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 $\lim\limits_{n\to\infty}(\text{product of areas})=L^2$.
Best Answer
Continuing your calculations
$$\left(\begin{array}{cc} n & P_n \\ 1000 & 19.93455668774900154012061 \\ 2000 & 19.94275333857622878761460 \\ 3000 & 19.94548671811436159136837 \\ 4000 & 19.94685362596789631782790 \\ 5000 & 19.94767384047956738149078 \\ 6000 & 19.94822067924005947638804 \\ 7000 & 19.94861129260183551122982 \\ 8000 & 19.94890426041492105340228 \\ 9000 & 19.94913212888703196970719 \\ 10000 & 19.94931442657381322499540 \\ 20000 & 19.95013479816578198778725 \\ 30000 & 19.95040826700053826290505 \\ 40000 & 19.95054500360003303052213 \\ 50000 & 19.95062704625801974143739 \\ 60000 & 19.95068174165430121991535 \\ \end{array}\right)$$
Slower would be difficult to find !
Edit
Using the last six values of the above table and the simplistic $$P_n=a+\frac b n$$ a curve fit predicts an asymptotic value of $19.9509552$ which is not recognized by the $ISC$.