<s>[INST] Summing up trigonometric series 
By considering $\sum_{r=1}^n z^{2r-1}$ where z= $\cos\theta + i\sin\theta$, show that if $\sin\theta$ $\neq$ 0, $$\sum_{r=1}^n \sin(2r-1)\theta=\frac{\sin^2n\theta}{\sin\theta}$$

I couldn't solve this at first but with some hints some of you gave, I was able to come up with my own solution. Here it is:
First, we want to consider what is given in the question,
$$\begin{align}
\sum_{r=1}^n z^{2r-1} & =z+z^3+z^5...\\
& = z + z(z^2)+z^3(z^2)+...\\
\end{align}$$
In Geometric Progression, sum is given by
$$\sum_{r=1}^n z^{2r-1}=S_n = \frac{a (1-r^n)}{1-r}=\frac{z(1-(z^2)^n)}{1-z^2} = \frac {z-z^{2n+1}}{1-z^2}=\frac {1-z^{2n}}{z^{-1}-z} $$
Now, substitute $z = \cos \theta +i \sin \theta$
$$\begin{align}
\sum_{r=1}^n z^{2r-1} & =\frac{1-(\cos (2n\theta) + i \sin(2n\theta))}{\cos\theta - i\sin\theta-(\cos\theta+i\sin\theta)} \\
 & = \frac{1-\cos (2n\theta) - i \sin(2n\theta)}{-2i\sin\theta} \cdot \frac{(i\sin\theta)}{(i\sin\theta)}\\ 
 & = \frac{i\sin\theta-i\sin\theta \cos (2n\theta) +  \sin\theta \sin(2n\theta)}{2\sin^2\theta} \\
\end{align}$$
Equating imaginary parts,
$$\sum_{r=1}^n \sin(2r-1)\theta = \frac{\sin\theta (1 - \cos(2n\theta))}{2\sin^2\theta} = \frac{\sin^2n\theta}{2\sin\theta} $$
 [/INST] An alternative (maybe easier) approach. Since:
$$ 2\sin(\theta)\sin((2r-1)\theta) = \cos((2r-2)\theta)-\cos(2r\theta), $$
the original sum, multiplied by $2\sin\theta$, becomes a telescopic sum, whose value equals $1-\cos(2n\theta) = 2\sin^2(n\theta).$
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