Bounding a polynomial with cosine roots Let $$f(x)=\prod\limits_{r=0}^{n}\left(x-\cos\left(\frac{2r+1}{2n+2}\cdot\pi\right)\right) $$
I want to put some type of bound on the modulus of this expression. In particular I want to show that $|f(x)|\le 1/2^n$ for all $x\in [-1,1]$. My thoughts were that these roots accumulate near zero, so $|f(x)|$ takes its max values at $|x|= 1$. From some trial cases, it seems to take the value $1/2^n$ here, but I am unsure how to prove this must always be the case.
 A: 
Warning: for convenience I have changed the definition of the function to:
  $$
f_n(x)=\prod_{k=1}^n\left(x-\cos\frac{2k-1}{2n}\pi\right)
$$

Introducing $\xi^n_k=\frac{2k-1}{2n}\pi$ we are going to prove the following lemma:

$$F_n(t):=\prod_{k=1}^n\left(\cos t-\cos\xi^n_k\right)=\frac{\cos n t}{2^{n-1}}.$$

Proof:
$$
F_n(t)=\frac{1}{2^n}\prod_{k=1}^n\left(e^{it}+e^{-it}-e^{i\xi_k^n}-e^{-i\xi_k^n}\right)=\frac{1}{2^nu^n}\underbrace{\prod_{k=1}^n\left(u^2+1-(e^{i\xi_k^n}+e^{-i\xi_k^n})u\right)}_{\equiv \Phi_n(u)},\tag{1}
$$
where substitution $u=e^{it}$ was used.
One observes by inspection that $u^\pm_k=e^{\pm i\xi_k^n}$ are the roots of the polynomial $\Phi_n(u)$. There are altogether $2n$ distinct values of $u^\pm_k$ corresponding to values of $k$ from $1$ to $n$. As the order of the polynomial $\Phi_n(u)$ is at most $2n$, the set $u^\pm_k$ represents all roots of the polynomial. 
Further one observes that $u^\pm_k$ are the roots of the equation:
$$
u^{2n}+1=0.
$$
It follows:
$$
\Phi_n(u)=u^{2n}+1.\tag{2}
$$
Substituting (2) in (1) one finally obtains:
$$
F_n(t)=\frac{u^n+u^{-n}}{2^n}=\frac{\cos nt}{2^{n-1}},
$$
as claimed.
Coming back to original question we set for $-1\le x\le1$: $x=\cos t$, so that:
$$
f_n(\cos t)=\prod_{k=1}^n\left(\cos t-\cos\xi^n_k\right)=\frac{\cos n t}{2^{n-1}},
$$
so that:
$$
|f_n(x)|\le\frac{1}{2^{n-1}},\quad (-1\le x\le1),
$$
as claimed (notice redefinition of $n$ in the beginning of the answer).
