Sum of a trigonometric series

Complex sum. Let $Z=\sum_{k=1}^ne^{ik\theta}=e^{i\theta}\frac{1-e^{in\theta}}{1-e^{i\theta}}$.

Imaginary part. $S_n=\operatorname{Im}(Z)$.

Computation.
$$Z=e^{i\theta}\frac{1-e^{in\theta}}{1-e^{i\theta}}=e^{i\theta}\cdot\frac{e^{in\theta/2}(e^{-in\theta/2}-e^{in\theta/2})}{e^{i\theta/2}(e^{-i\theta/2}-e^{i\theta/2})}=e^{i(n+1)\theta/2}\frac{\sin(n\theta/2)}{\sin(\theta/2)}.$$

Imaginary part. $S_n=\frac{\sin(n\theta/2)\sin((n+1)\theta/2)}{\sin(\theta/2)}$.

Conclusion. $S_n=\frac{\sin(n\theta/2)\sin((n+1)\theta/2)}{\sin(\theta/2)}$ (for $\sin(\theta/2)\neq0$).