I ask you to determine whether my equation describing area of $n$-pointed stars. Note that $n=2k-1$ and $k \in \mathbb{N}_3$ and that I describe the area with a radius of a circle in which my star is inscribed. For example
this is a star with properties $n = 11$, $k = 6$ and $r \doteq 6 \; \text{cm}$ (I measured this approximation with a ruler). My deduction was: we know the length of a side of a regular polygon:
$$ a = 2r\sin \frac{180^{\circ}}{n} $$
$r$ being the radius, and $n$ being the number of sides the chosen regular polygon has. Now we have to get the length of the side of the $n$-star side. Derived from the original radius $r$, it is such:
$$ a_1 = r\cfrac{\sin \cfrac{180^{\circ}}{2k-1}}{\sin \cfrac{270^{\circ}}{2k-1}}. $$
Note that I replaced $n$ with $2k-1$ because I only want to examine odd-pointed stars (I have a better intuitive grip regarding these). To be more specific, I'll be looking only into examples like $\{11/5\}$ and not $\{11/2\}$, $\{11/6\}$ or alike (so $\{n/k-1\}$) — in other words, I get my stars via connecting each of the points on the circle with the two opposite that are the furthest away.
So the objective now is to figure out the distance between the midpoint of $a$ and the point where two lines first meet; let's call it $l$. This is my deduction:
$$ l = r\sin\cfrac{180^{\circ}}{2k-1} \tan\cfrac{180^{\circ}(k-2)}{2k-1}. $$
Now we just have to obtain the area of the triangles determined by two points on the circle and the point from where $l$ originates. Then, we'll multiply the obtained area by $n$ and subtract it from the area of the regular polygon (taking into account that the area of a regular polygon $S_n = r^2 n \sqrt{1 - \sin^2 \frac{180^{\circ}}{n}} \sin \frac{180^{\circ}}{n}$):
\begin{align*} S_{n_1} &= S_n - n\cfrac{al}{2} = \\ &= r^2(2k-1) \left( \sqrt{1-\sin^2 \cfrac{180^{\circ}}{2k-1}} \sin \cfrac{180^{\circ}}{2k-1} - \sin^2 \cfrac{180^{\circ}}{2k-1} \tan \cfrac{180^{\circ}(k-2)}{2k-1} \right). \end{align*}
Is my equation incorrect or incomplete? Too generalised? I'd be glad to know. Thank you in advance.
Your equation is correct! Nice derivation. In your final formula, some simplification could be done using the trig identities $$1-\sin^2 (\theta) = \cos^2 (\theta) \quad \text{ and } \quad \sin(2\theta)=2\sin(\theta)\cos(\theta)$$ to obtain $$A_n=r^2 n\left(\frac{1}{2}\sin\left(\frac{360^\circ}{n}\right)-\sin^2 \left(\frac{180^{\circ}}{n}\right)\tan\left(\frac{180^{\circ}(k-2)}{n}\right)\right)$$ With your initial constraints $n=2k-1$, $k \in \mathbb{N}$ and $k>2$