<s>[INST] Showing $1+\cos(\theta)+\dots+\cos (n\theta)=\frac12+\frac{\sin((n+\frac12)\theta)}{2\sin(\frac\theta2)}$ 
Suppose that $n \geq 1$ is a natural number.
(i) Show that for $z$ in $\mathbb{C}$, provided $z \neq 1$,
$$ \frac{1-z^{n+1}}{1-z} = 1+z+...+z^n.$$
(ii) Using de Moivre’s Theorem, part (b)(i) and standard trigonometric identities, show that for $\theta$ in $(0,2\pi)$,
$$ 1+\cos(\theta) + \dots + \cos (n\theta) = \frac{1}{2} + \frac{\sin((n+\frac12)\theta)}{2\sin(\frac\theta2)}$$

I have set $z=a+bi$ and subbed this into the top equation but dont seem to be getting anwhere with this problem. Any guidance will be great!
 [/INST] Part (i) should be straightforward if you are familiar with finding the sum of a geometric series:
\begin{align}
S &= 1+z+z^2+\ldots+z^{n-1}+z^n \\
zS &= z+z^2+z^3+\ldots + z^n+z^{n+1} \\
zS - S &= z^{n+1}-1 \\
S(z-1) &= z^{n+1}-1 \\
S &= \frac{z^{n+1}-1}{z-1} = \frac{1-z^{n+1}}{1-z} \, . \\ 
\end{align}
For part (ii), you should first consider the geometric series
$$
1+e^{i\theta}+e^{2i\theta}+\ldots+e^{ni\theta} \, .
$$
Since $e^{ik\theta}=\cos k\theta+i\sin k\theta$ for $k \in \mathbb{Z}$, we know that
\begin{align}
&1+e^{i\theta}+e^{2i\theta}+\ldots+e^{ni\theta}\\=&1+(\cos\theta+i\sin\theta)+(\cos 2\theta+i\sin 2\theta)+\ldots+(\cos n\theta+i\sin n\theta)
\end{align}
Hence,
$$
1+\cos\theta+\cos 2\theta+\ldots+\cos n \theta=\Re(1+e^{i\theta}+e^{2i\theta}+\ldots+e^{ni\theta}) \, .
$$
</s>