Calculating $\int_{\partial B_8(1)} \frac{1}{1 - \cos(z)} \ dz$

Question

Using the residue theorem we know that $$\int_{\partial B_8(1)}f(z) \ dz = 2\pi i \sum_{z_0 \text{ pole of } f \text{ inside } B_8(1)} \mathrm{res}_{z_0}(f)$$ where $f(z) = \frac{1}{1- \cos(z)}$. I tried calculating the residue for z = 0 by using taylor expansion $$\frac{1}{1-\cos(z)} = \frac{1}{1 - \sum_{n = 0}^{\infty} \frac{(-1)^n z^{2n}}{(2n)!}} = \frac{1}{ - \sum_{n = 1}^{\infty} \frac{(-1)^n z^{2n}}{(2n)!}}$$ but got stuck here.