TITLE: Evaluate $\prod_{p=1}^{\lfloor {n/2} \rfloor} \cos\left( \frac{\pi pk}{n} \right)$ with $k \in \mathbb{N}$
QUESTION [6 upvotes]: I have found similar cosine products, but mainly in the case when $k = 1$. I'm asking this since it is related to a question I asked which has led to this product:
$$S_{n,k} =\prod_{p=1}^{\lfloor {n/2} \rfloor} \cos\left( \frac{\pi p k}{n} \right)$$
Where both $k,n \in \mathbb{N}$.
I would do this problem for some time before asking but given how long the previous question has taken to get to this product, I would like ask for some help on how to evaluate it.

EDIT:
With some testing in Sage (to get an idea on what answer could look like):
If $n$ is odd:
$$|S_{n,k}| = 2^{-((n-1)/2) + (\gcd(k,n)-1)/2}$$
If $n$ is even, let $n=2^\alpha n''$. Then:
$$|S_{n,k}| = \begin{cases}
0,  & \text{if $\gcd(k, 2^\alpha) \neq 2^\alpha$} \\
2^{-(n/2) + \gcd(k,n)/2}, & \text{otherwise}
\end{cases}$$
Not sure how to find the sign of $S_{n,k}$ yet.

REPLY [5 votes]: Here is my attempt. This is motivated by Jack D'Aurizio's trick os using Fourier series expansion of $\log\cos(x)$.  I need some more eyes to go through and see if I made any mistakes. Please feel free to edit/fix/add if you find it helpful. 
\begin{eqnarray*}
\log \cos(x) &=& \log \frac{1}{2} -\sum_{m\ge 1}^{\infty}{\frac{(-1)^{m} \cos(2 m x)}{m}}
\end{eqnarray*}
\begin{eqnarray*}
\sum_{k=1}^{\left\lfloor \frac{n}{2}\right\rfloor}{\cos \frac{2\pi p k}{n}} &=& \begin{cases} -\frac{1}{2}, & mod(n,p) \ne 0 \\ \left \lfloor \frac{n}{2} \right\rfloor, & mod(n,p) =0 \end{cases} 
\end{eqnarray*}
Generically,
\begin{eqnarray*}
\sum_{k=1}^{\left\lfloor \frac{n}{2}\right\rfloor}{\cos \frac{2\pi p k}{n}} &=& -\frac{1}{2} \left[1-\frac{\sin\left(\pi p \frac{1+2\left\lfloor \frac{n}{2}\right\rfloor}{n}\right)}{\sin \frac{\pi p}{n}} \right] \\
&=& \begin{cases} \frac{n-1}{2} , & mod(p,n) = 0, n, odd \\ -\frac{1}{2} , & mod(p,n) \ne 0, n, odd \\ -\frac{1}{2} \left[1-(-1)^{p}\right], & n, even\end{cases} \\
&=& \begin{cases}  \left\lfloor\frac{n}{2}\right\rfloor , & mod(p,n) = 0, n, odd \\ -\frac{1}{2} , & mod(p,n) \ne 0, n, odd  \\ -1, & n, even, p, odd \\ 0, & n, even, p, even \end{cases} 
\end{eqnarray*}
Note that,
\begin{eqnarray*}
 \frac{1+2\left\lfloor \frac{n}{2}\right\rfloor}{n} &=& \begin{cases} 1, & n, odd \\ 1+\frac{1}{n}, & n, even \end{cases} \\
\sin\left(\pi p \frac{1+2\left\lfloor \frac{n}{2}\right\rfloor}{n}\right) &=&  \begin{cases} 0, & n, odd \\ (-1)^{p}\sin \frac{\pi p}{n}, & n, even \end{cases} \\
\end{eqnarray*}
For $x=\frac{\pi p k}{n}$,
\begin{eqnarray*}
\sum_{k=1}^{\left\lfloor \frac{n}{2}\right\rfloor} \sum_{m\ge 1}^{\infty}{\frac{(-1)^{m} \cos \left(\frac{2 \pi p k m}{n}\right)}{m}} &=&  \sum_{m\ge 1}^{\infty} \sum_{k=1}^{\left\lfloor \frac{n}{2}\right\rfloor}{\frac{(-1)^{m} \cos \left(\frac{2 \pi p k m}{n}\right)}{m}} \\
 &=& \sum_{\underset{m\ge 1}{m \ne \frac{n}{(n,p)} \mathbb{Z} }}^{\infty} \sum_{k=1}^{\left\lfloor \frac{n}{2}\right\rfloor}{\frac{(-1)^{m} \cos \left(\frac{2 \pi p k m}{n}\right)}{m}} + \sum_{\underset{m\ge 1}{m = \frac{n}{(n,p)} \mathbb{Z} }}^{\infty} \sum_{k=1}^{\left\lfloor \frac{n}{2}\right\rfloor}{\frac{(-1)^{m} \cos \left(\frac{2 \pi p k m}{n}\right)}{m}}  \\
  &=& \sum_{\underset{m\ge 1}{m \ne \frac{n}{(n,p)} \mathbb{Z} }}^{\infty}  \frac{(-1)^{m}}{m} \sum_{k=1}^{\left\lfloor \frac{n}{2}\right\rfloor}{ \cos \left(\frac{2 \pi p k m}{n}\right)} +  \sum_{\underset{m\ge 1}{m = \frac{n}{(n,p)} \mathbb{Z} }}^{\infty}  \frac{(-1)^{m}}{m} \sum_{k=1}^{\left\lfloor \frac{n}{2}\right\rfloor}{ \cos \left(\frac{2 \pi p k m}{n}\right)} \\
   &=& \sum_{\underset{m\ge 1}{m \ne \frac{n}{(n,p)} \mathbb{Z} }}^{\infty}  \frac{(-1)^{m}}{m} \left(-\frac{1}{2}\right) +  \sum_{\underset{m\ge 1}{m = \frac{n}{(n,p)} \mathbb{Z} }}^{\infty}  \frac{(-1)^{m}}{m} \left\lfloor \frac{n}{2}\right\rfloor \\
   &=& \sum_{m\ne N_{p}}^{\infty}  \frac{(-1)^{m}}{m} \left(-\frac{1}{2}\right) + \sum_{m= i N_{p},\forall i}^{\infty}  \frac{(-1)^{m}}{m} \left\lfloor \frac{n}{2}\right\rfloor \\
   &=& \sum_{m\ne N_{p}}^{\infty}  \frac{(-1)^{m}}{m} \left(-\frac{1}{2}\right) + \sum_{i=1}^{\infty}  \frac{(-1)^{ i N_{p}}}{ i N_{p}} \left\lfloor \frac{n}{2}\right\rfloor \\
   &=& -\frac{1}{2} \sum_{m=1}^{\infty}  \frac{(-1)^{m}}{m} +  \frac{1}{2} \sum_{i=1}^{\infty}  \frac{(-1)^{ i N_{p}}}{ i N_{p}} \left(1+2 \left\lfloor \frac{n}{2}\right\rfloor\right)
 \end{eqnarray*}
Here $N_p \triangleq \frac{n}{gcd(n,p)}$.
