Can you simplify \prod_{k=1}^{N-1} \sin{\frac{k\pi}{N}}?

[zcd]

How would I go about showing $$\prod_{k=1}^{N-1} \sin{\frac{k\pi}{N}} = \frac{N}{2^{N-1}}$$

I've tried using Euler's equation to substitute sin, but it just gets messy.

 

[micromass]

Do start with

$$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$$

and try to get things into the following shape

$$\prod_{k=1}^{n-1}{1-\xi^k}$$

where $\xi$ is the n-th root of unity. Then you can use the identity

$$z^n-1=\prod_{k=1}^n{z-\xi^k}$$

somehow.

 

[epenguin]

"Be a country boy!"

What is sinΠ, what is sin2Π, what is sin3Π,... ?

Are you sure this is the question?

 

[HallsofIvy]

"Be a country boy!"

What is sinΠ, what is sin2Π, what is sin3Π,... ?

Are you sure this is the question?

Well, the question you ask is not the question posed in the original post because you have ignored the "N" in the denominator.

 

[epenguin]

Ah you're right.

 

[zcd]

Do start with

$$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$$

and try to get things into the following shape

$$\prod_{k=1}^{n-1}{1-\xi^k}$$

where $\xi$ is the n-th root of unity. Then you can use the identity

$$z^n-1=\prod_{k=1}^n{z-\xi^k}$$

somehow.

I substituted it into get
$$\prod_{k=1}^{N-1} \frac{e^{i\frac{k\pi}{N}}-e^{-i\frac{k\pi}{N}}}{2i}$$
after a bit of manipulation I arrived at
$$\prod_{k=1}^{N-1} \frac{-1}{2i e^{i\frac{k\pi}{N}}}\prod_{k=1}^{N-1} 1-(e^{i\frac{\pi}{N}})^k$$
which remotely resembles the form. From here, how would I apply the identity and how do I get rid of that other product?

 

[micromass]

I substituted it into get
$$\prod_{k=1}^{N-1} \frac{e^{i\frac{k\pi}{N}}-e^{-i\frac{k\pi}{N}}}{2i}$$
after a bit of manipulation I arrived at
$$\prod_{k=1}^{N-1} \frac{-1}{2i e^{i\frac{k\pi}{N}}}\prod_{k=1}^{N-1} 1-(e^{i\frac{\pi}{N}})^k$$
which remotely resembles the form. From here, how would I apply the identity and how do I get rid of that other product?

To get rid of the other product is easy. Just apply your exponential laws: $e^xe^y=e^{x+y}$.

To apply the other identity, you have that

$$\prod_{k=1}^{N-1}{z-(e^{i\pi /N})^k}=\frac{z^k-1}{z-1}=...$$

Our ultimate goal is to put z=1 in the above equality.

 

[zcd]

To get rid of the other product is easy. Just apply your exponential laws: $e^xe^y=e^{x+y}$.

To apply the other identity, you have that

$$\prod_{k=1}^{N-1}{z-(e^{i\pi /N})^k}=\frac{z^k-1}{z-1}=...$$

Our ultimate goal is to put z=1 in the above equality.

I managed to change
$$\prod_{k=1}^{N-1} \frac{-1}{2i e^{i\frac{k\pi}{N}}}=\frac{1}{2^{N-1}}\prod_{k=1}^{N-1}\frac{i}{\exp(i\frac{k\pi}{N})}=\frac{1}{2^{N-1}}(\frac{i}{i})^{N-1}=\frac{1}{2^{N-1}}$$
but I'm still trying to see how the roots of unity identity can be applied. Also, it seems that if
$$z^N-1=\prod_{k=1}^N{z-(e^{i\pi /N})^k}$$ then
$$z^N-1=({z-(e^{i\pi /N})^N})\prod_{k=1}^{N-1}{z-(e^{i\pi /N})^k}$$ so
$$\prod_{k=1}^{N-1}{z-(e^{i\pi /N})^k}=\frac{z^N-1}{z-(-1)}$$

 

[micromass]

I missed a 2, there. It has to be

$$z^N-1=\prod_{k=1}^N{z-(e^{2i\pi /N})^k}$$I think you're missing a 2 as well in your form. So it would have to be

$$\prod_{k=1}^{N-1}{z-(e^{2i\pi /N})^k}=\frac{z^N-1}{z-1}$$

Now, try to use the identity

$$z^N-1=(z-1)(z^{N-1}+z^{N-2}+...+z+1)$$

 

[zcd]

I took $$z=1+\epsilon$$ and then took $$\lim_{\epsilon\to 0} \frac{(1+\epsilon)^N-1}{(1+\epsilon)-1}=\frac{d}{dx}(x^N)=N$$for x=1

 

[zcd]

or I could do
$$\prod_{k=1}^{N-1}{z-(e^{2i\pi /N})^k}=\frac{z^N-1}{z-1}=\frac{(z-1)(z^{N-1}+z^{N-2}+...+z+1)}{z-1}=z^{N-1}+z^{N-2}+...+z+1$$ which sums up to N

 

[micromass]

Indeed, it seems you've got it!

 

[zcd]

Thanks for the help :)