Knowing $\cos\theta$, can $\cos(n\theta)=\cos(\pi k)$? $\cos(\pi k)=1$ or $-1$. After expressing $\cos(n\theta)$ in terms of $\cos\theta$, I have found that $\cos(n\theta)=\sum^{\lfloor{\frac{n}{2}}{\rfloor}}_{l=0}{n\choose2l}(-\frac{8}{9})^{l}(-\frac{1}{3})^{n-2l}$ but I cannot proceed. Is this the right approach? Thanks very much.
 A: Without lost of generality the question can be formulated

If $\cos\theta=-\frac{1}{3}$, is there some positive integer $n$ such that $\cos(n\theta)=1$?

This in turn implies that there exists integer number $k$ such that $\theta=\frac{2k}n\pi$. But this is impossible because by the Niven's theorem the only rational values of the sine or cosine of an angle which is a rational fraction of $\pi$ are $0, \pm\frac12$, and $\pm1$. The value $-\frac13$ is not in the list.
A: I think I found another way to prove the result for the particular value of $\cos x=-\frac13$.
Generally we have the following recurrence relation:
$$
\cos(n+2)x=2\cos (n+1)x\cos x-\cos nx\tag1
$$
If $\cos x=\frac pq$ with $p,q\in\mathbb Z, q>0, (p,q)=1$, then  $$\cos nx=\frac{a_n}{q^n},\tag2$$
where $a_n$ is an integer. Substituting $(2)$ into $(1)$ one obtains:
$$
\frac{a_{n+2}}{q^{n+2}}=2\frac{a_{n+1}}{q^{n+1}}\frac pq-\frac{a_{n}}{q^{n}}
\implies a_{n+2}=2p\cdot a_{n+1}-q^2\cdot a_{n}\tag3
$$
in the case $\cos x=-\frac13$ the equation $(3)$ reads:
$$
a_{n+2}=-2a_{n+1}-9\cdot a_{n},\quad\text{with}\quad a_0=1,\ a_1=-1.\tag4
$$
It follows from $(4)$ immediately by induction:
$$
a_n\mod 3=\begin{cases}
1,& n\text{ even}\\
2,& n\text{ odd}
\end{cases}\tag5$$
This means that $\cos nx=\frac{a_n}{3^n}$ cannot be integer, particularly it cannot be equal to $1$.
