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Evaluate the following product: $$\newcommand{\T}[1]{\frac{\sin\frac{\theta}{#1}}{\tan^2\frac{\theta}{#1}\tan\frac{2\theta}{#1} + \tan\frac{\theta}{#1}}} \\ P(\theta) = \T{2} \times \T{2^2} \times \T{2^3} \times .... \infty$$ For $\theta = \frac \pi 4$

Simplified, $P(\theta)$ is $$P(\theta) = \lim_{n \to \infty}\prod_{r=1}^n T(\theta,r)= \lim_{n \to \infty}\prod_{r=1}^n\T{2^r}$$ The denominator can be simplified as follows: $$D = \tan\frac{\theta}{2^r}\left( \tan\frac{\theta}{2^r}\tan\frac{\theta}{2^{r-1}} + 1\right) \\ = \tan\frac{\theta}{2^{r-1}} - \tan\frac{\theta}{2^{r}}$$ After this, $P(\theta)$ becomes $$P(\theta) = \lim_{n \to \infty}\prod_{r=1}^n \frac{\sin\frac{\theta}{2^r}}{\tan\frac{\theta}{2^{r-1}}- \tan\frac{\theta}{2^r}}$$

One more detail I found out is that $\lim_{n \to \infty} T(\theta,n) = 1$, but I couldn't proceed further from here. Any hints/solutions are appreciated.

EDIT: After the hints in the comments, $T(\theta, r)$ resolves to $\cos \frac \theta {2^{r-1}} \cos \frac \theta {2^r}$ as follows (assuming $\frac \theta {2^r} = t$) $$\begin{gather} T(\theta, n) = \frac{\sin t}{\tan^2t\tan 2t + \tan t} \\ = \frac{\cos t}{\tan t \tan 2t + 1} \\ = \frac{\cos t(1-\tan^2t)}{1+\tan^2t} \\ = \cos t \cos 2t \\ = \cos \frac \theta {2^{r-1}} \cos \frac \theta {2^r} \end{gather}$$

Now, $$P(\theta) = \lim_{n \to \infty} \frac{ \left( \cos\theta\cos\frac\theta2... \cos \frac{\theta}{2^n} \right)^2 }{\cos\theta} = \frac{\sin^2\theta}{2^{2n}\sin^2 \frac \theta {2^n}\cos \theta} = \frac{\sin^2 \theta}{\theta^2 \cos \theta}$$ Therefore, $$\boxed{P(\pi/4) = \frac{8\sqrt2}{\pi^2}}$$

However, the answer mentioned in the textbook is $\frac{2}{\pi}$. Where am I going wrong? (I think there's a silly mistake somewhere here; just not able to find it :(

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    $\begingroup$ The rth term in product is $\cos \frac{\theta} {2^{r-1}}\cos\frac {\theta} {2^r}$. $\endgroup$
    – Paramanand Singh
    Aug 10, 2020 at 5:22
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    $\begingroup$ And you know that $\frac{\sin x} {x} =\prod_{r\geq 1}\cos(x/2^r)$. $\endgroup$
    – Paramanand Singh
    Aug 10, 2020 at 5:25
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    $\begingroup$ The answer should come out $\frac{\sin \theta} {\theta} \frac{\sin 2\theta}{2\theta}$ which simplifies to $\frac{\sin^2\theta\cos\theta}{\theta^2}$ which equals $4\sqrt{2}/\pi^2$ if $\theta=\pi/4$. $\endgroup$
    – Paramanand Singh
    Aug 10, 2020 at 15:10
  • $\begingroup$ The book answer is wrong by the way. $\endgroup$
    – Paramanand Singh
    Aug 10, 2020 at 15:11
  • $\begingroup$ @ParamanandSingh You're right, I found the error. Thanks for helping me! (Can you post an answer so that I can accept it or should I go ahead and post one?) $\endgroup$ Aug 10, 2020 at 16:13

1 Answer 1

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Carrying on from the answer: $$\begin{gather} T(\theta, n) = \frac{\sin t}{\tan^2t\tan 2t + \tan t} \\ = \frac{\cos t}{\tan t \tan 2t + 1} \\ = \frac{\cos t(1-\tan^2t)}{1+\tan^2t} \\ = \cos t \cos 2t \\ = \cos \frac \theta {2^{r-1}} \cos \frac \theta {2^r} \end{gather}$$

Now, $$P(\theta) = \cos\theta \cos \frac \theta 2 \cdot \cos\frac\theta2 \cos\frac\theta{2^2}\cdot ... = \left( \cos\theta \cos\frac\theta2...\right) \left( \cos\frac\theta2\cos\frac\theta{2^2}...\right)$$ Let $$\begin{gather} S = \lim_{n \to \infty}\cos\frac\theta2...\cos\frac\theta{2^n}\\ S\sin\frac\theta{2^n} = \lim_{n \to \infty} \frac{\sin\theta}{2^n} \\ S = \lim_{n \to \infty} \frac{\sin\theta}{2^n \sin\frac{\theta}{2^n}} = \lim_{t \to 0} \frac{t\sin\theta}{\sin(\theta t)}\\ S = \frac{\sin\theta}{\theta} \end{gather}$$

Therefore, $P$ reduces to $$P(\theta) = \frac{\sin2\theta\sin\theta}{2\theta^2}$$ And the value of $P(\pi/4)$ would be $$P(\pi/4) = \frac{16}{2\sqrt2\pi^2}\\ \boxed{P(\pi/4) = \frac{4\sqrt2}{\pi^2}}$$

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  • $\begingroup$ You may accept this answer to remove this question from unanswered queue. $\endgroup$
    – Paramanand Singh
    Aug 11, 2020 at 9:57
  • $\begingroup$ The site says "You can accept your own answer in 17 hours" - Will do after that. $\endgroup$ Aug 11, 2020 at 11:58

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