This question already has an answer here:
Let $z=\cos\theta+i\sin\theta$ i.e. $z=e^{i\theta}$
Your sum:$$e^{i\theta}+e^{3i\theta}+e^{5i\theta}+…e^{(2n-1)i\theta}$$
This is a GP with common ratio $e^{2i\theta}$
Therefore sum is $$\frac{a(r^n-1)}{r-1}$$
$$\frac{e^{i\theta}(e^{2ni\theta}-1)}{e^{2i\theta}-1}$$
$$\frac{(\cos \theta+i\sin\theta)(\cos(2n\theta)+i\sin\theta-1)}{\cos(2\theta)+i\sin(2\theta)-1}$$
Computing it’s real part should give you the answer
Acknowledgement:Due credits to @LordShark Idea
Following @TheDeadLegend’s answer I found this telescoping technique. Turns out that you need a similar identity to make it work:
$$ \sin(\alpha + \beta) – \sin(\alpha – \beta) =2\cos \alpha \sin \beta$$
\begin{align}
g(x) &= \sum_{k=1}^n \cos(2k-1)x \\
&= \frac{1}{2\sin x}\sum_{k=1}^n 2\cos(2k-1)x \cdot \sin x \\
&= \frac{1}{2\sin x}\sum_{k=1}^n \left[\sin 2kx – \sin2(k-1)x \right] \\
&= \frac{1}{2\sin x}(\sin 2nx-0) = \frac{\sin 2nx}{2\sin x}.
\end{align}