Complex number - sum of imaginary part Question:

Let $z = \cos \theta + i \sin \theta$. Then what is the value of $$\sum^{15}_{m=1}\text{Im}(z^{2m-1})$$ at $\theta = 2^o$

Not sure how I should attempt this question. Obviously, as it's only the sum of the imaginary part, it will simply be:
$$\sin \theta + \sin 3\theta + \sin 5\theta + ... + \sin 29\theta$$
Not sure how to proceed from here though. 
 A: We can write $$z=\cos \theta+i\sin \theta = e^{i\theta}$$
Now we have to calculate $\Im(S)$ where $$S=\sum^{15}_{m=1}z^{2m-1} =  \sum^{15}_{m=1}e^{i(2m-1)\theta}$$
Now Let $$\displaystyle S=e^{i\theta}+e^{3i\theta}+...........+e^{29i\theta}..................(1)\times e^{2i\theta}$$
So $$S\cdot e^{2i\theta} = e^{i3\theta}+............+e^{i29\theta}+e^{i31\theta}.....(2)$$
So we get $$\displaystyle S = \frac{e^{i\theta}-e^{i31\theta}}{1-e^{i\theta}} = \frac{1-e^{i30\theta}}{e^{-i\theta}-e^{i\theta}}$$
Now Using $$\bullet\; \cos \theta +i\sin \theta = e^{i\theta}$$ and $$\bullet\; \cos \theta -i\sin \theta = e^{-i\theta}$$
and $$\displaystyle \bullet\;  e^{in\theta} = \cos n\theta+i\sin n\theta.$$
We get $$\displaystyle e^{-i\theta}-e^{i\theta}=-2i\sin \theta$$
So we get $$\displaystyle S = -\left[\frac{1-(\cos 30\theta+i\sin 30\theta)}{2i\sin \theta}\right]=i\left(\frac{1-\cos 30\theta}{2\sin \theta}\right)-\frac{\sin 30\theta}{2\sin \theta}$$
Now We have To Calculate $\Im(S)$  Which is $$\displaystyle \left(\frac{1-\cos 30\theta}{2\sin \theta}\right)$$
Now at $\theta = 2^0\;,$ We get $$\displaystyle \Im(S) = \left(\frac{1-\cos 60^0\theta}{2\sin 2^0}\right)=\frac{1}{4\sin 2^0}$$
A: Let $C=\cos\theta+\cos3\theta+\cos5\theta+...+\cos29\theta$
$S=\sin\theta+\sin3\theta+\sin5\theta+...+\sin29\theta$
$C+iS=(\cos\theta+i\sin\theta)+(\cos3\theta+i\sin3\theta)+...+(\cos29\theta+i\sin29\theta)$
$C+iS=e^{i\theta}+e^{3i\theta}+e^{5i\theta}+...+e^{29i\theta}$
$C+iS=\left\{\frac{e^{i\theta}(e^{30i\theta}-1)}{e^{2i\theta}-1}\right\}=\left\{\frac{(e^{30i\theta}-1)}{e^{i\theta}-e^{-i\theta}}\right\}=\left(\frac{\cos30\theta+i\sin30\theta-1}{2i\sin\theta}\right)$
Compare real and imaginary parts on both sides:$S=\left(\frac{1-\cos30\theta}{2\sin\theta}\right)$
Put $\theta=2^0$ and you get $S=(1-1/2)/(\sin2^0)=1/(4\sin2^0)$
