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I'm having trouble with this proof

  1. Sep 21, 2014 #1
    I'm having trouble completing this proof for homework.

    1. The problem statement, all variables and given/known data


    Prove that cos θ + cos 3θ + cos 5θ + ... + cos (2n-1)θ = (sin 2nθ)/(2 sin θ).
    Prove that sin θ + sin 3θ + sin 5θ + ... + sin (2n-1)θ = (sin nθ)^2/(sin θ).

    Use Euler's formula and the geometric progression formula.


    2. Relevant equations

    Euler's formula is c9f2055dadfb49853eff822a453d9ceb.png .
    The geometric progression formula is f827ba026ce248a9d33ef69e89ec68de.png , where a is the first term and r is the constant that each term is multiplied by to get the next term.

    3. The attempt at a solution

    I've tried a number of ways, none of which I have been able to simplify. Can anybody who can see the right way to complete this proof perhaps give me some direction? For example, what are the starting points for a and r? I feel like I have tried everything I can think of, but I couldn't simplify it.

    How can I do the geometric progression formula if the series isn't exponential? I know I can use Euler's formula to transfer the cosines to exponentials, but how can I do this is the series consists only of cosines? How can I use the formula without the i sin(θ) terms? It seems that I can't substitute for the formula then... Hmm....

    If anybody knows any tricks or guidance, it would be greatly appreciated! Thanks in advance

    - Leo
     
    Last edited by a moderator: Apr 19, 2017
  2. jcsd
  3. Sep 21, 2014 #2

    Mark44

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    I would try mathematical induction.
     
    Last edited by a moderator: Apr 19, 2017
  4. Sep 21, 2014 #3

    kreil

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    Admittedly I haven't completed the problem before saying this, but my first line of attack would be to use Taylor series expansions for sin and cos.

    Hint:
    [tex]
    e^{ix} = \sum_{n=0}^{\infty} \frac{i^n x^n}{n!} = 1 + ix - \frac{x^2}{2} - i\frac{x^3}{3!} + \frac{x^4}{4!} + i\frac{x^5}{5!} .... = \left(1 - \frac{x^2}{2} + \frac{x^4}{4!} .... \right) + i \left( x-\frac{x^3}{3!} + \frac{x^5}{5!} ... \right)[/tex]
     
  5. Sep 21, 2014 #4

    Simon Bridge

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    Why not just convert the two series to a sum of exponentials using the Euler formula, then applying the geometric series formula to each?
    Alternatively - take the first and add i times the second, so the combined series is now a sum of exponentials, then apply the geometric series formula... simplify and see what falls out.
     
  6. Sep 22, 2014 #5

    vela

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    Hint: ##\cos\theta = \text{real part of }e^{i\theta}## and ##\cos 3\theta = \text{real part of }e^{i3\theta}##, so ##\cos \theta + \cos 3\theta = \text{real part of }(e^{i\theta} + e^{3i\theta})##.
     
  7. Sep 22, 2014 #6
    Thanks to all for the response. I ended up taking the exponential series and dividing it into two parts. One corresponds to e^+iθ and the other corresponds to e^-iθ. I can then easily take these and use the geometric progression formula. At the end of the day, I get REALLY close... I end up getting sin(4n+2)θ / 2 sin θ.

    Any way I can simplify this to sin(2nθ) / 2 sin θ?
     
  8. Sep 22, 2014 #7

    vela

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    Please show your work.
     
  9. Sep 22, 2014 #8
  10. Sep 22, 2014 #9
    EDIT: in the second picture, the exponentials in the denominator should both have θ.
     
  11. Sep 22, 2014 #10

    vela

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  12. Sep 22, 2014 #11
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