For what $k$ is $P(k):=\prod_{j=1}^{13}\cos\frac{\pi kj}{13}$ negative?

Question

Let $k>0$ be an integer. For what $k$ is $P(k):=\prod_{j=1}^{13}\cos\frac{\pi kj}{13}$ negative? Since $13$ is prime, and for $\gcd(m,13)=1$, $P(2m)=P(2)=2^{-12}$ (can be shown by considering the real part of $(\cos\theta+i\sin\theta)^{13}=1$ and Vieta's relation), we know $|P(k)|=2^{-12}\quad\forall k\in\mathbb{N}$ and $\gcd(k,13)=1$.

Answer 1

Hint: If $k$ is divisible by $13$, things are quite easy. Otherwise, pair the terms $\cos\left(\frac{\pi kj}{13}\right)$ and $\cos\left(\frac{\pi k(13-j)}{13}\right)$. The case $k$ even will collapse immediately. For odd $k$ one may want to look at a product to sum identity.