1

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!

  • 1
    Have you done any Field Theory, MathEnthu? It's the product of conjugates of algebraic numbers, so there's no surprise that it's rational. – Gerry Myerson Aug 05 '20 at 08:53
  • no, I just completed high school. I suspected the product to be most likely irrational since individual sine values are irrational. –  Aug 05 '20 at 09:33
  • @mathcounterexamples.net I think i didn't effectively search, sorry for asking again –  Aug 05 '20 at 09:38

1 Answers1

1

$x = (\prod\limits_{n = 1}^4 \sin(\frac{n \pi}{10}))^2$

$\prod\limits_{n = 1}^4 \sin(\frac{n \pi}{10}) = \prod\limits_{n = 1}^2 \sin(\frac{n \pi}{10}) \prod\limits_{n = 1}^2 \cos(\frac{n \pi}{10}) = \prod\limits_{n = 1}^2 \sin(\frac{n \pi}{10}) \cos(\frac{n \pi}{10}) = \prod\limits_{n = 1}^2 \frac{\sin(\frac{n \pi}{5})}{2} = \frac{1}{4} \sin(\frac{\pi}{5}) \sin(\frac{2 \pi}{5})$

$x = \frac{1}{16} \sin^2(\pi/5) \sin^2(2\pi/5) = \frac{1}{64} (1 - \cos(2\pi / 5))(1 - \cos(4\pi / 5))$

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 $x = \frac{1}{64} \frac{5}{4} = \frac{5}{2^8}$.

Doctor Who
  • 3,471
  • Thanks, I had this problem when I was solving another question, I got stuck with this, so whatever I tried looped to the same one. I didn't expect that a relatively simple trignometric solution would exist. –  Aug 05 '20 at 09:54
  • Ah! The solution is not purely trig, the fourth line has used the sum of n roots of unity result. I think that result is hard to prove without de movier's theorem –  Aug 05 '20 at 09:59