$\sin(ωt) + \sin(ωt+\Delta\theta) + \sin(ωt+2\,\Delta\theta) + \cdots + \sin(ωt+\overline{n-1}\,\Delta\theta) ={}$? How will we add this series? 
Any simple process?
This is one of the equation derived from the equation to find intesity of ray at any particular point in Fraunhofer diffraction.
I think this can be solved using simple series but my teacher told any process involving Imaginary part iota, which I couldn't understand.
 A: 
Find the value of $\sum_{k=0}^{n-1}\sin(a+kt)$.

$$\begin{align}\sum_{k=0}^{n-1}\sin(a+kt)=\sum_{k=0}^{n-1}\frac{e^{i(a+kt)}-e^{-i(a+kt)}}{2i}\\=\sum_{k=0}^{n-1}\frac{e^{i(a+kt)}}{2i}-\sum_{k=0}^{n-1}\frac{e^{-i(a+kt)}}{2i}\\=\frac{e^{ia}}{2i}\sum_{k=0}^{n-1}e^{ikt}-\frac{e^{-ia}}{2i}\sum_{k=0}^{n-1}e^{-ikt}\\=\frac{e^{ia}}{2i}\frac{1-e^{int}}{1-e^{it}}-\frac{e^{-ia}}{2i}\frac{1-e^{-int}}{1-e^{-it}}\end{align}$$
Can you do next steps?
A: The hint.
Thy to multiply it by $2\sin\frac{\Delta\theta}{2}$ and use $$2\sin\alpha\sin\beta=\cos(\alpha-\beta)-\cos(\alpha+\beta).$$
\begin{align}
& \sin\omega t + \sin(\omega t+∆θ) + \sin(ωt+2∆θ) + \cdots + \sin(ωt+(n-1)∆θ) \\[10pt]
= {} & \frac{2\sin\frac{\Delta\theta}{2}\sin\omega t +2\sin\frac{\Delta\theta}{2} \sin(\omega t+∆θ)+ \cdots + 2\sin\frac{\Delta\theta}{2}\sin(ωt+(n-1)∆θ)}{2\sin\frac{\Delta\theta}{2}} \\[10pt]
= {} & \tfrac{\cos\left(\omega t-\frac{\Delta\theta}{2}\right)-\cos\left(\omega t + \frac{\Delta\theta}{2}\right) +\cos\left(\omega t+\frac{\Delta\theta}{2} \right) - \cos\left(\omega t+\frac{3\Delta\theta}{2} \right) + \cdots+\cos\left(\omega t+\left(n-\frac{3}{2}\right)\Delta\theta\right)-\cos\left(\omega t+\left(n-\frac{1}{2}\Delta\right)\theta\right)}{2\sin\frac{\Delta\theta}{2}} \\[10pt]
= {} & \frac{\cos\left(\omega t-\frac{\Delta\theta}{2}\right)-\cos\left(\omega t + \left(n-\frac{1}{2}\right)\Delta\theta\right)}{2\sin\frac{\Delta\theta}{2}}
\end{align}
