Product of areas in a circle

The limit is, as suspected, $$4\cosh^2\left(\frac{\pi}{2\sqrt 3}\right).$$ First, define $\epsilon_n=1-\frac n\pi\sin(\pi/n)$, so that $$\epsilon_n=\frac{\pi^2}{6n^2}-O(n^{-3})$$ and \begin{align*} a_n &:=\sum_{k=1}^n\log\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)\\ &=\sum_{k=1}^n\log\left(2-\frac n\pi\left[\sin\left(\frac{2k\pi}n\right)-\sin\left(\frac{2(k-1)\pi}n\right)\right]\right)\\ &=\sum_{k=1}^n\log\left(2-2\frac n\pi\sin\left(\frac\pi n\right)\cos\left(\frac{(2k-1)\pi}n\right)\right)\\ &=\sum_{k=1}^n\log\left(2-2\cos\left(\frac{(2k-1)\pi}n\right)+2\cos\left(\frac{(2k-1)\pi}n\right)\epsilon_n\right). \end{align*} Define $$b_n:=\sum_{k=1}^n\log\left(2-2\cos\left(\frac{(2k-1)\pi}n\right)\right),$$ so that $$c_n:=a_n-b_n=\sum_{k=1}^n\log\left(1+\frac{\epsilon_n\cos\left(\frac{(2k-1)\pi}n\right)}{1-\cos\left(\frac{(2k-1)\pi}n\right)}\right).$$ The summand here is symmetric about $k=\frac{n+1}2$, and is $\log(1-\epsilon_n/2)=o(1)$ at $k=\frac{n+1}2$, so $$c_n=o(1)+2\sum_{k=1}^{\lfloor n/2\rfloor}\log\left(1+\frac{\epsilon_n\cos\left(\frac{(2k-1)\pi}n\right)}{1-\cos\left(\frac{(2k-1)\pi}n\right)}\right).$$ Write $$c_{n,K}=\sum_{k=1}^K\log\left(1+\frac{\epsilon_n\cos\left(\frac{(2k-1)\pi}n\right)}{1-\cos\left(\frac{(2k-1)\pi}n\right)}\right).$$ We have, for large $n$ (being liberal with constant factors), that \begin{align*} |c_n-2c_{n,K}| &\leq 2\sum_{k=K+1}^{\lfloor n/2\rfloor}\left|\log\left(1+\frac{\epsilon_n\cos\left(\frac{(2k-1)\pi}n\right)}{1-\cos\left(\frac{(2k-1)\pi}n\right)}\right)\right|\\ &\leq 4\sum_{k=K+1}^{\lfloor n/2\rfloor}\frac{\epsilon_n}{1-\cos\left(\frac{(2k-1)\pi}n\right)}\\ &=2\epsilon_n\sum_{k=K+1}^{\lfloor n/2\rfloor}\frac1{\sin^2\left(\frac{(2k-1)\pi}{2n}\right)}\\ &\leq 4\epsilon_n\sum_{k=K+1}^{\lfloor n/2\rfloor}\frac1{\left(\frac{(2k-1)\pi}{2n}\right)^2}\\ &=\frac{16\epsilon_nn^2}{\pi^2}\sum_{k=K+1}^{\lfloor n/2\rfloor}\frac1{(2k-1)^2}\\ &\leq\frac 83\sum_{k=K+1}^\infty \frac1{(2k-1)^2}\leq \frac2{3K}. \end{align*} On the other hand, for $n$ large and $k<n^{1/4}$, \begin{align*} \log\left(1+\frac{\epsilon_n\cos\left(\frac{(2k-1)\pi}n\right)}{1-\cos\left(\frac{(2k-1)\pi}n\right)}\right) &=\log\left(1+\frac{\left(\frac{\pi^2}{6n^2}+O(n^{-3})\right)\left(1-O(n^{-1})\right)}{2\sin^2\left(\frac{(2k-1)\pi}{2n}\right)}\right)\\ &=\log\left(1+\frac{\left(\frac{\pi^2}{6n^2}+O(n^{-3})\right)\left(1-O(n^{-3/2})\right)}{\frac{(2k-1)^2\pi^2}{2n^2}+O(n^{-3})}\right)\\ &=\log\left(1+\frac1{3(2k-1)^2}+O(n^{-1})\right)\\ &=\log\left(1+\frac1{3(2k-1)^2}\right)+O\left(\frac 1n\right). \end{align*} This means that, setting $K=n^{1/4}$, $$c_n=2c_{n,K}+o(1)=o(1)+2\sum_{k=1}^K\log\left(1+\frac1{3(2k-1)^2}\right),$$ and thus $$\lim_{n\to\infty} c_n=2\sum_{k=1}^\infty \log\left(1+\frac1{3(2k-1)^2}\right).$$ Now, \begin{align*} e^{b_n} &=\prod_{k=1}^n\left(2-2\cos\left(\frac{(2k-1)\pi}n\right)\right)\\ &=\prod_{k=1}^n\left|1-e^{\frac{(2k-1)\pi i}n}\right|^2\\ &=\prod_{z^n=-1}\left|1-z\right|^2\\ &=(1-(-1))^2=4. \end{align*} This shows that $$\lim_{n\to\infty} a_n=2\ln 2+\lim_{n\to\infty}c_n=2\ln 2+2\sum_{k=1}^\infty \log\left(1+\frac1{3(2k-1)^2}\right).$$ So, we need only evaluate this infinite series, which we call $S$. We first expand the $\log$ to get $$S=\sum_{k=1}^\infty\sum_{m=1}^\infty \frac{(-1)^{m-1}}{m3^m(2k-1)^{2m}}=\sum_{m=1}^\infty \frac{(-1)^{m-1}}{m3^m}\sum_{k=1}^\infty \frac1{(2k-1)^{2m}},$$ where the interchange of summation is justified since the sum is absolutely convergent. Using that $$\sum_{k=1}^\infty \frac1{(2k-1)^{2m}}=\left(1-2^{-2m}\right)\zeta(2m),$$ we get $$S=\sum_{m=1}^\infty \frac{(-1)^{m-1}}m\left(3^{-m}-12^{-m}\right)\zeta(2m).$$ Write $$f(x)=\sum_{m=1}^\infty\frac{(-1)^{m-1}x^m\zeta(2m)}m.$$ We compute $$f'(x)=\sum_{m=1}^\infty(-x)^{m-1}\zeta(2m).$$ An identity here gives that $$\sum_{m=1}^\infty t^{2m}\zeta(2m)=\frac{1-\pi t\cot(\pi t)}{2},$$ and so $$f'(x)=-\frac{1-\pi i\sqrt{x}\cot(\pi i\sqrt x)}{2x}=\frac{\pi\sqrt x\coth(\pi\sqrt x)-1}{2x}.$$ This gives \begin{align*} S &=f\left(\frac13\right)-f\left(\frac1{12}\right)\\ &=\int_{1/12}^{1/3}\frac{\pi\sqrt x\coth(\pi\sqrt x)-1}{2x}dx\\ &=\int_{\frac1{2\sqrt3}}^{\frac1{\sqrt3}}\frac{\pi y\coth(\pi y)-1}ydy\\ &=\int_{\frac1{2\sqrt3}}^{\frac1{\sqrt3}}\pi\coth(\pi y)dy-\log 2\\ &=\log(\sinh(\pi y))\bigg|_{\frac1{2\sqrt3}}^{\frac1{\sqrt3}}-\log 2\\ &=\log\left(\frac{\sinh\left(\frac{\pi}{\sqrt 3}\right)}{2\sinh\left(\frac{\pi}{2\sqrt 3}\right)}\right)\\ &=\log\cosh\frac{\pi}{2\sqrt 3}. \end{align*} So, the limit is $$e^{2\ln 2+2S}=4\cosh^2\left(\frac{\pi}{2\sqrt 3}\right),$$ as desired.