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I already know that $$ a_n = \cos\left(\frac{\pi}{2^{n+1}}\right) = \overbrace{\frac{\sqrt{2+\sqrt{2+\ldots + \sqrt{2}}}}{2}}^{n\text{ roots}}$$ Also I know that $$\lim_{n\to\infty} 2\cos\left(\frac{\pi}{2^n}\right) = 2 \text{ and if } a_n \xrightarrow {n\to\infty} a \text{ then } \sqrt[n]{a_1 a_2 \ldots a_n} \xrightarrow{n\to\infty} a $$

With that method I only got indeterminate form

$$ \lim_{n\to\infty} \cos\left(\frac{\pi}{4}\right) \cos\left(\frac{\pi}{8}\right)\ldots \cos\left(\frac{\pi}{2^n}\right) = \Big(\frac{\sqrt[n]{a_1 a_2 \ldots a_n}}{2}\Big)^n = 1^\infty $$ Anyone knows a working solution?

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    $\begingroup$ multiply and divide the entire thing by $\sin\dfrac{\pi}{2^n}$ and see what happens. $\endgroup$ – dezdichado Dec 12 '18 at 16:46
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    $\begingroup$ Look at page 1 of gibbs.if.usp.br/~marchett/estocastica/… $\endgroup$ – saulspatz Dec 12 '18 at 16:54
  • $\begingroup$ Thank you, these two answers helped me solve this. $\endgroup$ – Victor Dec 12 '18 at 17:12
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If $x_n=\cos(\frac{\pi}{4}) \cos(\frac{\pi}{8})\ldots \cos(\frac{\pi}{2^n}) $ then $\ $ $$x_n\sin (\frac{\pi}{2^n})= \cos(\frac{\pi}{4}) \cos(\frac{\pi}{8}) \ldots \cos(\frac{\pi}{2^n}) \sin (\frac{\pi}{2^n}) $$ $$=\frac{1}{2^1} \cos(\frac{\pi}{4}) \cos(\frac{\pi}{8}) \ldots \cos(\frac{\pi}{2^{n-1}}) \sin (\frac{\pi}{2^{n-1}}) $$ $$ =\ldots= \frac{1}{2^{n-1}} $$ So $$x_n=\frac{1}{2^{n-1}\sin (\frac{\pi}{2^n})} $$ So $\lim_{n\to \infty }x_n=\frac{2}{\pi} $

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$$\sin(x)=2\sin(\frac{x}{2})\cos(\frac{x}{2})=2\cos(\frac{x}{2})2\sin(\frac{x}{4})\cos(\frac{x}{4})$$ By repeated application of the half angle formula we find $$\sin(x)=\lim_{n\to\infty}2^n\sin(\frac{x}{2^n})\prod_{k=1}^{n}\cos(\frac{x}{2^{k}})$$ By expanding $\sin(\frac{x}{2^n})$ into its taylor series we can easily derive

$$ \sin(x)=x\prod_{k=1}^{\infty} \cos(\frac{x}{2^k})$$

So $$\prod_{k=2}^{\infty}\cos(\frac{\pi}{2^k})=\lim_{x\to\pi}\frac{\sin(x)}{x\cos(\frac{x}{2})}=\lim_{x\to\pi}\frac{2\sin(\frac{x}{2})}{x}=\frac{2}{\pi}$$

See also Viete's formula

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What you are trying to proof is Viete's formula. What he did was trying to compare area's of regular polygons that are inscribed in a unit circle. The area of a regular polygon with $n$ sides is given by

$$ A_n = \frac12 n \sin\left(\frac\pi n\right)$$

If you compute now the ratio between two regular polygons, one with $2^n$ sides, and one with $2^{n-1}$ sides, then you get:

$$B_n=\frac{A_{2^{n-1}}}{A_{2^n}} = \frac{2^{n-1} \sin\left(\frac{\pi}{2^{n-1}}\right)}{2^{n} \sin\left(\frac{\pi}{2^{n}}\right)} = \cos\left(\frac{\pi}{2^{n}}\right)$$

This now implies that the product the OP tries to compute is equal to

$$C_n=B_3 B_4 ... B_n = \frac{A_4}{A_8}\cdot\frac{A_8}{A_{16}}\cdot\cdots\cdot\frac{A_{2^{n-1}}}{A_{2^n}}=\frac{A_4}{A_{2^n}}$$

Sine a regular polygon with an infinite amount of sides is equivalent to a circle, you have $$\lim_{n\rightarrow\infty}A_n=\pi$$. In essence, the complete product is nothing more than comparing the size of a circle with respect to its inscribed square. Hence,

$$\prod_{n=2}^\infty\cos\left(\frac{\pi}{2^n}\right)=\lim_{n\rightarrow\infty}C_n=\frac 2 \pi$$

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