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{\bf Question}

Prove that $$\cos \theta + \cos 3\theta + \cos 5\theta + ... +
\cos(2n-1)\theta = \frac{\sin2n\theta}{2\sin \theta}$$ $$\sin
\theta + \sin 3\theta + \sin 5\theta + ... + \sin(2n-1)\theta =
\frac{sin^2 n\theta}{\sin\theta}$$

\vspace{.25in}

{\bf Answer}

\begin{eqnarray*} e^{i\theta} + e^{3i\theta} + ... +
e^{(2n-1)i\theta} & = & \frac{e^{i\theta}(1 -
e^{2ni\theta})}{1-e^{2i\theta}} \\ & = &
\frac{1-2^{2ni\theta}}{e^{-i\theta} - e^{i\theta}} \\ & = &
\frac{1-e^{2ni\theta}}{-2i\sin\theta} \\ & = & \frac{i(1 -
\cos2n\theta - i\sin2n\theta)}{2\sin\theta} \\\end{eqnarray*}
Similarly \begin{eqnarray*} e^{i\theta} + e^{3i\theta} + ... +
e^{(2n-1)i\theta} & = & -\frac{i(1 - \cos2n\theta +
i\sin2n\theta)}{2\sin\theta} \\\end{eqnarray*} So
\begin{eqnarray*} \cos \theta +  ... +
\cos(2n-1)\theta & = & \frac{1}{2}(e^{i\theta} + e^{-i\theta} +
... + e^{(2n-1)i\theta} + e^{-(2n-1)i\theta}\\ & = & \frac{1}{2}
\left(\frac{i(1-\cos 2n\theta - i\sin2n\theta)}{2\sin\theta}
\right. \\ & & \hspace{.3in}\left.- \frac{(1-\cos 2n\theta +
i\sin2n\theta)}{2\sin\theta}\right) \\ & = &
\frac{1}{4\sin\theta}(i - i\cos 2n\theta + \sin2n\theta - \\ & &
\hspace{.7in}i + i\cos 2n\theta + \sin2n\theta) \\ & = &
\frac{2\sin2n\theta}{4\sin \theta} \\ & = &
\frac{\sin2n\theta}{2\sin \theta} \\ { } \\ \sin \theta +  ... +
\sin(2n-1)\theta & =& \frac{e^{-i\theta}(1 - e^{-2n\theta})}{1 -
e^{-2i\theta}} \\ & = & \frac{1}{ai\sin\theta}(i - i\cos 2n\theta
+ \sin 2n\theta + i \\ & & \hspace{.7in} - i\cos 2n\theta - \sin
2n\theta)
\\ & = & \frac{i - i\cos 2n\theta}{2i\sin\theta} \\ & = & \frac{1
- \cos 2n \theta}{2\sin\theta}
\\ & = & \frac{sin^2n\theta}{\sin\theta}\end{eqnarray*}

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