sequences and series - Proof of $cos theta+cos 2theta+cos 3theta+cdots+cos ntheta=frac{sinfrac12ntheta}{sinfrac12theta}cosfrac12(n+1)theta$

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.

Answer

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}
$$