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Math Help - Series expressions

  1. #1
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    Series expressions

    By expressing 2\sin\theta\cos 2r\theta as the difference between two sines, show that
    \sin\theta \displaystyle \sum^n_{r=1}\cos 2r\theta=\sin n\theta\cos (n+1)\theta
    Evaluate \displaystyle \sum^{100}_{r=1} \cos^2 \left(\frac{r \pi}{100}\right)

    My attempt:
    2\sin\theta\cos 2r\theta
    =2\sin\theta(\cos^2 r\theta-\sin^2 r\theta)
    =\sin (r+1)\theta\cos r\theta-2\sin\theta\sin^2 r\theta
    =\frac{1}{2}\sin (2r+1)\theta -2\sin\theta(1-\cos^2 r\theta)
    =\frac{1}{2}\sin (2r+1)\theta+\sin (r+1)\theta\cos r\theta -2\sin\theta
    =\frac{1}{2}\sin (2r+1)\theta+\frac{1}{2}\sin (2r+1)\theta-2\sin\theta
    =\sin (2r+1)\theta-2\sin\theta

    \sin\theta \displaystyle \sum^n_{r=1}\cos 2r\theta=\sin n\theta\cos (n+1)\theta=\frac{1}{2} \displaystyle \sum^n_{r=1}\left(\sin (2r+1)\theta-2\sin\theta\right)
    =\frac{1}{2}\left(\displaystyle \sum^n_{r=1} \sin (2r+1)\theta-2n\sin\theta\right)
    This is as far as I got.
    Thanks!
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  2. #2
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    I would start by

    2\sin{\theta}\cos{2r\theta} = 2\sin{\theta}(1 - 2\sin^2{r\theta})

     = 2\sin{\theta} - 4\sin{\theta}\sin^2{r\theta}.
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  3. #3
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    Thanks, that is a much shorter way, but I'm still not sure how to continue other than
    \sin\theta \displaystyle \sum^n_{r=1} \cos 2r\theta=\sum^n_{r=1} \left(\sin\theta-2\sin\theta\sin^2r\theta\right)
    then
    \sin\theta\left(n - \displaystyle \sum^n_{r=1}\sin^2r\theta\right)
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