Value of $\lim\limits_{n\rightarrow\infty}\prod\limits_{k=3}^n\left(1-\tan^4\frac{\pi}{2^k}\right)=\frac{\pi^3}{m}$

Question

Find the value of m for the following $$\lim\limits_{n\rightarrow\infty}\prod_{k=3}^n\left(1-\tan^4\frac{\pi}{2^k}\right)=\frac{\pi^3}{m}$$

Answer 1

Long, but detailed $$\prod_{k=3}^n\left(1-\tan^4\frac{\pi}{2^k}\right)= \prod_{k=3}^n\left(1-\tan^2\frac{\pi}{2^k}\right)\left(1+\tan^2\frac{\pi}{2^k}\right)=\\ \prod_{k=3}^n\left(\frac{\cos^2{\frac{\pi}{2^k}}-\sin^2{\frac{\pi}{2^k}}}{\cos^2{\frac{\pi}{2^k}}}\right)\left(\frac{\cos^2{\frac{\pi}{2^k}}+\sin^2{\frac{\pi}{2^k}}}{\cos^2{\frac{\pi}{2^k}}}\right)\overset{\cos{2x}=\cos^2{x}-\sin^2{x}}{=}\\ \prod_{k=3}^n\frac{\cos{\frac{\pi}{2^{k-1}}}}{\cos^4{\frac{\pi}{2^k}}}= \frac{\cos{\frac{\pi}{4}}}{\cos{\frac{\pi}{2^n}}}\cdot\prod_{k=3}^n\frac{1}{\cos^3{\frac{\pi}{2^k}}}=\\ \frac{\cos{\frac{\pi}{4}}}{\cos{\frac{\pi}{2^n}}}\cdot\prod_{k=3}^n\frac{\color{red}{\sin^3{\frac{\pi}{2^k}}}}{\color{red}{\sin^3{\frac{\pi}{2^k}}}\cdot\cos^3{\frac{\pi}{2^k}}}\overset{\sin{2x}=2\sin{x}\cos{x}}{=}\\ \frac{\cos{\frac{\pi}{4}}}{\cos{\frac{\pi}{2^n}}}\cdot\prod_{k=3}^n\frac{\sin^3{\frac{\pi}{2^k}}}{\frac{1}{2^3}\cdot\sin^3{\frac{\pi}{2^{k-1}}}}= 2^{3(n-2)}\cdot\frac{\cos{\frac{\pi}{4}}}{\cos{\frac{\pi}{2^n}}}\cdot\prod_{k=3}^n\frac{\sin^3{\frac{\pi}{2^k}}}{\sin^3{\frac{\pi}{2^{k-1}}}}=\\ 2^{3(n-2)}\cdot\frac{\cos{\frac{\pi}{4}}}{\cos{\frac{\pi}{2^n}}}\cdot\frac{\sin^3{\frac{\pi}{2^n}}}{\sin^3{\frac{\pi}{4}}}= 2^{3n-5}\cdot\frac{1}{\cos{\frac{\pi}{2^n}}}\cdot\sin^3{\frac{\pi}{2^n}}=\\ \frac{\pi^3}{2^5}\cdot\frac{1}{\cos{\frac{\pi}{2^n}}}\cdot\left(\frac{\sin{\frac{\pi}{2^n}}}{\frac{\pi}{2^n}}\right)^3\to\frac{\pi^3}{2^5}, n\to\infty$$ given that $$\cos{\frac{\pi}{2^n}}\to \cos{0}=1, n\to\infty$$ and $$\frac{\sin{\frac{\pi}{2^n}}}{\frac{\pi}{2^n}}\to 1, n\to\infty$$

Answer 2

It is possible to find a closed form for your product, if fact we have for any $ k\geq 3 $ :

\begin{aligned}1-\tan^{4}{\left(\frac{\pi}{2^{k}}\right)}&=\frac{1-\tan^{2}{\left(\frac{\pi}{2^{k}}\right)}}{\cos^{2}{\left(\frac{\pi}{2^{k}}\right)}}\\ &=\frac{\cos{\left(\frac{\pi}{2^{k-1}}\right)}}{\cos^{4}{\left(\frac{\pi}{2^{k}}\right)}}\\ 1-\tan^{4}{\left(\frac{\pi}{2^{k}}\right)}&=\frac{8\sin^{3}{\left(\frac{\pi}{2^{k}}\right)}}{\sin^{3}{\left(\frac{\pi}{2^{k-1}}\right)}}\times\frac{\cos{\left(\frac{\pi}{2^{k-1}}\right)}}{\cos{\left(\frac{\pi}{2^{k}}\right)}}\end{aligned}

Thus, if $ n\geq 3 $, then : \begin{aligned}\prod_{k=3}^{n}{\left(1-\tan^{4}{\left(\frac{\pi}{2^{k}}\right)}\right)}&=8^{n-2}\left(\prod_{k=3}^{n}{\frac{\sin^{3}{\left(\frac{\pi}{2^{k}}\right)}}{\sin^{3}{\left(\frac{\pi}{2^{k-1}}\right)}}}\right)^{3}\left(\prod_{k=3}^{n}{\frac{\cos{\left(\frac{\pi}{2^{k-1}}\right)}}{\cos{\left(\frac{\pi}{2^{k}}\right)}}}\right)\\ &=2^{3n-5}\frac{\sin^{3}{\left(\frac{\pi}{2^{n}}\right)}}{\cos{\left(\frac{\pi}{2^{n}}\right)}}\end{aligned}