Prove $\sum_{k=0}^{n-1} {n-1 \choose k}cos(\frac{2\pi k}{n}+\frac{2\pi}{n})(-1)^k \neq 0$ Let $f(x)$ be a polynomial such that $f(N)=cos((N-1)\frac{2\pi}{n})$ and $N\in\{1,2,3,...n\}$. I am interested in finding the degree of $f(x)$. Using the method of successive differences, I found the $n-1$ difference to be $$D_{n-1}=\sum_{k=0}^{n-1} {n-1 \choose k}cos(\frac{2\pi k}{n}+\frac{2\pi}{n})(-1)^k$$
I am attempting to prove $D_{n-1}\neq 0$ for $n=1,2,3,...$.
So far I have attempted to take the limit as $n\rightarrow\infty$ and checked $n$ values up to $30$. Neither of these have been useful.
 A: Let's abbreviate $\omega_n = e^{2\pi i/n}$, so that $D_{n-1}$ is the real part of $$A_{n-1} = \sum_{k=0}^{n-1} \binom{n-1}{k} \omega_n^{k+1} (-1)^k = \omega_n \sum_{k=0}^{n-1} \binom{n-1}{k} (-\omega_n)^k.$$
Now by the binomial theorem, this is just $\omega_n (1-\omega_n)^{n-1}$.
For any number $t$ we have $(1 + e^{it})^2 = (2 + 2 \cos t) e^{it} = 4 \cos^2 (t/2) e^{it}$, so if $-\pi \le t \le \pi$ you get $1 + e^{it} = 2 \cos(t/2)  e^{it/2} $. In particular, $-\omega_n = e^{i(2\pi/n-\pi)}$, so
$$ 1 - \omega_n = 2 \sin \frac{\pi}{n} \exp\left[ \left(\frac{1}{n} - \frac{1}{2} \right) \pi i \right]. $$
Thus,
\begin{align}
A_{n-1}
&= \left (2 \sin \frac{\pi}{n} \right)^{n-1} \exp\left[ \frac{2\pi i}{n} + (n-1) \left( \frac{1}{n} - \frac{1}{2} \right) \pi i \right] \\
&= \left (2 \sin \frac{\pi}{n} \right)^{n-1} \exp\left[ \left( \frac{3-n}{2} + \frac{1}{n} \right) \pi i \right]. \\
\end{align}
Finally,
$$ D_{n-1} = \operatorname{Re} A_{n-1} = \left (2 \sin \frac{\pi}{n} \right)^{n-1} \cos \left[ \left( \frac{3-n}{2} + \frac{1}{n} \right) \pi \right] $$
which is always nonzero.
