Square root of 3
 

The [URL="http://en.wikipedia.org/wiki/Square_root_of_2"]square root of 2 [/URL] can be written as an infinite product in the form:

[TEX]\sqrt{2} = \left(1+\frac{1}{1} \right)\left(1-\frac{1}{3} \right)\left(1+\frac{1}{5} \right)\left(1-\frac{1}{7} \right) \ldots[/TEX]

I wonder if there is an analogous infinite product for [TEX]\sqrt{3}[/TEX], does anyone know it, if there it is one?

Thanks,
Damián

If you come up with some sine-cosine identity involving √3 (analogous to the cos(pi/4) = sin(pi/4) = 1/√2 identity shown in the Wikipedia article for √2), you've got it made.

For instance,
[IMG]http://upload.wikimedia.org/math/a/6/8/a687595cf642d0dcb288e6daed90ca43.png[/IMG]

or, more simply,

cos(pi/6) = sin(pi/3) = √3/2

- -

[URL]http://en.wikipedia.org/wiki/Exact_trigonometric_constants[/URL] and

[url]http://mathworld.wolfram.com/TrigonometryAngles.html[/url]

have others.

[QUOTE=cheesehead;200484]If you come up with some sine-cosine identity involving √3 (analogous to the cos(pi/4) = sin(pi/4) = 1/√2 identity shown in the Wikipedia article for √2), you've got it made.

For instance,
[IMG]http://upload.wikimedia.org/math/a/6/8/a687595cf642d0dcb288e6daed90ca43.png[/IMG]

or, more simply,

cos(pi/6) = sin(pi/3) = √3/2

- -

[URL]http://en.wikipedia.org/wiki/Exact_trigonometric_constants[/URL] and

[url]http://mathworld.wolfram.com/TrigonometryAngles.html[/url]

have others.[/QUOTE]

Thank you!
Following your advice, and using the productory for the sine function, I could derive the following identities:

[TEX]\sqrt{2} = \frac{\pi}{2} \Pi_{n=1}^{\infty} \left( 1 - \frac{1}{16n^2} \right) [/TEX]

[TEX]\sqrt{3} = \frac{2\pi}{3} \Pi_{n=1}^{\infty} \left( 1 - \frac{1}{9n^2} \right) [/TEX]

[TEX] \displaystyle \sqrt{5} = 1 + \frac{2\pi}{5} \Pi_{n=1}^{\infty} \left( 1 - \frac{1}{100 n^2} \right) [/TEX]

I'll see if I can find a way to generalize those to a productory for [TEX]\sqrt{n}[/TEX] for any integer [TEX]n[/TEX]

Any hint on that?
Thanks,
Damián.

You will probably get to the same n's as in [URL="http://mersenneforum.org/showthread.php?t=12613"]triangulation[/URL]