Let $x_1,\dots,x_n$ be angles with $x_j \in (0, 2\pi)$ so that $x_1+\dots +x_n=2 \pi$
Let $P_j=e^{i\sum_{l=1}^{j-1}x_l}, j=1,\dots,n$ where $i$ denotes the imaginary unit.
Then $\overline{P_1}\overline{P_2},\dots,\overline{P_{n-1}P_n},\overline{P_nP_1}$ is a polygon with $n$ vertexes (corners).
How can I determine the angle $x_j$ so that the area of that polygon becomes maximal using Lagrange multipliers?
What I tried
So $a$ has length $1$ (I'm not even sure if this first step is correct). The area of each of those triangles is $A_i=\frac{ah}{2}$. Using $\sin(x_i)=h/a \Leftrightarrow h=a \sin(x_i)=\sin(x_i)$ we get $A_i=\frac{1}{2}\sin(x_i)$.
Therefore the combined area $A=\sum_{i=1}^{n}\sin(x_i)$
So my functions are:
$f(x_1,\dots,x_n)=\sum_{i=1}^{n}\sin(x_i)$
and
$g(x_1,\dots,x_n)=\sum_{i=1}^{n}\sin(x_i)-2\pi$
And I will maximize $f$ subjec to to $g=0$
I end up with $x_i=\frac{2 \pi}{n}$. Is that correct? Do I need to show something else to show that this choice of $x_i$ truly maximizes the area of the polygon?
I think that your constraint $g$ is rather $\sum_{k=1}^n x_i - 2\pi$. Other than that your reasoning is correct. You get the following Lagrangian: $$ \mathcal L(x_1, \dotsc, x_n, \lambda) = 2\pi\lambda + \sum_{k=1}^n \sin(x_i) - \lambda x_i $$ Then, computing its gradient and setting to zero: $$ \begin{align*} 0 & = \nabla_{x_i, \lambda} \mathcal L \\ & = \begin{pmatrix} -\cos(x_1) - \lambda\\ \vdots \\ -\cos(x_n) - \lambda \\ 2\pi - \sum_{k=1}^n x_i \end{pmatrix} \end{align*} $$ This immediately gives $\cos(x_1) = \cdots = \cos(x_n)$ from which $x_1 = \cdots = x_n$, and you conclude that $x_1 = \cdots = x_n = 2\pi /n$.
We found a critical point of $\mathcal L$ and in principle we must check that this critical point is indeed an extremum. This is usually done by computing the Hessian determinant, i.e. the determinant of the matrix of second derivatives of $\mathcal L$.
Doing so here will confirm you indeed have a maximum, i.e., that a regular polygon realises the maximal area.