TITLE: $\prod_{k=3}^{\infty} 1 - \tan( \pi/2^k)^4$
QUESTION [5 upvotes]: So I found this
$$\prod_{k=3}^{\infty} 1 - \tan( \pi/2^k)^4$$ 
here.
I have only ever done tests for convergence of infinite sums.
At this link it shows a way to convert but
in this case an is less than one for all k. I can see that $ f(k) = 1 - \tan( \pi/2^k)^4 $ is a strictly increasing function and in fact it is increasing very quickly however it is bound [ $ 1 - \tan( \pi/2^3)^4$, 1) where the lower bound is quite nearly one. 
This seems to mean it will converge to something slightly less than the lower bound of the function. 
I see that is converges. I want to know if there analytic method to provide a closed from of the solution such as with an alternating series. I figure I can simply do it computationally with MatLab and it would be very simple but I thought there must another way.

REPLY [9 votes]: Write $\displaystyle 1- \tan^4 \theta$ as $\dfrac{\cos^2 \theta - \sin^2 \theta}{\cos^4 \theta} = \dfrac{\cos 2\theta}{\cos^4 \theta}$ and make repeated use of the following trick:
To evaluate $\displaystyle \prod_{k=1}^{n}\cos \dfrac{\theta}{2^k}$, multiply and divide by $\sin \dfrac{\theta}{2^n}$ and use $\displaystyle 2\sin \theta \cos \theta = \sin 2\theta$.
This should give you a closed form formula for the product of first $\displaystyle n$ terms, which I believe can be evaluated easily as $\displaystyle n \to \infty$.
For more information on this you can refer: http://en.wikipedia.org/wiki/Vi%C3%A8te's_formula