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I am looking to evaluate

$$\prod_{k=1}^{n-1}\sin\frac{k \pi}{n}$$

without using complex numbers. I can show the result if $n$ is a power of $2$, but if $n$ is anything else I reach a point where I can no longer proceed.

My current method (for when $n$ is a power of $2$) is to note that all of the angle that sin is being evaluated on are evenly distributed between $0$ and $\pi$, but more importantly, one of the angles will always be $\frac{\pi}{2}$ and all angles which are greater than $\frac{\pi}{2}$ can be re-written as the cos of an angle that is less than $\frac{\pi}{2}$ (using the complementary angle). When this is done, we get

$$\prod_{k=1}^{n-1}\sin\frac{k \pi}{n} = \prod_{k = 1}^{\lfloor\frac{n - 1}{2}\rfloor}\frac{1}{2}\sin\frac{2k\pi}{n} = \frac{1}{2^{\lfloor\frac{n - 1}{2}\rfloor}}\prod_{k = 1}^{\lfloor\frac{n - 1}{2}\rfloor}\sin\frac{k\pi}{\frac{n}{2}}$$

This process can then be repeated recursively until we have just one $\sin$ term left in the product (which will be $\sin\frac{\pi}{2}$), then what is left is equivalent to

$$\frac{n}{2^{n - 1}}\;.$$

Can anyone suggest a way of generalizing this to when $n$ is not a power of $2$?

Thanks in advance.

share|cite|improve this question
Someone voted to close this as an exact duplicate of…. (Strangely the usual automatic comment about the possible duplicate is missing -- did the close-voter delete it?) I don't think it's a proper duplicate, since a) the present question is merely tacked on to the other question and b) it doesn't specify that complex numbers shouldn't be used, and in fact the only answer so far uses complex numbers. It's unlikely that the present question is going to be answered over there. – joriki Oct 12 '12 at 8:21
Also, this question presents a specific method for solving a special case and asks how to generalize this method, which isn't mentioned in the other question. – joriki Oct 12 '12 at 8:28

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