Product of areas in a circle A circle (or disk) of area $2n$ is divided into $n$ regions, as shown below with example $n=8$. The points are evenly spaced around the circle. (If the image doesn't load for you, just imagine $n$ evenly spaced points on a circle, with line segments joining one point with all the other points, like a seashell.)

What is the limit of the product of the areas of the regions, as $n$ approaches $\infty$ ?
Using basic trigonometry, I've got:
$$\lim_{n\to\infty}\prod_{k=1}^{n}\left(2-\frac{n}{\pi}\sin\left(\frac{2k\pi}{n}\right)+\frac{n}{\pi}\sin\left(\frac{2(k-1)\pi}{n}\right)\right)$$
I do not know how to evaluate this limit. Wolfram does not evaluate the limit, but tells me that when $n=10000$ the product is approximately $8.3$.
UPDATE1:
I am fairly confident that the limit is $4\cosh^2\left({\frac{\pi}{2\sqrt{3}}}\right)=8.29674...$
Here's why. I was trying to answer a similar question: A ball is divided into $n$ concentric shells of equal thickness. Can the average volume of the shells be fixed so that the product of the volumes converges to a positive number as $n\to\infty$? The answer turns out to be yes. If we fix the average volume of the shells to be $\frac{e^2}{3}$ then the product of the volumes converges to $2\cosh\left({\frac{\pi}{2\sqrt{3}}}\right)=2.8804...$ I noticed that this number seemed to be (to many decimal places) the square root of the answer to the question here. I think there must be a connection.
UPDATE2:
Another similar question is this. A circle (or disk) is divided into $n$ regions by equally spaced parallel lines (the circle is tangent to two of the lines). Can the average area of the regions be fixed so that the product of the areas converges to a positive number as $n\to\infty$? The answer seems to be yes. If we fix the average area of the regions to be $\frac{{\pi}e}{8}$ then the product of the areas seems to converge to $2\cos\left(\frac{\pi}{2\sqrt{3}}\right)$ (notice this is cos, not cosh). Again, I think there must be a connection.
(As for the source of the original question, I thought of the question by myself. It was inspired by an IB (high school) May 2021 exam question, which goes like this. A unit circle has $n$ evenly distributed points. Line segments are drawn joining one point with all the other points (like a seashell). The exam question led students to find the product of the lengths of the line segments, using complex numbers. The answer turns out to be $n$. Then I wondered, what is the product of the areas of the enclosed regions? Obviously the product approaches $0$ as $n\to\infty$ (since all the areas approach $0$). But could the average area be fixed so that the product of the areas converges to a positive number as $n\to\infty$? Definitely beyond the course syllabus. After experimenting on desmos, I found that the answer seems to be yes: if we fix the average area to be $2$, then the product of the areas seems to converge to a positive number. The question in this post is, what does it converge to?)
 A: I'll answer my own question.
Using $\sin{A}-\sin{B}=2\left(\sin{\frac{A-B}{2}}\right)\left(\cos{\frac{A+B}{2}}\right)$,
$$L=\lim_{n\to\infty}\prod_{k=1}^{n}\left(2-\frac{n}{\pi}\sin\frac{2k\pi}{n}+\frac{n}{\pi}\sin\frac{2(k-1)\pi}{n}\right)$$
$$=\lim_{n\to\infty}\prod_{k=1}^{n}\left(2-\left(\frac{2n}{\pi}\sin{\frac{\pi}{n}}\right)\cos{\frac{(2k-1)\pi}{n}}\right)$$
Using the Maclaurin series for sine,
$$L=\lim_{n\to\infty}\prod_{k=1}^{n}\left(2-\left(2-\left(\frac{\frac{\pi}{\sqrt3}}{n}\right)^2+O\left(\frac{1}{n^4}\right)\right)\cos{\frac{(2k-1)\pi}{n}}\right)$$
Since we are taking the limit, I think we can ignore the $O\left(\dfrac{1}{n^4}\right)$, but I'm not sure how to to justify this; feel free to comment or edit.
Then this answer to one of my other questions shows that
$$\lim_{n\to\infty}\prod_{k=1}^{n}\left(2-\left(2-\left(\frac{x}{n}\right)^2\right)\cos{\frac{(2k-1)\pi}{n}}\right)=4\cosh^2{\left(\frac{x}{2}\right)}$$
$$\therefore L=4\cosh^2{\left(\frac{\pi}{2\sqrt{3}}\right)}$$
