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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.
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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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$\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}}$ = $2sin(\frac{\pi}{96})$ = $2sin(\frac{\pi}{3*2^5})$
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

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:
$$
\lim_{n\to\infty}\frac{\sin\frac{\pi}{2^n}}{\sin\frac{\pi}{3\cdot 2^n}}
= \lim_{n\to\infty}\left(\frac{\sin\frac{\pi}{2^n}}{\sin\frac{\pi}{3\cdot 2^n}}
\cdot \frac{\frac{\pi}{2^n}}{\sin\frac{\pi}{2^n}}
\cdot\frac{\sin\frac{\pi}{3\cdot 2^n}}{\frac{\pi}{3\cdot 2^n}} \right)
= \lim_{n\to\infty}\frac{\frac{\pi}{2^n}}{\frac{\pi}{3\cdot 2^n}}
= \lim_{n\to\infty}3 = 3
$$
