I have worked something out like the following
https://math.stackexchange.com/a/1172454/626039 - reproduced here: $$\begin{align} \sum_{k=1}^{n} \sin (k\theta)&=\Im \sum_{k=1}^{n} e^{ik\theta}\\\\ &=\Im\left( e^{i\theta}\frac{e^{in\theta}-1}{e^{i\theta}-1}\right)\\\\ &=\Im\left( e^{i\theta}\frac{e^{in\theta/2}\left(e^{in\theta/2}-e^{-in\theta/2}\right)}{e^{i\theta/2}\left(e^{i\theta/2}-e^{-i\theta/2}\right)}\right)\\\\ &=\Im\left( e^{i\theta}\frac{e^{in\theta/2}\left(2i\sin(n\theta/2)\right)}{e^{i\theta/2}\left(2i\sin(\theta/2)\right)}\right)\\\\ &=\Im\left( e^{i(n+1)\theta/2}\frac{\sin(n\theta/2)}{\sin(\theta/2)}\right)\\\\ &=\Im\left( \left(\cos ((n+1)\theta/2)+i\sin ((n+1)\theta/2)\right)\frac{\sin(n\theta/2)}{\sin(\theta/2)}\right)\\\\ &=\frac{\sin ((n+1)\theta/2)}{\sin(\theta/2)}\sin(n\theta/2). \end{align}$$
The only problem is that i want to do this for when the sequence starts at k = 0 NOT k = 1 . I can't seem to achieve the same result but it should do according to my question.
$$\begin{align} \sum_{k=0}^{n} \sin (k\theta)&=\Im \sum_{k=0}^{n} e^{ik\theta}\\\\ &=\Im\left(\frac{e^{i(n+1)\theta}-1}{e^{i\theta}-1}\right)\\\\ &=\Im\left(\frac{e^{i(n+1)\theta/2}\left(e^{i(n+1)\theta/2}-e^{-i(n+1)\theta/2}\right)}{e^{i\theta/2}\left(e^{i\theta/2}-e^{-i\theta/2}\right)}\right)\\\\ &=\Im\left(\frac{e^{i(n+1)\theta/2}\left(2i\sin((n+1)\theta/2)\right)}{e^{i\theta/2}\left(2i\sin(\theta/2)\right)}\right)\\\\ &=\Im\left(e^{in\theta/2}\frac{\sin((n+1)\theta/2)}{\sin(\theta/2)}\right)\\\\ &=\Im\left( \left(\cos (n\theta/2)+i\sin (n\theta/2)\right)\frac{\sin((n+1)\theta/2)}{\sin(\theta/2)}\right)\\\\ &=\frac{\sin ((n+1)\theta/2)}{\sin(\theta/2)}\sin(n\theta/2). \end{align}$$
As one would expect, as the term added for $k = 0$ is just $0$.