Simplify $\prod_{n=1}^{17}\sin n\frac \pi {18}$

Question

Evaluate $$\prod_{n=1}^{17}\sin n\frac \pi {18}.$$

Source: Based on AOPS #10.

My work so far: Let $a = \frac \pi {18}$. The factor at $n = 9$ is $1$, and the other factors pair, giving us $$\prod_{n=1}^{8}\sin^2 na.$$

There are many relevant identities, but none seem to simplify things:

Each factor can be replaced with $(1 - \cos^2 na)$, but this doesn't seem to simplify things
The powers of $2$ can be replaced with double angle formulas, but this doesn't seem to help matters either
And $\sin na = \sin a \cos(n-1)a + \sin(n-1)a \cos a$, but again, what have I gained?

What strategy (other than brute force) can be used to simplify this?

Note: This problem is solvable via precalculus. Solutions using complex analysis aren't in scope.

This is a "well-known" fact. Did you try searching around? If so, what did you not understand from the solutions? Did you try plugging it into Wolfram to see what the final value is? Also, it might be helpful to provide some context for what you're comfortable with. — Calvin Lin, Jun 20 '23 at 01:19
@CalvinLin The link I provided gives an answer, but no method. Google wasn't helpful either, perhaps because of notation. Regardless, my question isn't "What's the answer?" but "Help me find the right strategy to use here". — SRobertJames, Jun 20 '23 at 01:21
@SRobertJames FWIW Searching Google for "prod sin npi/k" gave Gary's linked answer as the first result. I agree Google isn't the best search for things like this, give Approach Zero a try next time. — Calvin Lin, Jun 20 '23 at 01:26
Side note: The reason I asked for context for what you're comfortable with, is that the complex number proof might be scary especially if you're tagging the question with algebra-precalculus. There is an alternative proof via Chebyshev polynomials that could be easier (less frightening) to understand. The idea is to find a polynomial whose roots are those values, and hence the product of roots is the constant term / leading coefficient. — Calvin Lin, Jun 20 '23 at 01:34
@Gary This question is from a precalculus test (see the link) and hence tagged precalculus, so (at least the test author implies) it can be solved without complex analysis! — SRobertJames, Jun 20 '23 at 01:42
@SRobertJames How about this approach? https://math.stackexchange.com/q/1680207 — Gary, Jun 20 '23 at 02:07
I wonder what you mean by "complex analysis". The source from which your question comes has another question about complex numbers as well. — Juliamisto, Jun 20 '23 at 02:14
The AoPS Precalculus book covers complex numbers. We're not talking about analytic functions or contour integrals here. — Brian Moehring, Jun 20 '23 at 02:16

score -1 · Answer 1 · edited Jun 20 '23 at 03:03

Recall that $$\cos a + b = \cos a \cos b - \sin a \sin b \\ \sin a \sin b = \cos (a + b) \cdot \cos a \cos b$$

The $a$ and $b$ pair into $\pi$, such that $\cos(a+b)=-1$. That leaves the $\cos a \cos b$ factors.

This is not a complete solution, but perhaps a start.

Calvin's edit follows

Here is a precalculus method that likely doesn't generalize to different denominators.

Let's focus on $A = \prod_{n=1}^8 \sin n \frac{\pi}{18}$, and the desired answer is $A^2$.
Let $B = \sin \frac{2\pi}{18} \sin \frac{4\pi}{18} \sin \frac{6\pi}{18} \sin \frac{8\pi}{18} = \cos \frac{\pi}{18} \cos \frac{3\pi}{18} \cos \frac{5\pi}{18} \cos \frac{7\pi}{18} $
Let $ C = \sin \frac{\pi}{18} \sin \frac{3\pi}{18} \sin \frac{5\pi}{18} \sin \frac{7\pi}{18} = \cos \frac{2\pi}{18} \cos \frac{4\pi}{18} \cos \frac{6\pi}{18} \cos \frac{8\pi}{18} $
Show that $A = BC$.
Show that $2^4 BC = B$, so $C = \frac{1}{2^4}$.
- On the LHS, Use the cosine formulation of $B$ and the sine formulation of $C$.
- Use $ \sin 2 \theta = 2 \sin \theta \cos \theta$ repeatedly.
Show that $B = \frac{3}{16}$.
- This is "well-known". Look it up online for numerous proofs if you are stuck.
Thus $ A = \frac{3}{2^8}$ and so $A^2 = \frac{ 9}{2^{16} }$.

Are you sure about the second equation? Maybe it's a minus sign instead of a product? — Calvin Lin, Jun 20 '23 at 02:17
Note: I don't usually do this, but ecause this question is now closed, I made an edit to your solution to provide a pre-calculus approach. You may undo the edit if you wish. $\quad$ All that it uses is the sum and product identities in a clever manner. This doesn't generalize to $ \prod \sin n pi / k$. — Calvin Lin, Jun 20 '23 at 02:56

Simplify $\prod_{n=1}^{17}\sin n\frac \pi {18}$

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