$2\sin\frac{\theta}{2}\cos r\theta=\sin\frac{(2r+1)\theta}{2} - \sin\frac{(2r-1)\theta}{2}$
Putting r=1,2,....,n,
$2\sin\frac{\theta}{2}\cos \theta=\sin\frac{3\theta}{2} - \sin\frac{\theta}{2}$
$2\sin\frac{\theta}{2}\cos 2\theta=\sin\frac{5\theta}{2} - \sin\frac{3\theta}{2}$
...
$2\sin\frac{\theta}{2}\cos n\theta=\sin\frac{(2n+1)\theta}{2} - \sin\frac{(2n-1)\theta}{2}$
Adding we get,
$\sum_{1≤r≤n}2\sin\frac{\theta}{2}\cos r\theta=\sin\frac{(2n+1)\theta}{2} - \sin\frac{\theta}{2}$
Divide both sides by $2\sin\frac{\theta}{2}$, we shall get,
$\sum_{1≤r≤n}\cos r\theta=\frac{\sin\frac{(2n+1)\theta}{2}}{2\sin\frac{\theta}{2}} - \frac{1}{2}$ (Assuming $\sin\frac{\theta}{2}≠0$ or $\theta≠2s\pi$ where s is any integer.)
Or, $1+\sum_{1≤r≤n}\cos r\theta=\frac{\sin\frac{(2n+1)\theta}{2}}{2\sin\frac{\theta}{2}} + \frac{1}{2}$ (adding 1 to both sides )
Just observe that for $\sum_{r}\cos(A+2rB)$ where A,B are constants and r is an integer, we need to multiply with $2\sin B$ as
$2\cos(A+2rB)\sin B=sin(A+(2r+1)B) - sin(A+(2r-1)B)$
Putting different ranges of values of r & adding them, we shall get their sums in the compact form.
In the current problem, $A=0, 2B=\theta, 1≤r≤n$
Also as, $2\sin B\sin(A+2rB) = cos(A+(2r-1)B) - cos(A+(2r+1)B)$
This can be used for $\sum_{r}\sin(A+2rB)$.
Also using DonAntonio's approach, we know
$$\sin x=\frac{e^{ix}-e^{-ix}}{2i} \implies e^{ix}-e^{-ix}=2i\sin x$$
So,
$$\begin{align}
\frac{e^{(n+1)i\theta}-1}{e^{i\theta}-1} &=\frac{e^{(n+1)i\theta/2}\left(e^{(n+1)i\theta/2}-e^{-(n+1)i\theta/2}\right)}{e^{i\theta/2}\left(e^{i\theta/2}-e^{-i\theta/2}\right)} \\[0.5em]
&=e^{ni\theta/2}\frac{2i\sin\frac{(n+1)\theta}{2}}{2i\sin\frac{\theta}{2}} \\[0.5em]
&=\frac{\sin\frac{(n+1)\theta}{2}}{\sin\frac{\theta}{2}}\left(\cos\frac{n\theta}{2}+i\sin\frac{n\theta}{2}\right)\end{align}$$
Its real part is
$$\begin{align}
\frac{\sin\frac{(n+1)\theta}{2}}{\sin\frac{\theta}{2}}\cos\frac{n\theta}{2}
&=\frac{2\sin\frac{(n+1)\theta}{2}\cos\frac{n\theta}{2}}{2\sin\frac{\theta}{2}} \\
&=\frac{\sin\frac{(2n+1)\theta}{2}+\sin\frac{\theta}{2}}{2\sin\frac{\theta}{2}}\\
&=\frac{\sin\frac{(2n+1)\theta}{2}}{2\sin\frac{\theta}{2}}+\frac{1}{2}
\end{align}$$