Let $S = e^{i\alpha} + \frac{e^{i3\alpha}}{3} + \frac{e^{i5\alpha}}{3^2} + \cdots$
Find Im$(S)$ and show that it is equal to the sum
$I = \sin(\alpha) + \frac{\sin(3\alpha)}{3} + \frac{\sin(5\alpha)}{3^2} + \cdots$
So, I found that $S = \frac{3(3e^{i\alpha} - e^{-i\alpha})}{10 - 6\cos(2\alpha)}$ using the formula for geometric series.
I have a provided answer of $\frac{6\sin(\alpha)}{5 - 3\cos(2\alpha)}$ which I can see that I get if I just take the $\sin(\alpha)$ terms out of the $e$ terms in my numerator; however, why don't I have to change the $\cos(2\alpha)$ term in the denominator? Isn't this part of Re$(S)$? I was confused about his part and was trying to change this term before I looked at the answer.