# Thread:

1. ## Evaluating a particular product of sines

I need to evaluate the product $\prod_{m=n-a}^{n-1} \sin \left( \pi m /n \right)$.
I've found this identity: $\prod_{m=1}^{n-1} \sin \left( \pi m /n \right) \equiv \frac{n}{2^{n-1}}$, 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 $n-a$?

2. $\displaystyle \prod^{n-1}_{n-a}(\sin(\pi m/n)) = \prod^{n-1}_{1}(\sin(\pi m/n)) \div \prod^{n-a}_{1}(\sin(\pi m/n))$ 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.

3. Thanks, I'd already thought of that (in fact the second product only runs to $n-a+1$.
But I'd like to express the product as a function not involving trig functions.

4. I think you mean $n - a - 1$, but as for the rest of your question, I'm not sure I understand. Call $m-1 = n - a - 1$ then $\displaystyle \prod^{n-1}_{n-a}(\sin(\pi m/n)) = \prod^{n-1}_{1}(\sin(\pi m/n)) \div \prod^{m-1}_{1}(\sin(\pi m/n)) = \frac{n}{2^{n-1}} \div \frac{m}{2^{m-1}} = \frac{n}{2^{n-1}} \cdot \frac{2^{n- a - 1}}{n - a} = \frac{n}{2^{a}(n-a)}$.

If that's not what you're looking for then I don't know what is!

5. Yes, sorry I meant -1 of course.

Originally Posted by ragnar
Call $m-1 = n - a - 1$ then $\displaystyle \prod^{n-1}_{n-a}(\sin(\pi m/n)) = \prod^{n-1}_{1}(\sin(\pi m/n)) \div \prod^{m-1}_{1}(\sin(\pi m/n)) = \frac{n}{2^{n-1}} \div \frac{m}{2^{m-1}}$.
This is confusing - it looks to me like you're using $m$ 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 $b=n-a$ so that $\displaystyle \prod_{m=1}^{n-a-1} \sin \left( \frac{m\pi}{n} \right) = \prod_{m=1}^{b-1} \sin \left( \frac{m\pi}{n} \right)$
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?

6. Originally Posted by wglmb
I need to evaluate the product $\prod_{m=n-a}^{n-1} \sin \left( \pi m /n \right)$.
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 $a=1$. There is then only one term in the product, and it is equal to $\sin(\pi/n)$. For most values of n, there is no explicit formula for $\sin(\pi/n)$ in terms of more elementary functions of n. So even in that special case, there is no satisfactory solution to the problem.

Originally Posted by wglmb
I've found this identity: $\prod_{m=1}^{n-1} \sin \left( \pi m /n \right) \equiv \frac{n}{2^{n-1}}$, 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 $\sin \left( \pi m /n \right)\ (1\leqslant m\leqslant n-1)$ 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.

7. I see, thank you - that was very enlightening!

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