Given real angles $\phi_1,\ldots,\phi_N$, there exist infinitely many integers $n$ such that $\cos(n\phi_k) > 0$ for all $k$.

Question

I wonder whether the following claim is correct:

Given angles $\phi_1,\ldots,\phi_N\in\Bbb R$, there exist infinitely many $n\in\Bbb N$ such that $\cos(n\phi_k) > 0$ for all $k=1,\ldots,N$.

If all $\phi_k$ are rational multiples of $\pi$, the claim is obviously true and also for $N=1$, but even for $N=2$ I cannot think of a proof.

Answer 1

Yes, this is correct. We can view $\phi_i/(2\pi)$ as an element of $\mathbb{R}/\mathbb{Z}$. Then $(\phi_1,\ldots,\phi_N)/(2\pi)$ is an element of $M := (\mathbb{R}/\mathbb{Z})^N$. Now consider the sequence $a_n := (n\phi_1,\ldots,n \phi_N)/(2\pi) \in M$ for $n=1,2,\ldots$. Now $M$ is a compact set. Hence, for any $\epsilon>0$ there must exist $i < j$ for which the distance between $a_i$ and $a_j$ is less than $\epsilon$. Then take $n = j-i$ and one sees that $a_n$ is within $\epsilon$ of the identity.

So these cosines can be made arbitrarily close to 1.