Proof of $\cos \theta+\cos 2\theta+\cos 3\theta+\cdots+\cos n\theta=\frac{\sin\frac12n\theta}{\sin\frac12\theta}\cos\frac12(n+1)\theta$ 
State the sum of the series $z+z^2+z^3+\cdots+z^n$, for $z\neq1$.
By letting $z=\cos\theta+i\sin\theta$, show that
$$\cos \theta+\cos 2\theta+\cos 3\theta+\cdots+\cos n\theta=\frac{\sin\frac12n\theta}{\sin\frac12\theta}\cos\frac12(n+1)\theta$$
Where $\sin\frac12\theta\neq0$.

I know the first part, The second part Im kind of stuck in showing that
My Attempt: 
$$\begin{align}\Re{(\cos\theta+i\sin\theta)+(\cos\theta+i\sin\theta)^2+\ldots+(\cos\theta+i\sin\theta)^n}\end{align}$$
I realized that this is a Geometric Progression, so its in the form: 
$a+ar+ar^2+....+ar^n$ , where $a=(\cos\theta+i\sin\theta)$ and $r=(\cos\theta+i\sin\theta)$
So I will apply the formula for the Sum of a G.P to my problem.
$$\begin{align}\Re \frac{(\cos\theta+i\sin\theta)(1-(\cos\theta+i\sin\theta)^n)}{1-(\cos\theta+i\sin\theta)}\end{align}$$
I applied the De Movire Theorem and simplified as follows: 
$$\begin{align}\Re \frac{\cos\theta+i\sin\theta-\cos(n+1)\theta+i\sin(n+1)\theta}{1-(\cos\theta+i\sin\theta)}\end{align}$$
$$\begin{align}\Re \frac{\cos\theta+i\sin\theta-\cos(n+1)\theta+i\sin(n+1)\theta}{1-(\cos\theta+i\sin\theta)}\frac{(1+(cos\theta+isin\theta))}{(1+(cos\theta+isin\theta))}\end{align}$$
$$\begin{align}\Re \frac{\cos\theta+i\sin\theta-\cos(n+1)\theta+i\sin(n+1)\theta+cos^2\theta+2icos\theta sin\theta-sin^2\theta-cos\theta cos(n+1)\theta- isin \theta cos(n+1)\theta+i cos\theta sin(n+1)\theta-sin\theta sin(n+1)\theta}{1-cos^2\theta-2i cos\theta sin\theta+sin^2\theta}\end{align}$$
$$\begin{align}\Re \frac{\cos\theta+i\sin\theta-\cos(n+1)\theta+i\sin(n+1)\theta+cos (2\theta) +i sin (2\theta)- cos(n+1)\theta(cos\theta+isin \theta)+(i cos\theta -sin \theta) sin(n+1)\theta}{1-cos^2\theta-2i cos\theta sin\theta+sin^2\theta}\end{align}$$
$$\begin{align}\Re \frac{\cos\theta+i\sin\theta+cos (2\theta) +i sin (2\theta)- cos(n+1)\theta(1+cos\theta+isin \theta)+(1+i cos\theta -sin \theta) sin(n+1)\theta}{2sin^2\theta-2i cos\theta sin\theta}\end{align}$$
$$\begin{align}\Re \frac{[\cos\theta+i\sin\theta+cos (2\theta) +i sin (2\theta)- cos(n+1)\theta(1+cos\theta+isin \theta)+(1+i cos\theta -sin \theta) sin(n+1)\theta]}{2sin\theta(sin\theta-i cos\theta)} \frac{(sin\theta+i cos\theta)}{(sin\theta+i cos\theta)}\end{align}$$
$$\begin{align}\frac{[-cos(n+1)\theta(1+cos\theta+isin \theta)+(1+i cos\theta -sin \theta) sin(n+1)\theta]}{2sin\theta} \frac{(sin\theta+i cos\theta)}{1}\end{align}$$
$$\begin{align}\frac{[-cos(n+1)\theta(sin\theta+cos\theta sin\theta - cos\theta sin \theta)+(sin\theta-cos^2\theta -sin^2 \theta) sin(n+1)\theta]}{2sin\theta} \end{align}$$
$$\begin{align}\frac{[-cos(n+1)sin\theta+(sin\theta-1) sin(n+1)\theta]}{2sin\theta} \end{align}$$
I have no idea right now where I am taking this, I just dont know what the next step I should take. Please dont send me the solution (at least yet). Can anyone give me a hint (a little boost to my little mind) as to what I should do next, (make sure its a small hint, Don't give me the full next step) Just the help in order for me to construct the next step and carry on.
 A: Using the appropriate prosthaphaeresis formula,
$$\begin{align}
\sin{\frac{\theta}{2}}\cos{k\theta}
&=\frac12\left[\sin{\left(\frac{\theta}{2}+k\theta\right)}+\sin{\left(\frac{\theta}{2}-k\theta\right)}\right]\\
&=\frac12\left[\sin{\left(\left(k+\frac{1}{2}\right)\theta\right)}-\sin{\left(\left(k-\frac{1}{2}\right)\theta\right)}\right].
\end{align}$$
In this form, summation of $k$ now transparently telescopes:
$$\begin{align}
\sum_{k=1}^{n}\sin{\frac{\theta}{2}}\cos{k\theta}
&=\sum_{k=1}^{n}\frac12\left[\sin{\left(\left(k+\frac{1}{2}\right)\theta\right)}-\sin{\left(\left(k-\frac{1}{2}\right)\theta\right)}\right]\\
&=\frac12\left[\sin{\left(\left(n+\frac{1}{2}\right)\theta\right)}-\sin{\left(\left(1-\frac{1}{2}\right)\theta\right)}\right]\\
&=\frac12\left[\sin{\left(\left(n+\frac{1}{2}\right)\theta\right)}-\sin{\left(\frac{\theta}{2}\right)}\right]\\
&=\sin{\left(\frac{n\theta}{2}\right)\cos{\left(\frac{(n+1)\theta}{2}\right)}}.
\end{align}$$
Now divide through by $\sin\frac{\theta}{2}$.
A: Hint : You have not taken the real part correctly as you have missed the $isin\theta$ in the denominator
A: Here are the main steps.
You may write
$$
\begin{align}
\sum_{k=1}^{n} \cos (k\theta)&=\Re \sum_{k=1}^{n} e^{ik\theta}\\\\
&=\Re\left( e^{i\theta}\frac{e^{in\theta}-1}{e^{i\theta}-1}\right)\\\\
&=\Re\left( e^{i\theta}\frac{e^{in\theta/2}\left(e^{in\theta/2}-e^{-in\theta/2}\right)}{e^{i\theta/2}\left(e^{i\theta/2}-e^{-i\theta/2}\right)}\right)\\\\
&=\Re\left( e^{i\theta}\frac{e^{in\theta/2}\left(2i\sin(n\theta/2)\right)}{e^{i\theta/2}\left(2i\sin(\theta/2)\right)}\right)\\\\
&=\Re\left( e^{i(n+1)\theta/2}\frac{\sin(n\theta/2)}{\sin(\theta/2)}\right)\\\\
&=\Re\left( \left(\cos ((n+1)\theta/2)+i\sin ((n+1)\theta/2)\right)\frac{\sin(n\theta/2)}{\sin(\theta/2)}\right)\\\\
&=\frac{\sin(n\theta/2)}{\sin(\theta/2)}\cos ((n+1)\theta/2).
\end{align}
$$
