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Use $ 2\cos{n\theta} = z^n + z^{-n} $ to express $\cos\theta + \cos3\theta + \cos5\theta + ... + \cos(2n-1)\theta $ as a geometric progression in terms of $z$. Hence find the sum of this progression in terms of $\theta$. Any tips/help would be appreciated. I have the common ratio as $z^2$ and the first term as $z^{1-2n},$ and I can put these into the original formula, but I can't seem to get the answer I'm looking for.

z= $\cos\theta + i\sin\theta$ where i is the imaginary unit

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  • $\begingroup$ To get multicharacter exponents, put them in braces. So z^{-n} gives $z^{-n}$, while z^-n gives $z^-n$. Also a backslash before cos will give you the proper font: $\cos$ instead of $cos$ $\endgroup$
    – Ross Millikan
    Commented Oct 9, 2013 at 16:07
  • $\begingroup$ It's a bit easier to use $z = e^{i \theta}$, sure that you don't want to use that? Otherwise the solution gets really ugly really fast. $\endgroup$
    – DanielV
    Commented Oct 9, 2013 at 16:45

3 Answers 3

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$\begin{align} answer & = \sum_{i = 1}^{n} {\cos(\theta (2i - 1))} \\ & = \sum_{i = 0}^{n - 1} {\cos(\theta (2i + 1))} \\ & = \sum_{i = 0}^{n - 1} {\frac{z^{2i + 1} + z^{-2i - 1}}{2}} \\ & = \frac z 2 \sum_{i = 0}^{n - 1} {z^{2i} + \frac 1 {2z} \sum_{i = 0}^{n - 1} z^{-2i}} \\ & = \frac z 2 \text{geometric sum in } z^2 + \frac 1 {2z} \text{geometric sum in }z^{-2} \\ & = \frac z 2 \frac {z^{2n} - 1} {z^2 - 1} + \frac 1 {2z} \frac {z^{-2n} - 1} {z^{-2} - 1}\\ & = \frac z 2 \frac {z^{2n} - 1} {z^2 - 1} + \frac 1 {2z} \frac {z^{-2n} - 1} {z^{-2} - 1} \frac {z^2}{z^2} \\ & = \frac z 2 \frac {z^{2n} - 1} {z^2 - 1} + \frac z 2 \frac {z^{-2n} - 1} {1 - z^{2}}\\ & = \frac z 2 \frac {z^{2n} - z^{-2n}} {z^2 - 1} \end{align}$

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$$ z^{1-2n} + z^{1-2n + 2} + \cdots + z^{2n-1} = z^{1-2n} \left(1 + z^2 + \cdots + z^{4n-2}\right) = z^{1-2n} \frac{1-z^{4n-2}}{1 - z^2}.$$

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Regroup sum into two geometric progressions: $$2\left(\cos\theta + \cos3\theta + \cos5\theta + ... + \cos(2n-1)\theta\right)\\ =z+\dfrac{1}{z}+z^3+\dfrac{1}{z^3}+\ldots+z^{2n-1}+\dfrac{1}{z^{2n-1}}\\= z+z^3+\ldots+z^{2n-1}+\dfrac{1}{z}+\dfrac{1}{z^3}+\ldots+\dfrac{1}{z^{2n-1}}$$

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  • $\begingroup$ This is helpful, and I can get the formula for the sum of the progression, but I still can't manage to express the sum in terms of theta? $\endgroup$
    – Arkel
    Commented Oct 9, 2013 at 17:05
  • $\begingroup$ Remember that $z=e^{i\theta}$ $\endgroup$
    – M. Strochyk
    Commented Oct 9, 2013 at 17:12

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