Almost same as Eric's answer (using $\log(1+x)$ instead of $\tan^{-1}(x)$. You can argue the convergence of the series to $\log(1+x)$ using generalized alternate series test even though $|e^{inx}| = 1$)
$\sin(n \theta) = \frac{\exp(i n \theta) - \exp(-i n \theta)}{2i}$.
$\displaystyle \sum_{n = \text{odd}} \frac{x^n}{n} = \frac{\log(1+x) - \log(1-x)}{2}$
$\displaystyle \sum_{n=\text{odd}}^{\infty} \frac{\exp(i n \theta) - \exp(-i n \theta)}{2in} = \displaystyle \frac1{2i} \left( \sum_{n=\text{odd}}^{\infty} \frac{\exp(i n \theta)}{n} - \sum_{n=\text{odd}}^{\infty} \frac{\exp(-i n \theta)}{n} \right)$
$= \frac1{4i} \left( \log(1 + \exp(i \theta)) - \log(1 - \exp(i \theta)) - \log(1 + \exp(-i \theta)) + \log(1 - \exp(-i \theta)) \right)$
$= \frac1{4i} \log \left(\frac{(1 + \exp(i \theta))(1 - \exp(- i \theta))}{(1 - \exp(i \theta))(1 + \exp(- i \theta))} \right) = \frac1{4i} \log \left( \frac{\exp(i \theta) - \exp(-i \theta)}{\exp(-i \theta) - \exp(i \theta)} \right) =\frac1{4i} \log(-1) = \frac{\pi i}{4i} = \boxed{\frac{\pi}{4}}$