20091231, 21:26  #1 
May 2005
Argentina
10111010_{2} Posts 
Square root of 3
The square root of 2 can be written as an infinite product in the form:
I wonder if there is an analogous infinite product for , does anyone know it, if there it is one? Thanks, Damián 
20091231, 22:00  #2 
"Richard B. Woods"
Aug 2002
Wisconsin USA
2^{2}×3×641 Posts 
If you come up with some sinecosine 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, http://upload.wikimedia.org/math/a/6...daed90ca43.png or, more simply, cos(pi/6) = sin(pi/3) = √3/2   http://en.wikipedia.org/wiki/Exact_t...tric_constants and http://mathworld.wolfram.com/TrigonometryAngles.html have others. Last fiddled with by cheesehead on 20091231 at 22:15 
20091231, 23:30  #3  
May 2005
Argentina
2·3·31 Posts 
Quote:
Following your advice, and using the productory for the sine function, I could derive the following identities: I'll see if I can find a way to generalize those to a productory for for any integer Any hint on that? Thanks, Damián. 

20100101, 01:56  #4 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
3^{2}×17×61 Posts 
You will probably get to the same n's as in triangulation

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