What is a closed form of this limit? (product of areas in circle with parabolas) I am looking for a closed form of
$L=\lim\limits_{n\to\infty}\prod\limits_{k=1}^n \left(\left(6+\frac{4n}{\pi}\left(\sin{\frac{\pi}{2n}}\right)\left(\cos{\frac{(2k-1)\pi}{2n}}\right)\right)^2-\left(\frac{32n}{\pi}\left(\sin{\frac{\pi}{4n}}\right)\left(\cos{\frac{(2k-1)\pi}{4n}}\right)\right)^2\right)$
The limit seems to exist. Calling the product $P_n$, we have
$P_1\approx10.062$
$P_{10}\approx18.409$
$P_{100}\approx19.788$
$P_{200}\approx19.869$
$P_{300}\approx19.896$
$P_{400}\approx19.909$
$P_{500}\approx19.917$
$P_{570}\approx19.919$
(Then desmos stops working.)
Context
I came up with the following question. $4n$ $(n\in\mathbb{Z}^+)$ points are uniformly distributed around a circle. $2n$ parabolas are drawn in the manner shown below with example $n=3$. The parabolas share a vertex, that is one of the designated points, and a line of symmetry. Each parabola passes through two more of the designated points on the circle. One of the parabolas is degenerate and looks like a line segment.

Assume that, for every value of $n$, the average (arithmetic mean) area of the $4n$ regions is exactly $6$.
Find $\lim\limits_{n\to\infty}(\text{product of areas})$.
I have worked out that the answer is $L^2$ where $L$ is the limit shown above. So I am looking for a closed form of $L$.
If the average area is $5.999$ the product approaches $0$; if the average area is $6.001$ the product diverges. I have tested my expression for $P_{n}$ with some small values of $n$, and I think it is correct.
This question was inspired by other questions about the product of areas in a circle, such as this, this and this. I have tried to apply the techniques from the answers to those questions, to no avail.
For the limits in those other questions, the closed forms (if not $1$) involve $\text{cosh}$, so I would not be surprised if the closed form of $L$ in this question also involves cosh. Maybe $L=2\cosh{\left(\pi e^{\frac{1}{\pi}-\frac{1}{e}}\right)}\approx19.929$ ?
UPDATE: From @Claude Leibovici's answer, it seems that $\left(\dfrac{P_{4n}}{P_{2n}}-1\right)\to\dfrac{1}{2}\left(\dfrac{P_{2n}}{P_{n}}-1\right)$ as $n\to\infty$. Then based on his value of $P_{60000}$, I project that $L\approx 19.950955$.
UPDATE2: I believe the number of points around the circle does not have to be a multiple of $4$. Any positive integer will do. If the average area of the regions is $6$, then $\lim\limits_{n\to\infty}(\text{product of areas})=L^2$.
 A: The answer is

$$L=2\cosh{\left(\pi\sqrt{\dfrac{\phi}{\sqrt5}}\right)}+2\cosh{\left(\pi\sqrt{\dfrac{1}{\phi\sqrt5}}\right)}=19.9505522212724487...$$

where $\phi=\dfrac{1+\sqrt5}{2}$, the golden ratio.
We have
$L=\lim\limits_{n\to\infty}\prod\limits_{k=1}^n \left(\left(6+\dfrac{4n}{\pi}\left(\sin{\dfrac{\pi}{2n}}\right)\left(\cos{\dfrac{(2k-1)\pi}{2n}}\right)\right)^2-\left(\dfrac{32n}{\pi}\left(\sin{\dfrac{\pi}{4n}}\right)\left(\cos{\dfrac{(2k-1)\pi}{4n}}\right)\right)^2\right)$
Let $\theta=\dfrac{(2k-1)\pi}{2n}$, and replace all $n$ with $\frac{n}{2}$. (If $n$ is odd, replace all $n$ with $\frac{n-1}{2}$.)
$L=\lim\limits_{n\to\infty}\prod\limits_{k=1}^{n/2} \left(\left(6+\dfrac{2n}{\pi}\left(\sin{\dfrac{\pi}{n}}\right)\left(2\cos^2{\theta}-1\right)\right)^2-\left(\dfrac{16n}{\pi}\left(\sin{\dfrac{\pi}{2n}}\right)\cos{\theta}\right)^2\right)$
Factorise the difference of squares. I will use $A$ and $B$ for shorthand.
$L=\lim\limits_{n\to\infty}\prod\limits_{k=1}^{n/2}(A+B)(A-B)$
$=\lim\limits_{n\to\infty}\prod\limits_{k=1}^{n/2}(A+B)\prod\limits_{k=1}^{n/2}(A-B)$
$=\lim\limits_{n\to\infty}\prod\limits_{k=1}^{n/2}(A+B)\prod\limits_{k=n/2+1}^{n}(A+B)$
$=\lim\limits_{n\to\infty}\prod\limits_{k=1}^{n}(A+B)$
$=\lim\limits_{n\to\infty}\prod\limits_{k=1}^{n}\left(6+\dfrac{2n}{\pi}\left(\sin{\dfrac{\pi}{n}}\right)\left(2\cos^2{\theta}-1\right)+\dfrac{16n}{\pi}\left(\sin{\dfrac{\pi}{2n}}\right)\cos{\theta} \right)$
(In fact, we could have arrived at this last expression for $L$ more directly by using another method to express the areas of the regions in the circle, i.e. on each side of the common line of symmetry, let $f(k)=$ area enclosed by the $k$th parabola from the bottom and its outer arc, then the areas enclosed by neighboring parabolas and the circle are $f(k)-f(k-1)$.)
Factorise the quadratic in $\cos{\theta}$, and use the Maclaurin series for sine. After a lot of simplification, we get
$L=\lim\limits_{n\to\infty}\prod\limits_{k=1}^{n}\left(2\cos\theta-2-\dfrac{\dfrac{5+2\sqrt5}{20}\pi^2}{n^2}+O\left(\dfrac{1}{n^4}\right)\right)\left(2\cos\theta-2-\dfrac{\dfrac{5-2\sqrt5}{20}\pi^2}{n^2}+O\left(\dfrac{1}{n^4}\right)\right)$
Using the fact that $\lim\limits_{n\to\infty}\prod\limits_{k=1}^n \left(2\cos\theta-2-\dfrac{\alpha^2}{n^2} \right)=2\cosh\alpha$ for all $\alpha\in\mathbb{R}$ (which can be proved using the technique shown in @metamorphy's answer here),
$L=4\cosh{\left(\pi\sqrt{\dfrac{5+2\sqrt5}{20}}\right)}\cosh{\left(\pi\sqrt{\dfrac{5-2\sqrt5}{20}}\right)}$
Using the identity for $\cosh{A}+\cosh{B}$,
$L=2\cosh{\left(\pi\sqrt{\dfrac{\phi}{\sqrt5}}\right)}+2\cosh{\left(\pi\sqrt{\dfrac{1}{\phi\sqrt5}}\right)}$
