<s>[INST] Find the Maximum Trigonometric polynomial coefficient $A_{k}$ 
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.
 [/INST] My second answer:
Edit 2021/03/11
According to NTstrucker@AoPS's result (https://artofproblemsolving.com/community/c6h2477208), I give the following conjecture.
Conjecture 2: $\max A_1 = 2\cos \frac{\pi}{n+2}$.
When $n = 2$, $\max A_1 = \sqrt{2} = 2 \cos \frac{\pi}{4}$.
When $n = 3$, $\max A_1 = \frac{\sqrt{5} + 1}{2} = 2\cos \frac{\pi}{5}$.
When $n = 4$, $\max A_1 = \sqrt{3} = 2\cos \frac{\pi}{6}$.
When $n = 5$, $\max A_1 = 2\cos \frac{\pi}{7}$.
When $n = 6$, $\max A_1 \approx 1.8478$ (I only get numerical result) should be $2\cos \frac{\pi}{8}
\approx 1.847759065$.
When $n = 7$, $\max A_1 \approx 1.8794$ (I only get numerical result) should be $2\cos \frac{\pi}{9}
\approx 1.879385242$.
When $n = 8$, $\max A_1 \approx 1.9021$ (I only get numerical result) should be $2\cos \frac{\pi}{10}
\approx 1.902113033$.
When $n = 9$, $\max A_1 \approx 1.9190$ (I only get numerical result) should be $2\cos \frac{\pi}{11}
\approx 1.918985947$.
$\phantom{2}$
Let us prove that $\max A_m = 1$ if $\lfloor \frac{n}{2}\rfloor + 1 \le m \le n$.
We need Theorem 1. The proof is given at the end.
Theorem 1: Let $n\ge 2$ be a given integer. Let $A_i, i=1, 2, \cdots, n$ be given real numbers such
that $A_1\cos x + A_2 \cos 2x + \cdots + A_n \cos n x \le 1, \ \forall x\in \mathbb{R}$. Then
$A_m \le 1$ for $m = \lfloor \frac{n}{2}\rfloor + 1,  \cdots, n$.
By Theorem 1, we have immediately $\max A_m = 1$ if $\lfloor \frac{n}{2}\rfloor + 1 \le m \le n$
(simply let $A_m = 1$ and $A_k = 0$ for all $k\ne m$
and the condition $A_m \cos m x \le 1, \forall x\in \mathbb{R}$ is clearly satisfied.).
$\phantom{2}$
Proof of Theorem 1: First, we have Fact 1. The proof of Fact 1 is easy and thus omitted.
Fact 1: Let $M\ge 2$ be an integer. Let $K$ be a positive integer. Then
$$\frac{1}{M} + \frac{\cos K\pi}{M}\cdot \frac{1 + (-1)^M}{2}
+ \frac{2}{M}\sum_{j=1}^{\lfloor \frac{M+1}{2}\rfloor - 1} \cos \frac{2jK\pi}{M} = \left\{\begin{array}{cc}
                                                              0 & M \nmid K \\[5pt]
                                                              1 & M \mid K.
                                                            \end{array}
\right.$$
Second, we have Fact 2.
Fact 2: Let $n\ge 2$ be a given integer. Let $m$ be an integer with $\lfloor \frac{n}{2}\rfloor + 1 \le m \le n$.
Let $A_i, i= 1, 2, \cdots, n$ be given real numbers.
Let $f(x) = \sum_{k=1}^n A_k\cos (kx)$. Then
$$\frac{f(0)}{m} + \frac{f(\pi)}{m}\cdot \frac{1 + (-1)^m}{2}
+ \frac{2}{m}\sum_{j=1}^{\lfloor \frac{m+1}{2}\rfloor - 1} f\left(\frac{2j\pi}{m}\right) = A_m. $$
(Proof of Fact 2: It suffices to prove that
$$\frac{1}{m} + \frac{\cos k\pi}{m}\cdot \frac{1 + (-1)^m}{2}
+ \frac{2}{m}\sum_{j=1}^{\lfloor \frac{m+1}{2}\rfloor - 1} \cos \frac{2j k \pi}{m} = \left\{\begin{array}{cc}
                                                              0 & k \ne m \\[5pt]
                                                              1 & k = m.
                                                            \end{array}
\right. $$
This is true by Fact 1.)
By Fact 2, Theorem 1 is proved. We are done.
</s>