<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] I'll just do a subcase of the $n=3$ case.
$A_{k}=1$ is the maximum for k=2, 3.
First notice that letting $A_{k}=1$ and taking the other values of $A_{j}=0$ we have
$$cos(kx)\leq 1$$
which holds for all x. So each $A_{k}$ is at least 1. Now we will show that each $A_{k}$ is at most 1 for $k=2,3$.
Plugging in $x=0$ gives us $A_{1}+A_{2}+A_{3} \leq 1$.
If $A_{3}>1$ then this forces $A_{1}+A_{2} \leq 1-A_{3} < 0$. However, plugging in $x=\frac{2\pi}{3}$ we have $A_{3} + \frac{-A_{1}-A_{2}}{2} \leq 1$
So $A_{3}\leq 1+\frac{A_{1}+A_{2}}{2} < 1$
Therefore $A_{3}\leq 1$ thus $A_{3}=1$.
If $A_{2}>1$ then $A_{1}+A_{3} \leq 1-A_{2} < 0$.
Plugging in $x=\pi$ we have $A_{2} + -A_{1}-A_{3} \leq 1$.
So $A_{2}\leq 1+ A_{1} + A_{3} < 1$.
Therefore $A_{2}\leq 1$ thus $A_{2}=1$.
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