Showing $\sin{\frac{\pi}{13}} \cdot \sin{\frac{2\pi}{13}} \cdot \sin{\frac{3\pi}{13}} \cdots \sin{\frac{6\pi}{13}} = \frac{\sqrt{13}}{64}$

Question

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.

Answer 1

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.

Answer 2

For positive integer $n$

If $\sin(2n+1)x=0, (2n+1)x=m\pi\iff x=\frac{m\pi}{2n+1} $ where $m$ is any integer

From $(3)$ of this, $\displaystyle \sin(2n+1)x=2^{2n}s^{2n+1}+\cdots+(2n+1)s=0$ where $s=\sin\frac{m\pi}{2n+1}$

So the roots of $\displaystyle 2^{2n}s^{2n+1}+\cdots+(2n+1)s=0 $ are $\sin\frac{m\pi}{2n+1}; 0\le m\le2n$

So the roots of $\displaystyle 2^{2n}s^{2n}+\cdots+(2n+1)=0 $ are $\sin\frac{m\pi}{2n+1}; 1\le m\le2n$

Using Vieta's formula $\displaystyle\prod_{m=1}^{2n}\sin\frac{m\pi}{2n+1}=\frac{2n+1}{2^{2n}}$

Now using $\displaystyle\sin(\pi-y)=\sin y,\sin\left(\pi-\frac m{2n+1}\pi\right)=\sin\frac{(2n+1-m)\pi}{2n+1}$

$\displaystyle\prod_{m=1}^{2n}\sin\frac{m\pi}{2n+1}=\prod_{m=1}^n\sin^2\frac{m\pi}{2n+1}$

Now for $\displaystyle, 1\le m\le n;0<\frac{m\pi}{2n+1}<\frac\pi2\implies\sin\frac{m\pi}{2n+1}>0$