We have $$n\in\mathbb{N}\quad k=1,...,n$$
we want to find
$$\max_k{\cos(\frac{k\pi}{n+1})}$$
As we don't have a continuous application , we have a set of $n$ points we cannot do the typical derivative and then find the zeros of the application.
I don´t know how to start, some hint, please!
On
It is for $k = 1$ since
1) $\cos(x) \leq 0$ if $x \in [\frac{\pi}{2},\pi]$ and thus $\cos\left(\frac{k\pi}{(n+1)} \right)\leq 0$ for $\frac{k}{n+1} \geq \frac{1}{2}$, i.e. $\frac{n+1}{2}\leq k \leq n$
2) for $x \in [0,\frac{\pi}{2}], \frac{d}{dx}\cos(x) = -\sin(x) \leq 0$, i.e. $\cos(x)$ is decreasing on this interval and thus $\cos\left(\frac{\pi}{n+1}\right)\geq \cos\left(\frac{k\pi}{n+1}\right)$ for $1\leq k <\frac{n+1}{2}$
Hints: You can still use information about the function $\cos x$ to help you solve the problem. If $k \in \{1,\ldots,n\}$ then $0 < k\pi/(n+1) < \pi$. Is $\cos x$ increasing, decreasing, or both on the interval $(0,\pi)$? This should be enough to reach an answer.