \cos(n \vartheta) - \cos((n-1) \vartheta)$$ can be written as $$a_{n+1} = 2 a_n - 9 a_{n-1}$$ and then it's obvious that if $$a_n$$ and $$a_{n-1}$$ are integers, so is $$a_{n+1}$$, and if $$a_{n} \not\equiv 0 \bmod 3$$ then $$a_{n+1} \not\equiv 0 \bmod 3$$.