Show that $$\frac{1}{2}+\sum_{k=1}^n \cos(kx)=\frac{\sin(nx+\frac{x}{2})}{2\sin\frac{x}{2}}$$
I try the next $$ \frac{1}{2}+\sum_{k=1}^n \cos(kx) = \frac{1}{2}+\Re \sum_{k=1}^n e^{ikx} = \frac{1}{2}+\Re \left[\frac{1-e^{i(n+1)x}}{{1-e^{ix}}}-1\right] $$ But i not sure to how take to $$\Re\left[\frac{1-e^{i(n+1)x}}{{1-e^{ix}}}-1\right]$$ Any hint or help i will be very grateful
