<s>[INST] Showing $\sin{\frac{\pi}{13}} \cdot \sin{\frac{2\pi}{13}} \cdot \sin{\frac{3\pi}{13}} \cdots \sin{\frac{6\pi}{13}} = \frac{\sqrt{13}}{64}$ I would like to show that 
$$ \sin{\frac{\pi}{13}} \cdot \sin{\frac{2\pi}{13}} \cdot \sin{\frac{3\pi}{13}} \cdots \sin{\frac{6\pi}{13}} = \frac{\sqrt{13}}{64} $$
I've been working on this for a few days. I've used product-to-sum formulas, writing the sines in their exponential form, etc. When I used the product-to-sum formulas, I'd get a factor of $1/64$, I obtained the same with writing the sines in their exponential form. I'd always get $1/64$ somehow, but never the $\sqrt{13}$.
I've come across this: http://mathworld.wolfram.com/TrigonometryAnglesPi13.html, (look at the 10th equation). It says that this comes from one of Newton's formulas and links to something named "Newton-Girard formulas", which I cannot understand. :(
Thanks in advance.
 [/INST] Use this formula (found here, and mentioned recently on MSE here):
$$\prod _{k=1}^{n-1}\,\sin \left({\frac {k\pi }{n}} \right)=\frac{n}{2^{n-1}} .$$
Let $n=13$, which gives $$\left(\sin{\frac{\pi}{13}} \cdot \sin{\frac{2\pi}{13}} \cdot \sin{\frac{3\pi}{13}} \cdots \sin{\frac{6\pi}{13}}\right)\left(\sin{\frac{7\pi}{13}} \cdot \sin{\frac{8\pi}{13}} \cdot \sin{\frac{9\pi}{13}} \cdots \sin{\frac{12\pi}{13}}\right) = \frac{13}{{2^{12}}}.$$
Use the fact that $\sin\dfrac{k\pi}{13}=\sin\dfrac{(13-k)\pi}{13}$ to see that this is the same as 
$$\left(\sin{\frac{\pi}{13}} \cdot \sin{\frac{2\pi}{13}} \cdot \sin{\frac{3\pi}{13}} \cdots \sin{\frac{6\pi}{13}}\right)^2 = \frac{13}{2^{12}}$$
and take the square root of both sides to get your answer.
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