Let $n \geq 3$ and consider the family of functions $\{f_k\}$ indexed by $k = 1,2,\ldots,n-1$, defined on the unit circle via $$f_k(\varphi) = \textrm{Re}\left[e^{-ik\varphi}\left(1-(1-e^{i\varphi})\frac{k}{n}\right)^n\right] = \sum_{j=0}^n {n\choose j} \left( \frac{k}{n} \right)^j \left( 1 - \frac{k}{n} \right)^{n-j} \cos(k-j)\varphi$$ Observe that $f_k(-\varphi) = f_k(\varphi) = f_{n-k}(\varphi)$. It is not hard to show that $\max_{k,\varphi} f_k(\varphi) = 1$, attained at $\varphi = 0$ for any $k$. However, it is the global minimum which interests me. Concretely, numerically one finds that $\min_{k,\varphi} f_k(\varphi) = -\left(1-\tfrac{2}{n}\right)^n$, attained only for $k \in \{1,n-1\}$ at $\varphi = \pm \pi$. My question is: How to prove that?
I've tried solving $f_k'(\varphi) = 0$, but unfortunately that lead me nowhere. Any hint would be much appreciated.
