Talyor Poly. of Degree 2n

$\displaystyle f(x) = cosx$, $\displaystyle n \in \mathbb{N}$

Find the Taylor Poly. of $\displaystyle P_{2n}:= {P_{2n}}^{f,0}$

I know $\displaystyle x_0 = 0$ and $\displaystyle f^{(k)}(x) = cosx$ and $\displaystyle f^{(k)}(0) = cos(0) = 1$ and $\displaystyle P_{2n} = \displaystyle\sum_{k=0}^{2n} \frac{f^{(k)}(x_0)}{k!} \cdot (x-x_0)^k = f(0) + \sum_{k=1}^{2n} \frac{f^{(k)}(0)}{k!} \cdot x^k = 1+ \sum_{k=1}^{2n} \frac{x^k}{k!}$

So I am confused on what I should do from here to continue (if what I did previously was correct) because I know this is not complete but I do not know where to go from here, give me a boost please. Thank you