Taylor remainder of $f(x,y)=\sin x\cdot \cos y$ Given $f\colon \mathbb R^2\rightarrow\mathbb R,(x,y)\mapsto\sin x\cdot\cos y$ I want to show that there exists $M>0$ such that $$|f(x,y)-T_2(x,y)|\leq M(|x|+|y|)$$ for all $(x,y)\in\mathbb R^2$. $T_2$ is the taylor-polynomial of order $2$.
I know I have to consider the remainder of $T_2$. So I calculated all the third partial derivatives:
\begin{align*}
\frac{\partial f^3}{\partial x^3}(x,y)&=-\cos(x)\cos(y)\\
\frac{\partial f^3}{\partial y^3}(x,y)&=\sin(x)\sin(y)\\
\frac{\partial f^3}{\partial y^2\partial x}(x,y)&=\frac{\partial f^3}{\partial x\partial y^2}(x,y)=\frac{\partial f^3}{\partial y\partial x\partial y}(x,y)=-\cos(x)\cos(y)\\
\frac{\partial f^3}{\partial x^2\partial y}(x,y)&=\frac{\partial f^3}{\partial x\partial y\partial x}(x,y)=\frac{\partial f^3}{\partial y\partial x^2}(x,y)=\sin(x)\sin(y)
\end{align*}
But now I am really stuck. I looked up the formula for the remainder on wikipedia but I never get the term $|x|+|y|$. So how can you show that $|R_3(x,y)|\leq M(|x|+|y|)$?

First compute $T_2(x,y)$ which is given by

$$ T_2(x,y) = x. $$

Now, we have
$$ \Big| f(x,y)-T_2(x,y)\Big|= \Big| \sin(x)\cos(y)-x\Big|= \Big| \sin(x)\cos(y)-\sin(x)+\sin(x)-x\Big|$$
$$ \leq |\sin(x)(\cos(y)-1)|+|\sin(x)|+|x| $$
$$ =|(\cos(y)-1)|+2|x| $$
$$ \leq |y|+2|x|  $$
$$ \leq 2(|x|+|y|). $$
Note: The following inequalities 
$$ |\sin(x)|\leq |x|,\quad |\cos(y)-1|\leq |y| $$
can be proved using the mean value theorem. 
