Product of sine

**TerenceCS** · August 10th 2010, 11:48 PM

Hi! I have stuck in this problem for almost an hour, it is from a textbook:
Prove that
$\displaystyle \sin \frac{\pi}{27}\sin \frac{2\pi}{27}\sin \frac{3\pi}{27} \cdots \sin \frac{13\pi}{27}=\frac{3\sqrt{3}}{2^{13}}$
Any help will be appreciated!!

**Vlasev** · August 11th 2010, 01:45 AM

Right now I am trying to figure it out as well.

What you have stated is
$\displaystyle \prod_{k=1}^{13} \sin\left(\frac{k\pi}{27}\right) = \frac{3\sqrt{3}}{2^{13}}$

What is interesting is that also

$\displaystyle \prod_{k=1}^{26} \sin\left(\frac{k\pi}{27}\right) = \left(\frac{3\sqrt{3}}{2^{13}}\right)^2$

EDIT:

Aparently, [LaTeX ERROR: Convert failed] from formula 23 on this page Sine identities

**Vlasev** · August 11th 2010, 05:36 AM

We need to consider a general term $\sin\left(\pi\left(\frac{k}{n}+z\right)\right) = \sin(\pi z_k)$ where $n$ is a positive integer, $k$ is a positive integer between $1$ and $n-1$ inclusive and $z$ is a positive real number. We will derive an identity that is similar to yours but features $z$ where we will take the limit as $z$ approaches $0$ to get your answer.

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

$ \displaystyle \sin(\pi z_k) = \frac{\pi}{\Gamma(z_k)\Gamma(1-z_k)}$

Now lets perform the following calculation

$\displaystyle \prod_{k=1}^{n-1} \sin(\pi z_k) = \prod_{k=1}^{n-1}\frac{\pi}{\Gamma(z_k)\Gamma(1-z_k)} = \pi^{n-1} \prod_{k=1}^{n-1}\frac{1}{\Gamma(z_k)\Gamma(1-z_k)} = \pi^{n-1} \prod_{k=1}^{n-1}\frac{1}{\Gamma\left(z+\frac{k}{n}\right)\Gamma\ left(1-z-\frac{k}{n}\right)}$

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

$\displaystyle \prod_{k=1}^{n-1} \sin(\pi ( k/n)) = n2^{1-n}$

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

$\displaystyle \prod_{k=1}^{n-1} \sin(\pi ( k/n)) = \prod_{k=1}^{2m} \sin\left(\pi\frac{k}{2m+1} \right) = \prod_{k=1}^{m} \sin\left(\pi\frac{k}{2m+1} \right) \sin\left(\pi\frac{2m+1-k}{2m+1} \right) = \prod_{k=1}^{m} \sin^2\left(\pi\frac{k}{2m+1} \right)$

Hence by taking the square root of both sides we get

$\displaystyle \prod_{k=1}^{m} \sin\left(\pi\frac{k}{2m+1} \right)= \frac{\sqrt{2m+1}}{2^{m}}$

Now when we plug in m = 13, we get

$\displaystyle \prod_{k=1}^{13} \sin\left(\pi\frac{k}{27}\right) = \frac{\sqrt{27}}{2^{13}} = \frac{3\sqrt{3}}{2^{13}}$

And this completes the proof!

**TerenceCS** · August 11th 2010, 08:14 PM

Thanks mattpi I just found another proof inspired from the cyclotomic equation.

Consider $x^{27}-1=0$ and let $w=e^{2\pi i/27}$ be one of its roots, then
$\displaystyle \frac{x^{27}-1}{x-1}=(x-w)(x-w^2)\cdots (x-w^{26})$ .

Evaluate $x=1$ by taking the limit on the LHS, we get
$\displaystyle 27=(1-w)(1-w^2)\cdots (1-w^{26})$ .

Now express the RHS using trigonometric functions and simplify by double-angle formula
$ \displaystyle 1-e^{2k\pi i/{27}} = 1-\cos \frac{2k\pi}{27}-i\sin \frac{2k\pi}{27}=2\sin \frac{k\pi}{27}\left (\sin \frac{k\pi}{27}-i\cos\frac{k\pi}{27}\right )= -2i\sin \frac{k\pi}{27}e^{\frac{k\pi i}{27}}. $

Combine all these together we get
$ \displaystyle \prod_{k=1}^{26} \left (1-w^{k}\right )=\prod_{k=1}^{26} \left (1-e^{2k\pi i/{27}}\right )=\prod_{k=1}^{26} -2i\sin \frac{k\pi}{27}e^{\frac{k\pi i}{27}} = 2^{26} \prod_{k=1}^{26} \sin \frac{k\pi}{27} $

Rearrange and notice the symmetric property of sine function, we get
$\displaystyle \prod_{k=1}^{26} \sin \frac{k\pi}{27}=\frac{27}{2^{26}}$

So the proof is archieved by taking square root on both sides:
$\displaystyle \prod_{k=1}^{13} \sin \frac{k\pi}{27}=\frac{3\sqrt{3}}{2^{13}}.$

**Vlasev** · August 11th 2010, 10:27 PM

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.

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