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Math Help - Cos i*Pi / 17 ?

  1. #1
    Junior Member darence's Avatar
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    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
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  2. #2
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    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).
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  3. #3
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    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.
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    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).
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  5. #5
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    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
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  6. #6
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    Re: Cos i*Pi / 17 ?

    Quote Originally Posted by darence View Post
    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.
    Last edited by Gusbob; May 7th 2013 at 11:40 PM.
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  7. #7
    Junior Member darence's Avatar
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    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.

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