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).

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- May 7th 2013, 07:12 AM #1

- May 7th 2013, 06:20 PM #2

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- May 7th 2013, 09:01 PM #3

- May 7th 2013, 09:13 PM #4

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- May 7th 2013, 09:19 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 7th 2013, 11:29 PM #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.

using with

Now the from to in the numerator is the same as the for even in the denominator, so we can cancel those out to get

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 to re-index the numerator and obtain

which finally gives us

This proof is easily generalised to an arbitrary number (here ) of odd products (You need to adjust the denominator of your cosines accordingly). The even case is uninteresting since when .

EDIT: You can also approach this question by splitting the polynomial of into linear factors (you'll need to know what all the roots are) and evaluating at .

- May 9th 2013, 11:42 AM #7