Sum of a complex, finite geometric series and its identity

I have the formula for summing a finite geometric series as $$1+z+z^2\cdots +z^n = \frac{1-z^{n+1}}{1-z},$$ where $z\in\mathbb{C}$ and $n=0,1,...$. I am asked to infer the identity $$1+\cos\theta+\cos 2\theta+\cdots+\cos n\theta = \frac{1}{2}+\frac{\sin(n+1/2)\theta}{2\sin\theta /2}.$$ Now, I understand that on the left hand side I'm going to get $$1+\cos\theta +\cdots + \cos n\theta + i[\sin\theta + \sin 2\theta +\cdots + \sin n\theta]$$
using $z=e^{i\theta}$ for any complex $z$. However, when I make that substitution on the right hand side, a monstrous expression occurs and I cannot simplify it down to the desired result. For instance, I get $$\frac{1-e^{i(n+1)\theta}}{1-e^{i\theta}}$$ and using identities I get $$\frac{1-\cos [(n+1)\theta]+i(\sin[(n+1)\theta])}{1-\cos n\theta -i\sin n\theta}.$$ From here I did a whole lot of manipulating, but never getting any closer to the identity asked.

If anyone could shed some light it would be greatly appreciated!
~Dom

Answer

Hint: factor out $e^{i (n+1) \theta/2}$ from the numerator and $e^{i \theta/2}$ from the denominator, then take the real part of the complex expression.