# Thread: Cos i*Pi / 17 ?

1. ## Cos i*Pi / 17 ?

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

I have no usefull ide

2. ## Re: 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).

3. ## Re: 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.

4. ## Re: 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).

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

6. ## Re: Cos i*Pi / 17 ?

Originally Posted by darence
Hello.
Please help me, how to calculate this: $\prod_{i=1}^{17}\cos\frac{i\pi}{17}$

I have no usefull ide
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.

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

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

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

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 $\sin(x+\pi)=-\sin(x)$ to re-index the numerator and obtain

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

which finally gives us

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

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

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

7. ## Re: Cos i*Pi / 17 ?

Thank you very much! I have found another solution:

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

Now, start from $\pi/17$.

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