trigonometry - Simple expressions for $sum_{k=0}^ncos(ktheta)$ and $sum_{k=1}^nsin(ktheta)$?

I'm curious if there is a simple expression for
$$
1+\cos\theta+\cos 2\theta+\cdots+\cos n\theta
$$
and

$$
\sin\theta+\sin 2\theta+\cdots+\sin n\theta.
$$
Using Euler's formula, I write $z=e^{i\theta}$, hence $z^k=e^{ik\theta}=\cos(k\theta)+i\sin(k\theta)$.
So it should be that
$$
\begin{align*}
1+\cos\theta+\cos 2\theta+\cdots+\cos n\theta &= \Re(1+z+\cdots+z^n)\\
&= \Re\left(\frac{1-z^{n+1}}{1-z}\right).
\end{align*}

$$
Similarly,
$$
\begin{align*}
\sin\theta+\sin 2\theta+\cdots+\sin n\theta &= \Im(z+\cdots+z^n)\\
&= \Im\left(\frac{z-z^{n+1}}{1-z}\right).
\end{align*}
$$
Can you pull out a simple expression from these, and if not, is there a better approach? Thanks!

Answer

Take the expression you have and multiply the numerator and denominator by $1-\bar{z}$, and using $z\bar z=1$:
$$\frac{1-z^{n+1}}{1-z} = \frac{1-z^{n+1}-\bar{z}+z^n}{2-(z+\bar z)}$$

But $z+\bar{z}=2\cos \theta$, so the real part of this expression is the real part of the numerator divided by $2-2\cos \theta$. But the real part of the numerator is $1-\cos {(n+1)\theta} - \cos \theta + \cos{n\theta}$, so the entire expression is:

$$\frac{1-\cos {(n+1)\theta} - \cos \theta + \cos{n\theta}}{2-2\cos\theta}=\frac{1}{2} + \frac{\cos {n\theta} - \cos{(n+1)\theta}}{2-2\cos \theta}$$

for the cosine case. You can do much the same for the case of the sine function.