Complex Numbers - Infinite Series 2

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Submitted by roamingfree on Fri, 06/25/2021 - 19:34
skills
You need to be able to move between trigonometric and exponential representations
Question
Let $S$ be the sequence $$ S = \sin(\theta) - \frac{1}{2}\sin(2\theta) + \frac{1}{4}\sin(3\theta) - \frac{1}{8}\sin(4\theta) + \dots $$ Determine the integers $a,b,c$ such that $$ S = \frac{a\sin\theta}{b + c\cos\theta} $$
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$$\frac{4\sin\theta}{5 + 4\cos\theta}$$
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The process here is similar to the previous question in the series. Recall that $$\sin(n\theta) = \frac{1}{2i}(e^{in\theta} - e^{-in\theta})$$

Consequently, $S$ can be rewritten as $$ \begin{align} S &= \frac{1}{2i}(e^{i\theta} - e^{-i\theta}) -\frac{1}{4i}(e^{2i\theta} - e^{-2i\theta}) +\frac{1}{8i}(e^{3i\theta} - e^{-3i\theta}) -\frac{1}{16i}(e^{4i\theta} - e^{-4i\theta}) +\dots \end{align}$$

We can now split this series into two distinct series (there are questions to be asked about the validity of such an operation, but as these series are convergent, its legitimate!)

$$ \begin{align} S^+ &= \frac{1}{2i}e^{i\theta} - \frac{1}{4i}e^{2i\theta} + \frac{1}{8i}e^{3i\theta} - \frac{1}{16i}e^{4i\theta} +\dots \\ &= \frac{\frac{1}{2i}e^{i\theta}}{1 + \frac{1}{2}e^{i\theta}}\\ S^- &= -\frac{1}{2i}e^{-i\theta} + \frac{1}{4i}e^{-2i\theta} - \frac{1}{8i}e^{-3i\theta} + \frac{1}{16i}e^{-4i\theta} + \dots \\ &= \frac{-\frac{1}{2i}e^{-i\theta}}{1 + \frac{1}{2}e^{-i\theta}} \end{align} $$

 

We can now write $S$ as 

$$ \begin{align} S &= S^+ + S^- \\ &= \frac{\frac{1}{2i}e^{i\theta}}{1 + \frac{1}{2}e^{i\theta}} + \frac{-\frac{1}{2i}e^{-i\theta}}{1 + \frac{1}{2}e^{-i\theta}} \\ &=\frac{\frac{1}{2i}e^{i\theta}(1+\frac{1}{2}e^{-i\theta}) - \frac{1}{2i}e^{-i\theta}(1 +\frac{1}{2}e^{i\theta})}{(1 + \frac{1}{2}e^{i\theta})(1 + \frac{1}{2}e^{-i\theta})} \\ &= \frac{\frac{1}{2i}(e^{i\theta} - e^{-i\theta})}{\frac{5}{4} + \frac{1}{2}(e^{i\theta} + e^{-i\theta})} \\ &= \frac{\sin\theta}{\frac{5}{4} + \cos \theta} \\ &= \frac{4\sin\theta}{5 + 4\cos\theta} \end{align} $$