
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!!

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, $\displaystyle \displaystyle \prod_{k=1}^{n1}\sin \left(\frac{k \pi}{n}\right) =n2^{1n}$ from formula 23 on this page Sine identities

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 $\displaystyle n1$ 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
$\displaystyle
\displaystyle \sin(\pi z_k) = \frac{\pi}{\Gamma(z_k)\Gamma(1z_k)}$
Now lets perform the following calculation
$\displaystyle \displaystyle \prod_{k=1}^{n1} \sin(\pi z_k) = \prod_{k=1}^{n1}\frac{\pi}{\Gamma(z_k)\Gamma(1z_k)} = \pi^{n1} \prod_{k=1}^{n1}\frac{1}{\Gamma(z_k)\Gamma(1z_k)} = \pi^{n1} \prod_{k=1}^{n1}\frac{1}{\Gamma\left(z+\frac{k}{n}\right)\Gamma\ left(1z\frac{k}{n}\right)}$
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.
$\displaystyle \displaystyle \prod_{k=1}^{n1}\frac{1}{\Gamma\left(1z\frac{k}{n}\right)} = \prod_{k=1}^{n1}\frac{1}{\Gamma\left(1z\frac{nk}{n}\right)} = \prod_{k=1}^{n1}\frac{1}{\Gamma\left(z+\frac{k}{n}\right)}$
Where the last two follow from flipping the sum around. Now we use the identity from here to get
[Math]\displaystyle \pi^{n1} \prod_{k=1}^{n1}\frac{1}{\Gamma\left(z+\frac{k}{n}\right)\Gamma\ left(1z\frac{k}{n}\right)} = \pi^{n1}\prod_{k=1}^{n1}\frac{1}{\Gamma\left(z+\frac{k}{n}\right)}\prod_ {k=1}^{n1}\frac{1}{\Gamma\left(z+\frac{k}{n}\right)} = [/tex]
$\displaystyle \displaystyle = \pi^{n1}\left(\frac{ (2\pi)^{(n1)/2}n^{nz+1/2}\Gamma(nz) }{\Gamma(z)} \frac{ (2\pi)^{(n1)/2}n^{nz+1/2}\Gamma(nz) }{\Gamma(z)}\right)^{1} [/tex]
$\displaystyle \displaystyle = \frac{\Gamma(z)\Gamma(z)}{n2^{n1}\Gamma(nz)\Gamma(nz)} = \frac{1}{n2^{n1}}\frac{\pi}{z \sin(\pi z)}\frac{nz \sin(\pi n z)}{\pi} = \frac{\sin(\pi n z)}{2^{n1}\sin(\pi z)}$
Thus we have an identity:
[tex]\displaystyle \prod_{k=1}^{n1} \sin(\pi (z + k/n)) = \frac{\sin(\pi n z)}{2^{n1}\sin(\pi z)}$
The limit on the right hand side is $\displaystyle n2^{1n}$ with L'Hopital's rule. Thus we get
$\displaystyle \displaystyle \prod_{k=1}^{n1} \sin(\pi ( k/n)) = n2^{1n}$
We are almost there. Consider n = 2m +1, that is it's odd. Then
$\displaystyle \displaystyle \prod_{k=1}^{n1} \sin(\pi ( k/n)) = \prod_{k=1}^{2m} \sin\left(\pi\frac{k}{2m+1} \right) = \prod_{k=1}^{m} \sin\left(\pi\frac{k}{2m+1} \right) \sin\left(\pi\frac{2m+1k}{2m+1} \right) = \prod_{k=1}^{m} \sin^2\left(\pi\frac{k}{2m+1} \right)$
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!

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
$\displaystyle \displaystyle \frac{x^{27}1}{x1}=(xw)(xw^2)\cdots (xw^{26})$.
Evaluate $\displaystyle x=1$ by taking the limit on the LHS, we get
$\displaystyle \displaystyle 27=(1w)(1w^2)\cdots (1w^{26})$.
Now express the RHS using trigonometric functions and simplify by doubleangle formula
$\displaystyle
\displaystyle 1e^{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 (1w^{k}\right )=\prod_{k=1}^{26} \left (1e^{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:
$\displaystyle \displaystyle \prod_{k=1}^{13} \sin \frac{k\pi}{27}=\frac{3\sqrt{3}}{2^{13}}.$

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.