I want to calculate the Taylor polynomial $T_{m;x_0}$ of order $m$ at $x_0=\left (\pi, \frac{\pi}{2}\right )$ for the function $f:\mathbb{R}^2\rightarrow \mathbb{R}$ defined by $f(x)=\cos (x_1+x_2)$.
We have the following:
\begin{align*}&f\left (\pi, \frac{\pi}{2}\right )=\cos \left (\frac{3\pi}{2}\right )=0\\ &\frac{\partial{f}}{\partial{x_1}}=\frac{\partial{f}}{\partial{x_2}}=-\sin (x_1+x_2) \Rightarrow \frac{\partial{f}}{\partial{x_1}}\left (\pi, \frac{\pi}{2}\right )=\frac{\partial{f}}{\partial{x_2}}\left (\pi, \frac{\pi}{2}\right )=1\\ &\frac{\partial^2{f}}{\partial{x_1}^2}=\frac{\partial{f}}{\partial{x_2}^2}=\frac{\partial^2{f}}{\partial{x_1x_2}}=-\cos (x_1+x_2) \Rightarrow \frac{\partial^2{f}}{\partial{x_1}^2}\left (\pi, \frac{\pi}{2}\right )=\frac{\partial{f}}{\partial{x_2}^2}\left (\pi, \frac{\pi}{2}\right )=\frac{\partial^2{f}}{\partial{x_1x_2}}\left (\pi, \frac{\pi}{2}\right )=0\\& \frac{\partial^3{f}}{\partial{x_1}^3}=\frac{\partial{f}}{\partial{x_2}^3}=\sin (x_1+x_2) \Rightarrow \frac{\partial^3{f}}{\partial{x_1}^3}\left (\pi, \frac{\pi}{2}\right )=\frac{\partial{f}}{\partial{x_2}^3}\left (\pi, \frac{\pi}{2}\right )=-1 \\ & \frac{\partial^4{f}}{\partial{x_1}^4}=\frac{\partial{f}}{\partial{x_2}^4}=\cos (x_1+x_2)\Rightarrow \frac{\partial^4{f}}{\partial{x_1}^4}\left (\pi, \frac{\pi}{2}\right )=\frac{\partial{f}}{\partial{x_2}^4}\left (\pi, \frac{\pi}{2}\right )=0\end{align*} etc.
At the even derivatives are equal to $0$ and the odd ones are $1$ or $-1$.
How can we use that fact to determine the Taylor polynomial of order $m$?
The general formula is $$T_{m;x_0}=\sum_{k\leq m}\frac{D^kf(x_0)\cdot (x-a_0)^k}{k!}$$
different trick: Use $\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y)$ and $\cos(\pi/2)=\sin(\pi)=0$