What is the product of the areas of every regular polygon inscribed in a circle of area $1$?

Question

What is a closed form of $P=\prod\limits_{k=3}^{\infty}\frac{k}{2\pi}\sin{\left(\frac{2\pi}{k}\right)}\approx 0.05934871...$ ?

This is the product of the areas of every regular polygon inscribed in a circle of area $1$.

I do not know how to evaluate this product. I have been trying to use complex numbers, to no avail.

Remarks:

$P$ is remarkably close to $\frac{1}{6\pi-2}\approx 0.0593487\color{red}{45}...$ But this is not a closed form, because the product decreases as the number of factors increases, and according to desmos, $$\prod\limits_{k=3}^{10^8}\frac{k}{2\pi}\sin{\left(\frac{2\pi}{k}\right)}<\frac{1}{6\pi-2}$$

There is a related question: What is the product of the circumferences of every regular polygon inscribed in a circle of circumference $1$? This is $P'=\prod\limits_{k=3}^{\infty}\frac{k}{\pi}\sin{\left(\frac{\pi}{k}\right)}\approx 0.51633595...$ The ratio of $P$ to $P'$ is $\prod\limits_{k=3}^{\infty}\cos{\left(\frac{\pi}{k}\right)}\approx 0.11494204...$ which is the Kepler-Boukamp constant and has no known closed form. But is there a closed form of $P$?

EDIT

The product of the areas of every odd-gon inscribed in a circle of area $1$, equals $\frac{\pi}{2}\times$ the Kepler-Boukamp constant, as shown here.

Answer 1

Not a closed form but some ideas.

$$P=\prod\limits_{k=3}^{\infty}\frac{k}{2\pi}\sin{\left(\frac{2\pi}{k}\right)}=\frac{135 \sqrt{\frac{1}{2} \left(5+\sqrt{5}\right)}}{16 \pi ^4}\,\,\prod\limits_{k=7}^{\infty}\frac{k}{2\pi}\sin{\left(\frac{2\pi}{k}\right)}$$ Considering now $$\log \left(\frac{k }{2 \pi }\sin \left(\frac{2 \pi }{k}\right)\right)=\sum_{n=1}^\infty (-1)^n \,\frac{ 4^{2 n-1} \,\pi ^{2 n}\, B_{2 n} }{n^2\, \Gamma (2 n)}k^{-2 n}$$ $$\sum_{k=7}^\infty \log \left(\frac{k }{2 \pi }\sin \left(\frac{2 \pi }{k}\right)\right)=\sum_{n=1}^\infty (-1)^n \frac{4^{2 n-1}\, \pi ^{2 n}\, B_{2 n} \,\Big[\zeta (2 n)-\sum_{k=1}^6 k^{-2 n}\Big]}{n^2\, \Gamma (2 n)}$$

At this point, I am stuck but notice that the partial sums $\sum_{n=1}^p$ converge quite fast $$\left( \begin{array}{cc} p &\sum_{n=1}^p \\ 10 & \color{red}{-1.021056747445}10376401183253045 \\ 20 & \color{red}{-1.02105674744521511310815}554805 \\ 30 & \color{red}{-1.02105674744521511310815627874} \\ 40 & \color{red}{-1.02105674744521511310815627874} \\ \end{array} \right)$$ as well as the partial products which can be computed as accurately as desired $$P \sim 0.0593487129451443884588576520019102071726674\cdots$$

Inverse symbolic calculators do not identify this number but, amuzing with no meaning at all, this is very close to the smallest root of $4753 x^2-9364 x+539=0$ (the difference is $1.95\times 10^{-15}$).