How do I evaluate this trigonometric limit involving infinite product?

Question

Tried a lot but couldn't solve it. How do I begin approaching this limit?

$ \displaystyle\lim _{n\to\infty} \displaystyle\prod_{k=3}^{n} 1 - \tan^4\left( \frac{\pi}{2^k}\right)$

Answer 1

There isn't that many trick to create a horrible looking and yet doable limit.

In school, if you are asked to evaluate a series or product which involve trigonometry functions whose arguments contain terms like $\frac{(\cdots)}{2^k}$. One thing you should do is lookup the double-angle formula for various trigonometry functions and see whether you can use them to turn your series/product into a telescoping one.

In this case, it do work.

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Let $t = \tan\theta$, we have

$$1 - t^4 = (1-t^2)^2\frac{1+t^2}{1-t^2} = \left(\frac{2t}{\frac{2t}{1-t^2}}\right)^2\frac{1+t^2}{1-t^2}$$

Recall $$\frac{2t}{1-t^2} = \tan(2\theta)\quad\text{ and }\quad \frac{1-t^2}{1+t^2} = \cos(2\theta) = \frac12\frac{\sin(4\theta)}{\sin(2\theta)} $$

We obtain $$1 - t^4 = 8\left(\frac{\tan(\theta)}{\tan(2\theta)}\right)^2\frac{\sin(2\theta)}{\sin(4\theta)} = 8\frac{\tan^2\theta\sin(2\theta)}{\tan^2(2\theta)\sin(4\theta)} $$ The product is telescoping and

$$\begin{align}\prod_{k=3}^n\left(1 - \tan^4\frac{\pi}{2^k}\right) &= 8^{n-2} \frac{\tan^2\frac{\pi}{2^n}\sin \frac{\pi}{2^{n-1}}}{\tan^2\frac{\pi}{4}\sin\frac{\pi}{2}}\\ &= \frac{1}{32} \left(2^n\tan \frac{\pi}{2^n}\right)^2\left(2^{n-1}\sin\frac{\pi}{2^{n-1}}\right)\end{align} $$

Using $$\lim_{N\to\infty} N\tan\frac{\pi}{N} = \lim_{N\to\infty} N\sin\frac{\pi}{N} = \pi$$ We obtain $$\lim_{n\to\infty} \prod_{k=3}^n\left(1 - \tan^4\frac{\pi}{2^k}\right) = \frac{\pi^3}{32}$$

Answer 2

I wouldn't quickly have an idea either, but WolframAlpha gives that $\prod_{k=3}^\infty\left(1-\tan^4\left(\frac{\pi}{2^k}\right)\right)\approx0.968946$ (https://www.wolframalpha.com/input/?i=product+of+(1-tan(pi%2F2%5En)%5E4)+from+3+to+infty).