Let $n$ be a natural number and let $z$ be a complex number. Consider the following proposition:
$P(n)$: If $\cos (nz)$ is bounded above by one in absolute value, then $\cos z$ itself is likewise bounded above.
In [AG, p,53 #4 , soln p.184] the authors prove that $P(n)$ is true for the specific value $n = 3$. Of course, this immediately establishes validity also for $n = 9$, $27$, etc. However, it appears to hold for other values as well.
Question: For which positive integers $n$ does $P(n)$ fail to hold?
Reference: [AG] Mathematical Olympiad Challenges by Andreescu & Gelca.
For any positive integer $n$, define $$\alpha_k = \cos\frac{\pi(2k+1)}{2n}\quad\text{ for }\quad k = 0, 1,\ldots, n-1.$$ For any $z \in \mathbb{C}$, we know
$$\cos(nz) = T_n(\cos z) = 2^{n-1}\prod_{k=0}^{n-1}\left(\cos(z) - \alpha_k\right)$$ where $T_n(x)$ is the $n^{th}$ Chebyshev's polynomial of the first kind.
Since $\alpha_k = -\alpha_{n-1-k}$, we have
$$\big|T_n(\cos z)\big| = 2^{n-1}\prod_{k=0}^{n-1}\left| \cos(z)^2 - \alpha_k^2 \right|^{1/2} \ge 2^{n-1} \prod_{k=0}^{n-1} \left| |\cos(z)|^2 - \alpha_k^2 \right|^{1/2} = \big|T_n(|\cos z|)\big|$$
As a result, whenever $|\cos nz| \le 1$, we will have $\big|T_n(|\cos z|)\big| \le 1$. Now $|\cos z|$ is a non-negative number and we know $$T_n(x) = \cosh(n\cosh^{-1}(x)) > 1\quad\text{ for }x > 1$$ This means whenever $|\cos nz| \le 1$, we will have $|\cos z| \le 1$ too. i.e. $P(n)$ is true for all positive integer $n$.