Is there convenient notation for Viète's formula? Is there a convenient way to write Viète's formula $\displaystyle \frac2\pi=
\frac{\sqrt2}2\cdot
\frac{\sqrt{2+\sqrt2}}2\cdot
\frac{\sqrt{2+\sqrt{2+\sqrt2}}}2\cdots$
using sigma and/or pi notation
a) without recursive notation;
b) even with recursive notation?
As many ways as possible would be appreciated. Thanks in advance!
 A: The RHS can be rewritten as $$\prod_{k=2}^{\infty}\cos \left(\frac{\pi}{2^k}\right)$$
Maybe there's no recurrence relation since $\pi$ is a transcendental number.
A: To reach the final result, you can follow these detailed steps:

*

*Write the double-angle formula
$$
    \sin x = 2 \sin\frac x2 \cos\frac x2
$$
and apply it repeatedly to finally get
$$
    \sin x = 2^n \sin\frac{x}{2^n} \prod_{i = 1}^{n} \cos\frac{x}{2^i} \quad (n \in \mathbb{Z}^*). \tag{$*$}
$$


*Divide both sides of $(*)$ by $x$ (given $x > 0$), then take the limit of both sides of the equation when $n \to \infty$. Note that the change of value of $n$ has no effects on LHS, so we get
$$
    \frac{\sin x}{x} 
  = \lim_{n \to \infty} \Big( 
        \frac{\sin(x / 2^n)}{x / 2^n} \prod_{i = 1}^{n} \cos\frac{x}{2^i}
    \Big) 
  = \prod_{i = 1}^{\infty} \cos\frac{x}{2^i}.
$$
This is the result Tien Kha Pham provided directly. (Note that $\prod_{i = 1}^{\infty} \cos(x / 2^i)$ makes sense only if the corresponding limit $\lim_{n \to \infty} \prod_{i = 1}^{n} \cos(x / 2^i)$ exists.)


*To relate this result to the form of Viète's formula you mentioned in the question, substituting $x = \pi / 2$ with formulas
$$
    \cos \Big(\frac{\pi}{4}\Big) = \frac{\sqrt{2}}{2}
    \quad\text{and}\quad
    \cos \Big(\frac{x}{2^k}\Big) = \frac 12 \sqrt{2 + 2\cos \Big(\frac{x}{2^{k - 1}}\Big)} \quad (k \ge 2),
$$
we finally conclude with
$$
    \frac{\pi}{2} = \prod_{k = 2}^{\infty} a_k,
$$
where $a_2 = \sqrt{2} / 2$, $a_3 = \sqrt{2 + 2a_2} / 2 = \sqrt{2 + \sqrt{2}} / 2$ and $a_{k + 1} = \sqrt{2 + 2a_k} / 2$ for $k \ge 3$.

Once there was a minor mistake in the answer made by Tien Kha Pham. The typo no longer existed. But another consequent mistake, in the comment to that answer, will remain. Therefore I keep the following words visible only as a reminder for those who may get confused by the inconsistency between formulas.

Just a correction to the answer of Tien Kha Pham (since I don't have
enough reputation to add a comment): it should be $$ \frac{2}{\pi} =
 \prod_{k = 2}^{\infty} \cos \Big( \frac{\pi}{2^k} \Big), $$ of which
the $\pi$ appears in numerator, not denominator.

