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Math Help - Summation of sine and cosine

  1. #1
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    Summation of sine and cosine

    I have a problem solving the following exercise.

    i have to prove that \sum_{j=0}^{n-1} cos (2j \pi /{n}})^2 = \sum_{j=0}^{n-1} sin (2j \pi /{n}})^2 = 1/2

    I have to prove that each sum is equal to 1/2 !!

    (n is odd and I know that both sums together are 1, but I cannot prove why each is 1/2)

    Help is appreciated!
    Last edited by sssitex; February 13th 2011 at 04:20 AM.
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  2. #2
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    is i natural?
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  3. #3
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    i'm sorry, i will edit it, it is not natural
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  4. #4
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    It's a simple application of Euler's Formula - \displaystyle r\,e^{i\theta} = r\cos{\theta} + i\,r\sin{\theta}...
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  5. #5
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    I have to prove that the sums are equal to 1/2, that's my only problem. I think I didn't explain it right, I'm sorry.
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    Quote Originally Posted by sssitex View Post
    I have a problem solving the following exercise.

    i have to prove that \sum_{j=0}^{n-1} cos (2j \pi /{n}})^2 = \sum_{j=0}^{n-1} sin (2j \pi /{n}})^2 = 1/2

    I have to prove that each sum is equal to 1/2 !!

    (n is odd and I know that both sums together are 1, but I cannot prove why each is 1/2)

    Help is appreciated!
    Shouldn't that be n/2 instead of 1/2?

    Hint: Can you show the first part? I.e.,

    \sum_{j=0}^{n-1} cos (2j \pi /{n}})^2 = \sum_{j=0}^{n-1} sin (2j \pi /{n}})^2
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  7. #7
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    I solved it, thanks
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  8. #8
    MHF Contributor chisigma's Avatar
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    The 'geometrical sum' of the n squared 'roots of unity' is...

    \displaystyle \sum_{k=0}^{n-1} e^{2\ j\ \frac{2\ k\ \pi}{n}} = \frac{1-e^{4\ j\ \pi}}{1-e^{4\ j\ \frac{\pi}{n}}} =0 (1)

    ... and from (1) we derive that is...

    \displaystyle \sum_{k=0}^{n-1} (\cos^{2} \frac{2\ k\ \pi}{n} - \sin^{2} \frac{2\ k\ \pi}{n}) =0 (2)

    Bur is also...

    \displaystyle \sum_{k=0}^{n-1} (\cos^{2} \frac{2\ k\ \pi}{n} + \sin^{2} \frac{2\ k\ \pi}{n}) =n (3)

    ... and from (2) and (3) we derive that is...

    \displaystyle \sum_{k=0}^{n-1} \cos^{2} \frac{2\ k\ \pi}{n} = \sum_{k=0}^{n-1} \sin^{2} \frac{2\ k\ \pi}{n} = \frac{n}{2} (4)

    Kind regards

    \chi \sigma
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