Proving the existence of $b$ such that $\prod_{k=1}^n(1-\cos(a_k-b))=\frac{1}{2^n}$ 
Let $n>0$ and $a_1,\ldots,a_n\in \mathbb R$. Prove there is some $b$ such that $\prod_{k=1}^n(1-\cos(a_k-b))=\frac{1}{2^n}$

This is motivated by this question Finding a point on the unit circle that achieves this equality
Numerical trials suggest it is true indeed, and that for any choice of $a_1,\ldots,a_n$, $b\to \prod_{k=1}^n(1-\cos(a_k-b))$ achieves a maximum quite greater that $\frac{1}{2^n}$.
EDIT: It may be worth noting that: $$\prod_{k=1}^n(1-\cos(a_k-b))=2^n\prod_{k=1}^n\sin^2(\frac{a_k-b}{2})$$
I attempted induction on $n$, to no avail.
 A: Let
$$F(x)=\prod_{i=1}^n \left(1-\cos(a_i-x)\right)$$
for any $a_i$. Now, noticing that $F(x)\ge0$ for all $x$, we look at
$$\ln F(x)=\sum_{i=1}^n \ln\left(1-\cos(a_i-x)\right)$$
and compute the average over all $x$:
$$
\frac{1}{2\pi}\int_0^{2\pi}\ln F(x)\,dx
=\sum_{i=1}^n \frac{1}{2\pi} \int_0^{2\pi}\ln\left(1-\cos(a_i-x)\right)\,dx.
$$
However, the value of the last integral does not depend on $a_i$ as we have
$$
\int_0^{2\pi}\ln\left(1-\cos(a_i-x)\right)\,dx
=\int_0^{2\pi}\ln\left(1-\cos u\right)\,du
=-2\pi\ln 2,
$$
resulting in
$$
\frac{1}{2\pi}\int_0^{2\pi}\ln F(x)\,dx=-n\ln 2.
$$
Now, since the average is $-n\ln 2$ and $F(x)$ is not constant, there must be some value of $x$ for which $\ln F(x)>-n\ln 2$, which makes $F(x)>1/2^n$.

On a side node, if the $a_i$ are evenly spread using $a_i=2\pi i/n$, we get $F(x)=2^{1-n}(1-\cos(nx))$ which has maximum $2^{2-n}$: i.e., four times what we have proven.
A natural follow-up question is if the maximum of $F(x)$ will always be $2^{2-n}$ or higher?
