# Is there an infinite product like this for $\cos x$?

There are infinite products of iterated square roots for $\log x$ and $\arccos x$ as functions of $x$. For example

$$\log x = \frac{x - 1}{\sqrt{x}\sqrt{\frac{1}{2} + \frac{1}{2}\left ( \frac{1 + x}{2\sqrt{2}} \right )}\sqrt{\frac{1}{2} + \frac{1}{2}\sqrt{\frac{1}{2} + \frac{1}{2}\left ( \frac{1 + x}{2\sqrt{2}} \right )\cdots}}} \tag{1}$$

$$\frac{\sqrt{1-x^2}}{\arccos x} = \frac{\sqrt{2+2x}}{2}\frac{\sqrt{2+\sqrt{2+2x}}}{2}\frac{\sqrt{2+\sqrt{2+\sqrt{2+2x}}}}{2}\cdots\tag{2}$$

Keeping this in mind, I wonder if there is an infinite product of square roots, similar to those products above, for $\cos x$, as a function of $x$.

Are the identities $(1)$ and $(2)$ helpful in obtaining a similar product for $\cos x$? If not, is there any infinite product with square roots, which can be derived in some way, or is there some demonstration that such a product can not exist?

References:

An infinite product of nested radicals for from the Archimedean algorithm log x

Thomas J. Osler, Walter Jacob and Ryo Nishimura.

• Cosine is an entire function and it will be very surprising if it was possible to represent it in terms of infinite product of nested radicals.
– Nemo
Nov 1, 2017 at 16:57
• The standard infinite product for the cosine is, IIRC, $\cos(\pi x)=\prod_{n=0}^\infty 1-\frac{(2x)^2}{(2n+1)^2}$ Nov 1, 2017 at 17:03
• A product formula is better if it comes with a domain of convergence.
– YCor
Nov 1, 2017 at 17:19