On calculating the limit of the infinite product $\prod_{k=3}^n (1-\tan^4\frac{\pi}{2^k})$ 
Let $S_n=\prod_{k=3}^n (1-\tan^4\frac{\pi}{2^k})$. What is the value of $\lim_{n \to \infty} S_n$ ? 

What I attempted:- 
$\log S_n=\sum_{k=3}^n \log (1-\tan^4\frac{\pi}{2^k})$.
Since $\lim_{x \to 0} \frac{\tan x}{x}=1$, $\tan^4\frac{\pi}{2^k}\approx \left(\frac{\pi}{2^k}\right)^4$
Thus, $\log S_n=\sum_{k=3}^n \log (1-\frac{\pi^4}{2^{4k}})\approx \sum_{k=3}^n\left( -\frac{\pi^4}{2^{4k}}\right) $ 
Taking limit as $n \to \infty$, $\lim_{n \to \infty} \log S_n=\frac{-\pi^4}{3840}$. 
Finally, $\lim_{n \to \infty}S_n=e^{\frac{-\pi^4}{3840}}\approx 1-\frac{\pi^4}{3840}$ 
Am I correct? Is there any other better method which yield the accurate limit? I was told that the exact limit should be one of the $4$ options:- $\frac{\pi^3}{4},\frac{\pi^3}{16}, \frac{\pi^3}{32},\frac{\pi^3}{256}$. I guess the third option is correct as it is giving almost the same value that I have obtained.
 A: You obtained an approximation of the exact value. In order to find such exact vale, note that
$$(1-\tan^4(\alpha/2))=(1+\tan^2(\alpha/2))(1-\tan^2(\alpha/2))=
\frac{4}{\cos(\alpha)}\left(\frac{\tan(\alpha/2)}{\tan(\alpha)}\right)^2.$$
Hence, as $n$ goes to infinity,
$$\prod_{k=3}^n \left(1-\tan^4(\pi/2^k)\right)=\frac{4^{n-2}}{\prod_{k=3}^n\cos(\pi/2^{k-1})}\cdot \left(\prod_{k=3}^n\frac{\tan(\pi/2^k)}{\tan(\pi/2^{k-1})}\right)^2\\=4^{n-2}\cdot 2^{n-2}\sin(\pi/2^{n-1})\cdot \left(\frac{\tan(\pi/2^n)}{\tan(\pi/2^{2})}\right)^2\to \frac{\pi^3}{32}$$
where we used the known fact that
$$\prod\limits_{k=2}^{n}\cos\left(\frac{\pi }{2^{k}}\right)= 
\frac{1}{2^{n-1}\sin(\pi/2^n)}$$
(see for example How to evaluate $\lim\limits_{n\to \infty}\prod\limits_{r=2}^{n}\cos\left(\frac{\pi}{2^{r}}\right)$).
A: What you're describing is how to arrive at an approximation of the actual product.
But you can compute the exact value for the product using trigonometric identities.
$$\begin{split}
S_n&=\prod_{k=3}^n (1-\tan^4\frac{\pi}{2^k})\\
&=\prod_{k=3}^n (1-\tan^2\frac{\pi}{2^k})(1+\tan^2\frac{\pi}{2^k})\\
&=\prod_{k=3}^n (\frac {\cos^2(\frac{\pi}{2^k}) - \sin^2(\frac{\pi}{2^k})}{\cos^2 \frac{\pi}{2^k}})(\frac 1 {\cos^2\frac{\pi}{2^k}})\\
&=\prod_{k=3}^n \frac {\cos(\frac {\pi}{2^{k-1}})}{\cos^4 \frac{\pi}{2^k}}\\
&= \frac{\cos\frac \pi 4}{\cos^4 \frac \pi 8}\frac{\cos\frac \pi 8}{\cos^4 \frac \pi {16}}\frac{\cos\frac \pi {16}}{\cos^4 \frac \pi {32}}...\\
&=\cos\frac \pi 4 \prod_{k=3}^n \frac 1 {\cos^3 \frac \pi {2^k}}
\end{split}$$
Now because $$\frac 1 {\cos \theta}=\frac {2 \sin \theta}{\sin 2\theta}$$
you can verify that, for any $0<\theta<\pi$,
(Viète's formula)
$$
\prod_{k=1}^{+\infty} \frac 1 {\cos \frac \theta {2^k}}=\frac \theta {\sin \theta}
$$
Which yields, for $\theta=\frac \pi 4$,
$$
\prod_{k=3}^{+\infty} \frac 1 {\cos \frac \pi {2^k}}=\frac \pi {4\sin \frac \pi 4}
$$
So finally,
$$\lim_{n\rightarrow+\infty}S_n=\left(\cos{\frac \pi 4}\right)\frac {\pi^3} {4^3\sin^3 \frac \pi 4}$$
In other words,
$$\lim_{n\rightarrow+\infty}S_n=\frac {\pi^3}{32}$$
