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, 07:12 AMdarenceCos i*Pi / 17 ?
Hello.

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

I have no usefull ide :( - May 7th 2013, 06:20 PMchiroRe: Cos i*Pi / 17 ?
Hey darence.

Hint: Note that cos(x) = [e^(i*x) + e^(-i*x)]/2. So if x = (i*pi)/17 then e^(ix) = e^(-pi/17). - May 7th 2013, 09:01 PMProve ItRe: Cos i*Pi / 17 ?
Sorry Chiro, but you have misunderstood what the OP wrote. The OP is using the symbol "i" as the counter in the product, NOT as the imaginary unit.

- May 7th 2013, 09:13 PMchiroRe: Cos i*Pi / 17 ?
Ohh thanks for the heads up Prove It.

Well in that case, just evaluate it using a computer or calculator unless you need some closed form answer (which is a different thing). - May 7th 2013, 09:19 PMProve ItRe: 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 7th 2013, 11:29 PMGusbobRe: 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, 11:42 AMdarenceRe: 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 $\displaystyle \pi/17$.

http://imageshack.us/a/img832/6514/30244448.png

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

http://img62.imageshack.us/img62/9982/26868458.png