# Evaluating a particular product of sines

• Feb 28th 2011, 05:15 AM
wglmb
Evaluating a particular product of sines
I need to evaluate the product $\displaystyle \prod_{m=n-a}^{n-1} \sin \left( \pi m /n \right)$.
I've found this identity: $\displaystyle \prod_{m=1}^{n-1} \sin \left( \pi m /n \right) \equiv \frac{n}{2^{n-1}}$, which looks useful but isn't quite what I need.

Does anyone know of a generalisation of this identity so that the lower range is $\displaystyle n-a$?
• Mar 2nd 2011, 01:01 PM
ragnar
$\displaystyle \displaystyle \prod^{n-1}_{n-a}(\sin(\pi m/n)) = \prod^{n-1}_{1}(\sin(\pi m/n)) \div \prod^{n-a}_{1}(\sin(\pi m/n))$ or something like that. I'm not thinking very carefully about the indices and have to run, but this is the idea. Correct it where appropriate.
• Mar 3rd 2011, 03:48 AM
wglmb
Thanks, I'd already thought of that (in fact the second product only runs to $\displaystyle n-a+1$.
But I'd like to express the product as a function not involving trig functions.
• Mar 3rd 2011, 09:14 AM
ragnar
I think you mean $\displaystyle n - a - 1$, but as for the rest of your question, I'm not sure I understand. Call $\displaystyle m-1 = n - a - 1$ then $\displaystyle \displaystyle \prod^{n-1}_{n-a}(\sin(\pi m/n)) = \prod^{n-1}_{1}(\sin(\pi m/n)) \div \prod^{m-1}_{1}(\sin(\pi m/n)) = \frac{n}{2^{n-1}} \div \frac{m}{2^{m-1}} = \frac{n}{2^{n-1}} \cdot \frac{2^{n- a - 1}}{n - a} = \frac{n}{2^{a}(n-a)}$.

If that's not what you're looking for then I don't know what is!
• Mar 3rd 2011, 09:58 AM
wglmb
Yes, sorry I meant -1 of course.

Quote:

Originally Posted by ragnar
Call $\displaystyle m-1 = n - a - 1$ then $\displaystyle \displaystyle \prod^{n-1}_{n-a}(\sin(\pi m/n)) = \prod^{n-1}_{1}(\sin(\pi m/n)) \div \prod^{m-1}_{1}(\sin(\pi m/n)) = \frac{n}{2^{n-1}} \div \frac{m}{2^{m-1}}$.

This is confusing - it looks to me like you're using $\displaystyle m$ as both the product-variable and as a constant used to define the product range.
What I think you're trying to do is this:

Let $\displaystyle b=n-a$ so that $\displaystyle \displaystyle \prod_{m=1}^{n-a-1} \sin \left( \frac{m\pi}{n} \right) = \prod_{m=1}^{b-1} \sin \left( \frac{m\pi}{n} \right)$
If that's what you mean, then your next part is wrong, since the identity in my first post doesn't apply.

Or did I misunderstand?
• Mar 3rd 2011, 12:52 PM
Opalg
Quote:

Originally Posted by wglmb
I need to evaluate the product $\displaystyle \prod_{m=n-a}^{n-1} \sin \left( \pi m /n \right)$.

I think you're out of luck here. What I mean by that is that I don't think there is a formula of the type that you are looking for.

Think about the particular case of the product that arises when $\displaystyle a=1$. There is then only one term in the product, and it is equal to $\displaystyle \sin(\pi/n)$. For most values of n, there is no explicit formula for $\displaystyle \sin(\pi/n)$ in terms of more elementary functions of n. So even in that special case, there is no satisfactory solution to the problem.

Quote:

Originally Posted by wglmb
I've found this identity: $\displaystyle \prod_{m=1}^{n-1} \sin \left( \pi m /n \right) \equiv \frac{n}{2^{n-1}}$, which looks useful but isn't quite what I need.

The fact that there is a good solution in this case is due to the fact that the numbers $\displaystyle \sin \left( \pi m /n \right)\ (1\leqslant m\leqslant n-1)$ are the roots of a fairly easily computable polynomial of degree n–1, and the product of the roots is given by the constant term in the polynomial. But that doesn't apply if you only have a subset of the roots.
• Mar 3rd 2011, 12:58 PM
wglmb
I see, thank you - that was very enlightening! :)