# Thread: Product of sine

1. ## Product of sine

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

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

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

What is interesting is that also

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

EDIT:

Aparently, $\displaystyle \displaystyle \prod_{k=1}^{n-1}\sin \left(\frac{k \pi}{n}\right) =n2^{1-n}$ from formula 23 on this page Sine identities

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

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

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

Now lets perform the following calculation

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

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.

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

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]

$\displaystyle \displaystyle = \pi^{n-1}\left(\frac{ (2\pi)^{(n-1)/2}n^{-nz+1/2}\Gamma(nz) }{\Gamma(z)} \frac{ (2\pi)^{(n-1)/2}n^{nz+1/2}\Gamma(-nz) }{\Gamma(-z)}\right)^{-1} [/tex]$\displaystyle \displaystyle = \frac{\Gamma(z)\Gamma(-z)}{n2^{n-1}\Gamma(nz)\Gamma(-nz)} = \frac{1}{n2^{n-1}}\frac{-\pi}{z \sin(\pi z)}\frac{nz \sin(\pi n z)}{-\pi} = \frac{\sin(\pi n z)}{2^{n-1}\sin(\pi z)}$Thus we have an identity: [tex]\displaystyle \prod_{k=1}^{n-1} \sin(\pi (z + k/n)) = \frac{\sin(\pi n z)}{2^{n-1}\sin(\pi z)}$

The limit on the right hand side is $\displaystyle n2^{1-n}$ with L'Hopital's rule. Thus we get

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

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

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

Hence by taking the square root of both sides we get

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

Now when we plug in m = 13, we get

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

And this completes the proof!

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

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

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

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

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

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

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

5. 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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