For even $N_p$, the harmonic series diverges. The only interesting case is $N_p$ being odd. In which case,'
\begin{eqnarray*}
 &=&-\frac{1}{2}\sum_{m=1}^{\infty}  \frac{(-1)^{m}}{m} + \frac{1}{2} \sum_{m=1}^{\infty}  \frac{(-1)^{ m N_{p}}}{ m N_{p}} \left(1+2 \left\lfloor \frac{n}{2}\right\rfloor\right) \\
 &=&\frac{1}{2} \sum_{m=1}^{\infty}  \frac{(-1)^{ m}}{ i} \left(\frac{1+2 \left\lfloor \frac{n}{2}\right\rfloor}{N_p} -1\right) \\
 &=&\frac{1}{2} \left(\frac{1+2 \left\lfloor \frac{n}{2}\right\rfloor}{N_p} -1\right) \sum_{m=1}^{\infty}  \frac{(-1)^{ m}}{ i} \\
 &=&\frac{1}{2} \left(\frac{1+2 \left\lfloor \frac{n}{2}\right\rfloor}{N_p} -1\right) \log(2).
 \end{eqnarray*}
\begin{eqnarray*}
\sum_{k=1}^{\left\lfloor \frac{n}{2}\right\rfloor}{\log \cos\left(\frac{\pi p k}{n}\right)} &=& \sum_{k=1}^{\left\lfloor \frac{n}{2}\right\rfloor} \log \frac{1}{2} - \sum_{k=1}^{\left\lfloor \frac{n}{2}\right\rfloor} \sum_{m\ge 1}^{\infty}{\frac{(-1)^{m}  \cos\left(\frac{2\pi p k m}{n}\right)}{m}} \\
&=&  \left\lfloor \frac{n}{2}\right\rfloor  \log \frac{1}{2} - \sum_{k=1}^{\left\lfloor \frac{n}{2}\right\rfloor} \sum_{m\ge 1}^{\infty}{\frac{(-1)^{m}  \cos\left(\frac{2\pi p k m}{n}\right)}{m}} \\
&=&  \left\lfloor \frac{n}{2}\right\rfloor  \log \frac{1}{2} - \frac{1}{2} \left(\frac{1+2 \left\lfloor \frac{n}{2}\right\rfloor}{N_p} -1\right) \log(2) \\
&=&  \left\lfloor \frac{n}{2}\right\rfloor  \log \frac{1}{2} + \frac{1}{2} \left(\frac{1+2 \left\lfloor \frac{n}{2}\right\rfloor}{N_p} -1\right)  \log \frac{1}{2} \\
&=&  \frac{1}{2} \left(2  \left\lfloor \frac{n}{2}\right\rfloor+\frac{1+2 \left\lfloor \frac{n}{2}\right\rfloor}{N_p} -1\right)  \log \frac{1}{2}  \\
&=&   \log \left[\frac{1}{2^{ \left( \left\lfloor \frac{n}{2}\right\rfloor+\frac{1+2 \left\lfloor \frac{n}{2}\right\rfloor}{2 N_p} -\frac{1}{2}\right)}} \right]
\end{eqnarray*}
Looking at the exponent, 
\begin{eqnarray*}
\prod_{k=1}^{\left\lfloor \frac{n}{2}\right\rfloor}{\cos\left(\frac{\pi p k}{n}\right)}
&=&  2^{- \left( \left\lfloor \frac{n}{2}\right\rfloor+\frac{1+2 \left\lfloor \frac{n}{2}\right\rfloor}{2 N_p} -\frac{1}{2}\right)} \\
&=&  2^{- \left( \left\lfloor \frac{n}{2}\right\rfloor+\frac{1+2 \left\lfloor \frac{n}{2}\right\rfloor}{2 \frac{n}{gcd(n,p)}} -\frac{1}{2}\right)}.
\end{eqnarray*}
A special case is when $p=1$ and $n$, $N_p=n/gcd(n,p)=n$, it becomes $2^{-n}$, the well known formula linked above.