Answer: \(\sum_{r=1}^{n}\sin((2r-1)\theta)=\frac{\sin^2(n\theta)}{\sin\theta}\), for \(\sin\theta\ne0\).
Also, \(\sum_{r=1}^{n}(2r-1)\cos\left(\frac{(2r-1)\pi}{2n}\right)=-\operatorname{cosec}\left(\frac\pi{2n}\right)\cot\left(\frac\pi{2n}\right)\).
Let \(z=\cos\theta+i\sin\theta=e^{i\theta}\). Then
\(\sum_{r=1}^{n}z^{2r-1}=z+z^3+\cdots+z^{2n-1}\).
This is a geometric series, so
\(\sum_{r=1}^{n}z^{2r-1}=\frac{z(1-z^{2n})}{1-z^2}=\frac{1-z^{2n}}{z^{-1}-z}\).
Since \(z^{-1}-z=-2i\sin\theta\),
\(\sum_{r=1}^{n}z^{2r-1}=\frac{1-z^{2n}}{-2i\sin\theta}\).
Using \(z^{2n}=\cos(2n\theta)+i\sin(2n\theta)\), the imaginary part is
\(\sum_{r=1}^{n}\sin((2r-1)\theta)=\frac{1-\cos(2n\theta)}{2\sin\theta}\).
Since \(1-\cos(2n\theta)=2\sin^2(n\theta)\),
\(\sum_{r=1}^{n}\sin((2r-1)\theta)=\frac{\sin^2(n\theta)}{\sin\theta}\).
Differentiate both sides with respect to \(\theta\):
\(\sum_{r=1}^{n}(2r-1)\cos((2r-1)\theta)=\frac{d}{d\theta}\left(\frac{\sin^2(n\theta)}{\sin\theta}\right)\).
The derivative on the right is
\(\frac{2n\sin(n\theta)\cos(n\theta)\sin\theta-\sin^2(n\theta)\cos\theta}{\sin^2\theta}\).
Now put \(\theta=\frac\pi{2n}\). Then \(n\theta=\frac\pi2\), so \(\sin(n\theta)=1\) and \(\cos(n\theta)=0\). Therefore
\(\sum_{r=1}^{n}(2r-1)\cos\left(\frac{(2r-1)\pi}{2n}\right)=-\frac{\cos(\pi/(2n))}{\sin^2(\pi/(2n))}\).
Since \(\frac{\cos x}{\sin^2x}=\operatorname{cosec}x\cot x\), this becomes
\(\sum_{r=1}^{n}(2r-1)\cos\left(\frac{(2r-1)\pi}{2n}\right)=-\operatorname{cosec}\left(\frac\pi{2n}\right)\cot\left(\frac\pi{2n}\right)\).
