sequences and series - Evaluating $sum_{k=0}^inftysin(kx)$ and $sum_{k=0}^inftycos(kx)$

Playing with sines I wanted to find
$$
S(x)=\sum_{k=0}^\infty\sin(kx)
$$

Writing it as
$$
S(x)=\mathrm{Im}\big(A(x)\big),\quad\text{where}\quad A(x)=\sum_{k=0}^\infty e^{ikx}
$$

and using $z=e^{ix}=\cos(x)+i\sin(x)$, cheating about $\|z\|=1$ in order to write
$$
A(x)=\frac{1}{1-z},
$$

I found that

$$
2S(x)=\frac{\sin(x)}{1-\cos(x)}

$$

as WolframAlpha does.

My question is: why with the same method I found
$$
\sum_{k=0}^\infty\cos(kx)=\frac{1}{2}
$$

while WolframAlpha and G.H.Hardy on the book "Divergent Series" (pg. 2) give $-1/2$?