Hi! I have stuck in this problem for almost an hour, it is from a textbook:

Prove that

Any help will be appreciated!!

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- August 11th 2010, 12:48 AMTerenceCSProduct of sine
Hi! I have stuck in this problem for almost an hour, it is from a textbook:

Prove that

Any help will be appreciated!! - August 11th 2010, 02:45 AMVlasev
Right now I am trying to figure it out as well.

What you have stated is

What is interesting is that also

EDIT:

Aparently, [LaTeX ERROR: Convert failed] from formula 23 on this page Sine identities - August 11th 2010, 06:36 AMVlasev
We need to consider a general term where is a positive integer, is a positive integer between and inclusive and is a positive real number. We will derive an identity that is similar to yours but features where we will take the limit as approaches to get your answer.

After plugging in in identity 41 Here and rearranging the sine and the gamma functions, we get

Now lets perform the following calculation

We want to eventually use the Gauss Multiplication formula. Now we break up the product in two pieces and we'll deal with the latter first.

[LaTeX ERROR: Convert failed]

Where the last two follow from flipping the sum around. Now we use the identity from here to get

[Math]\displaystyle \pi^{n-1} \prod_{k=1}^{n-1}\frac{1}{\Gamma\left(z+\frac{k}{n}\right)\Gamma\ left(1-z-\frac{k}{n}\right)} = \pi^{n-1}\prod_{k=1}^{n-1}\frac{1}{\Gamma\left(z+\frac{k}{n}\right)}\prod_ {k=1}^{n-1}\frac{1}{\Gamma\left(-z+\frac{k}{n}\right)} = [/tex]

[LaTeX ERROR: Convert failed]

Thus we have an identity:

[tex]\displaystyle \prod_{k=1}^{n-1} \sin(\pi (z + k/n)) = \frac{\sin(\pi n z)}{2^{n-1}\sin(\pi z)}[/Math]

The limit on the right hand side is [LaTeX ERROR: Convert failed] with L'Hopital's rule. Thus we get

We are almost there. Consider n = 2m +1, that is it's odd. Then

Hence by taking the square root of both sides we get

Now when we plug in m = 13, we get

And this completes the proof! - August 11th 2010, 09:14 PMTerenceCS
Thanks mattpi I just found another proof inspired from the cyclotomic equation.

Consider and let be one of its roots, then

.

Evaluate by taking the limit on the LHS, we get

.

Now express the RHS using trigonometric functions and simplify by double-angle formula

Combine all these together we get

Rearrange and notice the symmetric property of sine function, we get

So the proof is archieved by taking square root on both sides:

- August 11th 2010, 11:27 PMVlasev
That's a lot better than going through all that Gamma function pain! I was thinking of doing this problem with complex exponentials because of their nice multiplicative properties but I ended up getting stuck on the fact that I started with the sine functions and not with the cyclotomic polynomial.