TITLE: Find the Maximum Trigonometric polynomial coefficient $A_{k}$
QUESTION [10 upvotes]: Let $n,k$ be given positive integers and $n\ge k$. Let $A_i, i=1, 2, \cdots, n$ be given real numbers. If for all real numbers $x$ we have $$A_{1}\cos{x}+A_{2}\cos{(2x)}+\cdots+A_{n}\cos{(nx)}\le 1$$
Find the maximum value of $A_{k}$.

I don't know if this question has been studied
If  $n=2$ it is easy to solve it.

REPLY [2 votes]: NTstrucker@AoPS's answer:
It is a known result:
$$\max A_k = 2 \cos \frac{\pi}{\lfloor \frac{n}{k} \rfloor + 2}, \ 1\le k \le n.$$
See e.g. Theorem 16.2.4 in [1].
So, $\max A_1 = 2\cos \frac{\pi}{n + 2}$,
$\max A_m = 1$ for $\lfloor \frac{n}{2}\rfloor + 1 \le m \le n$,
$\max A_{\lfloor \frac{n}{2}\rfloor} = \sqrt{2}$.
[1] Qazi Ibadur Rahman and Gerhard Schmeisser, “Analytic Theory of Polynomials”, 2002.