Division of 2 different Infinite Nested square roots Has anyone come across the following type of nested square roots problem?
$\sqrt{2-{\sqrt{2+{\sqrt{2+...n times {\sqrt{2}}}}}}}$ divided by 
$\sqrt{2-{\sqrt{2+{\sqrt{2+...(n+1)times {\sqrt{3}}}}}}}$
Converging towards 3 as the 'n' increases
Are there any theorem or formulas to calculate multiplication or division of infinite nested square roots?
Note: 2nd sum done in calculator has same $\sqrt3$ at its end which is not visible. Just one term of nested square root is increased which is shown in picture
$\sqrt{2}$ = $2cos(\frac{\pi}{4})$
$\sqrt{2+\sqrt{2}}$ = $2cos(\frac{\pi}{8})$
$\sqrt{2+\sqrt{2+\sqrt{2}}}$ = $2cos(\frac{\pi}{16})$
.
.
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$\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2}}}}$ = $2sin(\frac{\pi}{32})$ or $2sin(\frac{\pi}{2^5})$
$\sqrt{3}$ = $2cos(\frac{\pi}{6})$
$\sqrt{2+\sqrt{3}}$ = $2cos(\frac{\pi}{12})$
$\sqrt{2+\sqrt{2+\sqrt{3}}}$ = $2cos(\frac{\pi}{24})$
$\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}$ = $2cos(\frac{\pi}{48})$
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.
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$\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}}$ = $2sin(\frac{\pi}{96})$ = $2sin(\frac{\pi}{3*2^5})$
This looks more interesting to note
$\frac{2sin(\frac{\pi}{2^5})}{2sin(\frac{\pi}{3*2^5})}$
For very small values. "x" and "sin(x)" are almost the same
(as long as "x" is in Radians!)
It is true that $\frac{(\frac{\pi}{2^5})}{{(\frac{\pi}{3*2^5})}}$ simplifying to 3?
Is there any other means like, using limits or integration to solve such problems?
Please throw light me someone
Thanks in advance
 A: Well, you seem to have basically proven it. We want to show that
$$
\lim_{n\to\infty}\frac{\sin\frac{\pi}{2^n}}{\sin\frac{\pi}{3\cdot 2^n}} = 3
$$
We would like to use that $\sin x \approx x$ for small $x$. To be rigorous, we can write $\sin x = x + xo(1)$, where $o(1)$ is a function that goes to $0$ as $x\to0$. Then
$$
\frac{\sin\frac{\pi}{2^n}}{\sin\frac{\pi}{3\cdot 2^n}}
= \frac{\frac{\pi}{2^n} + \frac{\pi}{2^n}o_1(1)}{\frac{\pi}{3\cdot 2^n} + \frac{\pi}{3\cdot 2^n}o_2(1)}
= \frac{3+3o_1(1)}{1+o_2(1)}
\underset{n\to\infty}\to 3
$$
Alternatively, we can use the well known fact that $\frac{\sin x}{x} \to 1$ for $x\to0$. Thus:
$$
\lim_{n\to\infty}\frac{\sin\frac{\pi}{2^n}}{\sin\frac{\pi}{3\cdot 2^n}}
= \lim_{n\to\infty}\left(\frac{\sin\frac{\pi}{2^n}}{\sin\frac{\pi}{3\cdot 2^n}}
\cdot \frac{\frac{\pi}{2^n}}{\sin\frac{\pi}{2^n}}
\cdot\frac{\sin\frac{\pi}{3\cdot 2^n}}{\frac{\pi}{3\cdot 2^n}} \right)
= \lim_{n\to\infty}\frac{\frac{\pi}{2^n}}{\frac{\pi}{3\cdot 2^n}}
= \lim_{n\to\infty}3 = 3
$$
A: I was going to post this as a comment, but my MathJax just isn't good enough to get in right in the comment box, without being able to see the rendered output. 
It looks like a triviality to prove the formulas you are hinting at for the finite nested radicals.   
As for 
$$\lim_{n\to\infty}\frac{2\sin\frac{\pi}{2^5}}{2\sin\frac{\pi}{3\cdot2^5}}=3,$$
that's an immediate consequence of $\lim_{x\to0}{\sin x\over x}=1$
