I must verify the following formula: $$ \sum _{i=1}^n \sin ^2 (2k-1) \theta = -\frac{n}{2} - \frac{\sin 4n \theta}{4 \sin 2 \theta} $$
I believe that I must do this by using the Euler's formula, $$ e^{i \theta} = \cos \theta + i \sin \theta $$ and taking only the imaginary parts of it. However, I am really stuck. Here is my work so far:
Let $x=e^{i \theta}$. Then, $$ e^{i \theta} = \cos \theta + i \sin \theta $$ $$ e^{2i \theta} = \cos ^2 \theta + i \sin \theta \cos \theta - \sin ^2 \theta $$ The sum we need to verify then becomes equivalent to the imaginary part of: $$ -e^{2i \theta} + \cos ^2 \theta-e^{6i \theta} + \cos ^2 3\theta -e^{10i \theta} + \cos ^2 5\theta -... +...-e^{(4k-2)i \theta} + \cos ^2 (2k-1)\theta. $$ This sum doesn't really seem to help me, especially because the exponent of the $e$ terms do not increase linearly. How should I approach this problem?
\begin{align*} \sum _{i=1}^n \sin ^2 (2k-1) \theta &= \frac{1}{2}\sum _{k=1}^n (1-\cos 2(2k-1)\theta)\\ &= \frac{n}{2} - \frac{1}{2}\sum _{k=1}^n \cos (4k - 2) \theta\\ &= \frac{n}{2} - {\text {Real part of }}\frac{1}{2}\sum _{k=1}^n e^{(4k - 2)i\theta}\\ &=\frac{n}{2} - {\text {Real part of }}\frac{1}{2}e^{2i\theta}(1+e^{4i\theta} + \cdots + e^{4(n-1)\theta})\\ &=\frac{n}{2} - {\text {Real part of }}\frac{1}{2}e^{2i\theta}\frac{1-e^{4ni\theta}}{1-e^{4i\theta}}\\ \end{align*} The computation can now be easily done.