By considering \(\sum_{r=1}^{n} z^{2 r-1}\), where \(z=\cos \theta+\mathrm{i} \sin \theta\), show that, if \(\sin \theta \neq 0\),
\(\sum_{r=1}^{n} \sin (2 r-1) \theta=\frac{\sin ^{2} n \theta}{\sin \theta}\)

Deduce that
\(\sum_{r=1}^{n}(2 r-1) \cos \left[\frac{(2 r-1) \pi}{2 n}\right]=-\operatorname{cosec}\left(\frac{\pi}{2 n}\right) \cot \left(\frac{\pi}{2 n}\right)\)