Show that $1+\cos\theta+\cos(2\theta)+\dots+\cos(n\theta)=\frac{1}{2}+\frac{\sin((n+\frac{1}{2})\theta)}{2\sin(\theta\mathbin{/}2)}$

Question

I have shown that $1+\cos(\theta)+\cos(2\theta)+\cos(3\theta)+\dots+\cos(n\theta)=\Re(\frac{1-e^{(n+1)i\theta}}{1-e^{i\theta}})=\frac{1-\cos(\theta)+\cos(n\theta)-\cos((n+1)\theta)}{2-2\cos(\theta)}=\frac{1}{2}+\frac{\cos(n\theta)-\cos((n+1)\theta)}{2-2\cos(\theta)}$

but I am unsure how to proceed from here. If possible, I would like to avoid using to many identities as this is an exercise in my complex analysis book.