Simple expressions for $\sum_{k=0}^n\cos(k\theta)$ and $\sum_{k=1}^n\sin(k\theta)$? 
Possible Duplicate:
How can we sum up $\sin$ and $\cos$ series when the angles are in A.P? 

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!
 A: 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.
A: The answer is "yes", but here are a few more details (absolutely not original with me):
Substitute $z = \exp(i \theta)$ and $z^{n+1} = \exp(i (n+1) \theta)$, use Euler's (not his, but what the heck) formula to get quotients involving $\sin(\theta)$, $\cos(\theta)$, $\sin((n+1))\theta)$, and $\cos((n+1) \theta)$, and then separate the real and imaginary parts.
