Hello.

Please help me, how to calculate this: $\displaystyle \prod_{i=1}^{17}\cos\frac{i\pi}{17}$

I have no usefull ide

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- May 7th 2013, 08:12 AM #1

- May 7th 2013, 07:20 PM #2

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- May 7th 2013, 10:01 PM #3

- May 7th 2013, 10:13 PM #4

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- May 7th 2013, 10:19 PM #5
## Re: Cos i*Pi / 17 ?

No problem, I'm trying to wrap my head around trying to find a closed form answer myself haha, and I very nearly made the same mistake you did. I generally prefer not to use the symbol "i" for anything but the imaginary unit for that very reason. There are 25 other letters to choose from in our alphabet alone :P haha

- May 8th 2013, 12:29 AM #6

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## Re: Cos i*Pi / 17 ?

As others have mentioned before, the i is confusing. I'm going to replace it with k if you don't mind. I would also like to point out that whoever set you this question without giving you the final answer or some hints must have high expectations indeed. I have sketched the proof without detailed working. That, you should be able to work out yourself.

$\displaystyle \prod_{k=1}^{17}\cos\left(\frac{k\pi}{17}\right) $

$\displaystyle = \cos(\pi) \cdot \prod_{k=1}^{16}\cos\left(\frac{k\pi}{17}\right) $

$\displaystyle = - \prod_{k=1}^{16}\frac{\sin\left(\frac{2k\pi}{17} \right)}{2\sin \left(\frac{k\pi}{17}\right)}$ using $\displaystyle \sin(2x)=2\sin(x)\cos(x)$ with $\displaystyle x=\frac{k\pi}{17}$

Now the $\displaystyle \sin\left(\frac{2k\pi}{17} \right)$ from $\displaystyle k=1$ to $\displaystyle 8$ in the numerator is the same as the $\displaystyle \sin\left(\frac{k\pi}{17} \right)$ for even $\displaystyle k\in \{1,2,...,17\}$ in the denominator, so we can cancel those out to get

$\displaystyle = -\frac{1}{2^{16}} \frac{ \displaystyle{\prod_{k=9}^{17}} \sin\left(\frac{2k\pi}{17} \right)}{\displaystyle{\prod_{k=1}^{8}} \sin\left(\frac{(2k-1)\pi}{17} \right)} $

Note the change in argument for the sines in the denominator. This was done to make the equation readable...

Next, we use the identity that $\displaystyle \sin(x+\pi)=-\sin(x)$ to re-index the numerator and obtain

$\displaystyle = -\frac{(-1)^{8}}{2^{16}} \frac{ \displaystyle{\prod_{k=1}^{8}} \sin\left(\frac{(2k-1)\pi}{17} \right)}{\displaystyle{\prod_{k=1}^{8}} \sin\left(\frac{(2k-1)\pi}{17} \right)} $

which finally gives us

$\displaystyle = -\frac{1}{2^{16}}$

This proof is easily generalised to an arbitrary number $\displaystyle n$ (here $\displaystyle n=17$) of odd products (You need to adjust the denominator of your cosines accordingly). The even case is uninteresting since $\displaystyle \cos\frac{k\pi}{n}=0$ when $\displaystyle k=\frac{n}{2}$.

EDIT: You can also approach this question by splitting the polynomial of $\displaystyle x^{2n}-1$ into linear factors (you'll need to know what all the roots are) and evaluating at $\displaystyle x=i$.

- May 9th 2013, 12:42 PM #7
## Re: Cos i*Pi / 17 ?

Thank you very much! I have found another solution:

$\displaystyle \cos(x)\cos(2x)\cos(4x)=\frac{(\sin(x)\cos(x))\cos (2x)\cos(4x)}{\sin(x)}=\frac{(\sin(2x)\cos(2x))\co s(4x)}{2\sin(x)}=\frac{\sin(4x)\cos(4x)}{4\sin(x)} =\frac{\sin(8x)}{8\sin(x)}$.

Now, start from $\displaystyle \pi/17$.

Now, start from $\displaystyle 3\pi/17$.