find sum of n terms of series $sum cos(ntheta)$

Use the result $1 + z + z^2...+z^n=\frac{z^{n+1}-1}{z-1}$ to sum the series to n terms

$1+\cos\theta+\cos2\theta+...$

also show that partial sums of series $\sum \cos (n\theta)$ is bounded when $0<\theta<\pi/2$

My attempt

so z can be written as $e^{i\theta}$ which means:

$1+ \cos \theta + \cos 2\theta ....+\cos n\theta + i(\sin \theta+\sin 2\theta+....+\sin n\theta)=\frac{z^{n+1}-1}{z-1}$

after this.. i dont know

Answer

Remember that
$$
e^{it}=\cos t+i\sin t\;\;\;\;\forall t\in\Bbb C
$$
and that

$$
\sum_{j=0}^{n}z^j=\frac{1-z^{n+1}}{1-z}\;\;\forall z\in\Bbb C,\;\;|z|<1.
$$
Thus
$$
\sum_{j=0}^{n}\cos(j\theta)=
\sum_{j=0}^{n}\Re{(e^{ij\theta})}=
\Re\left(\sum_{j=0}^{n}(e^{ij\theta})\right)=
\Re\left(\frac{1-e^{i\theta(n+1)}}{1-e^{i\theta}}\right)
$$

The last term I wrote can be handled easily in order to be written explicitly and get the results you wanted.