Show that the series sum_r=0^infinity (1)/(4^r) 2rtheta is a convergent geometric progression.

Here's the solution for question 6b:

Part (i)
To show that the series $\sum_{r=0}^{\infty} \frac{1}{4^r} \sin 2r\theta$ is a convergent geometric progression, we consider a related complex geometric series.
Let $w = \frac{1}{4}(\cos 2\theta + i\sin 2\theta)$.
By De Moivre's theorem, $w^r = \left(\frac{1}{4}\right)^r (\cos 2r\theta + i\sin 2r\theta) = \frac{1}{4^r} \cos 2r\theta + i \frac{1}{4^r} \sin 2