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Math Help - Deduce Q

  1. #1
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    Deduce Q

    Given \cos\theta=\frac{1}{2}(e^{i\theta}+e^{-i\theta}) and \sin\theta=\frac{1}{2i}(e^{i\theta}-e^{-i\theta})

    Deduce 1+2\cos\theta+2\cos2\theta+...+2\cos n\theta=\frac{\sin(n+\frac{1}{2})\theta}{\sin\frac  {1}{2}\theta}

    Use this result to show 1+(2\cos\theta)^2+(2\cos2\theta)^2+...+(2\cos n\theta)^2=\frac{\sin(2n+1)\theta}{\sin\theta}

    I was thinking geometric series but cant seem to figure it out. It can easily be done using sums to products but i was wondering if there was a way which makes use of the given result.

    thanks
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  2. #2
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    Quote Originally Posted by vuze88 View Post
    Given \cos\theta=\frac{1}{2}(e^{i\theta}+e^{-i\theta}) and \sin\theta=\frac{1}{2i}(e^{i\theta}-e^{-i\theta})

    Deduce 1+2\cos\theta+2\cos2\theta+...+2\cos n\theta=\frac{\sin(n+\frac{1}{2})\theta}{\sin\frac  {1}{2}\theta}

    Use this result to show 1+(2\cos\theta)^2+(2\cos2\theta)^2+...+(2\cos n\theta)^2=\frac{\sin(2n+1)\theta}{\sin\theta}\ {\color{red}{}+2n}

    I was thinking geometric series but cant seem to figure it out. It can easily be done using sums to products but i was wondering if there was a way which makes use of the given result.
    1+2\cos\theta+2\cos2\theta+...+2\cos n\theta= e^0 + (e^{i\theta}+ e^{-i\theta}) + \ldots + (e^{in\theta} + e^{-in\theta}) = \sum_{k=-n}^ne^{ik\theta}. This is a geometric series with common ratio e^{i\theta} and sum \frac{e^{-in\theta}(e^{i(2n+1)\theta}-1)}{e^{i\theta}-1}. Divide top and bottom of that by 2ie^{i\theta/2} and you will have something that you can express in terms of sines.

    For the second part (which should have an extra term +2n as shown), take the formula for the first part, replace \theta by 2\theta and use the formula \cos2\phi = 2\cos^2\phi-1.
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