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The question is to evaluate $$\prod_{r=1}^{\infty} \frac{\sin \frac {a}{2^r}}{\tan^2 \frac {a}{2^r} \tan \frac {a}{2^{r-1}}+\tan \frac {a}{2^{r}}}$$

I could rewrite the denominator as $$\tan\frac {a}{2^r}\left(\tan \frac {a}{2^r}\tan \frac{a}{2^{r-1}}+1\right)$$which is same as $$\tan \frac{a}{2^r}-\tan\frac {a}{2^{r-1}}$$And so the product transforms to $$\prod_{r=1}^{\infty} \frac{\sin \frac {a}{2^r}}{\tan \frac{a}{2^r}-\tan\frac {a}{2^{r-1}}}$$ I have no idea on how to proceed after this. Any help is appreciated. Thanks.

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    Is the multiplicand essentially equivalent to limit as n -> infinity of $(\frac{sin\frac{a}{2} }{tan^2\frac{a}{2}tana + tan\frac{a}{2}})^n$ – 2017-02-23

2 Answers 2

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The product can be rewritten as

$$\prod_{r=1}^{\infty} \frac{\cos{\left ( 2^{-r} a \right )}}{1+\tan{\left ( 2^{-r} a \right )} \tan{\left ( 2^{-r+1} a \right )}} $$

Use the tangent half-angle formula to simplify the denominator:

$$\begin{align}1+\tan{\left ( 2^{-r} a \right )} \tan{\left ( 2^{-r+1} a \right )} &= 1+\sqrt{\frac{1-\cos{\left ( 2^{-r} a \right )}}{1+\cos{\left ( 2^{-r} a \right )}}} \frac{\sqrt{1-\cos^2{\left ( 2^{-r} a \right )}}}{\cos{\left ( 2^{-r} a \right )}}\\ &= \frac1{\cos{\left ( 2^{-r} a \right )}} \end{align}$$

The product is then equal to

$$\left [ \prod_{r=1}^{\infty} \cos{\left ( 2^{-r} a \right )} \right ]^2 $$

Note that

$$ \prod_{r=1}^{\infty} \cos{\left ( 2^{-r} a \right )} = \frac{\sin{a}}{a}$$

Therefore, the product is equal to

$$\frac{\sin^2{a}}{a^2} $$

ADDENDUM

The product in the square brackets may be evaluated with ease using the sine double angle forumla. The product of the cosines is equal to

$$\lim_{n \to \infty} \frac{\sin{a}}{2 \sin{(a/2)}} \frac{\sin{(a/2)}}{2 \sin{(a/2)}} \cdots \frac{\sin{(a/2^{n-1})}}{2 \sin{(a/2^n)}} = \lim_{n \to \infty} \frac{\sin{a}}{2^n \sin{(a/2^n)}}$$

The result follows.

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    Thanks for your answer.Can you please explain how did you get $\cos a$ in the product. – 2017-02-23
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    Error in the original analysis. Fixed and numerically verified. – 2017-02-23
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    I think in your first line it should be $\tan (2^{-r+1}a)$ instead of $\tan (2^{-r-1}a)$ – 2017-02-23
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    @navinstudent: yes, that was a typo, thanks. – 2017-02-23
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    Now it is all fine.Thanks. – 2017-02-23
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Lets start with:

$\tan \frac {a}{2^{r-1}} = \frac {2\tan \frac {a}{2^{r}}}{1-\tan^2 \frac {a}{2^{r}}}$

and that gets everything in terms of $\frac {a}{2^{r}}$

$u = \frac {a}{2^{r}}$

$\frac {\sin u}{\tan^2 u\frac {2 \tan u}{1-\tan^2 u}+\tan u}\\ \frac {\sin u (1-\tan^2 u)}{\tan u(1 +\tan^2 u)}\\ \frac {\sin u (1-\tan^2 u)}{\tan u(\sec^2 u)}\\ \cos^2 \cot u\sin u(1-\tan^2 u)\\ \cos u (\cos^2u-\sin^2 u)\\ \cos u (\cos 2u)$

$\prod_\limits{r=1}^{\infty} \cos \frac {a}{2^{r}} \cos \frac {a}{2^{r-1}} = \cos a\prod_\limits{r=1}^{\infty} \cos^2 \frac {a}{2^{r}}\\ \frac {\sin^2 a}{a^2} \cos {a}$