Product of sines when the angles form a sequence I am wondering how to find the following value of $x$
$$x=\prod_{n=1}^{9}\sin\left(\frac{n\pi}{10}\right)$$
I notice that it has something to do with the de moivre's theorem as the angles are root angles of $1^\frac{1}{10}$
To my surprise, the value of the above product seems to be a rational number.
The solution is given as
$$x=\frac{10}{2^9}$$
I attempted to solve this product in different ways but none seems to give out such an elegant solution. Thank you for any help!

$x = (\prod\limits_{n = 1}^4 \sin(\frac{n \pi}{10}))^2$
$\prod\limits_{n = 1}^4 \sin(\frac{n \pi}{10}) = \prod\limits_{n = 1}^2 \sin(\frac{n \pi}{10}) \prod\limits_{n = 1}^2 \cos(\frac{n \pi}{10}) = \prod\limits_{n = 1}^2 \sin(\frac{n \pi}{10}) \cos(\frac{n \pi}{10}) = \prod\limits_{n = 1}^2 \frac{\sin(\frac{n \pi}{5})}{2} = \frac{1}{4} \sin(\frac{\pi}{5}) \sin(\frac{2 \pi}{5})$
$x = \frac{1}{16} \sin^2(\pi/5) \sin^2(2\pi/5) = \frac{1}{64} (1 - \cos(2\pi / 5))(1 - \cos(4\pi / 5))$
Now consider the fact that $\sum\limits_{j = 0}^4 \cos(2j \pi / 5) = 0$. In particular, we have $2 \cos(2 \pi / 5) + 2 \cos(4 \pi / 5) = -1$. Then, letting $w = \cos(2\pi / 5)$, we have $2 w + 4 w^2 - 2 = -1$. Then $2w + 4w^2 = 1$.
We have $x = \frac{1}{64} (1 - w)(2 - 2w^2)$. Now we have $4(1 - w)(2 - 2 w^2) = 2(1 - w)(4 - 4w^2) = 2(1 - w)(3 + 2w) = 2 (3 - 2w^2 - w) = 6 - 4w^2 - 2w = 6 - 1 = 5$. Then $x = \frac{1}{64} \frac{5}{4} = \frac{5}{2^8}$.
