<s>[INST] Find the extrema of $\sin(x)\cos(y)$ with the Hessian I've got the following function: $f(x,y) = \sin(x)\cos(y)$.
I performed the Hessian matrix correctly, with the second derivatives:
\begin{bmatrix}
-\sin(x)\cos(y) & -\cos(x)\sin(y)\\
-\cos(x)\sin(y) & -\sin(x)\cos(y)
\end{bmatrix}
But the trouble comes when I want to find the extrema (critical points and saddle points). By solving $\cos(x)\cos(y) = 0$ and $-\sin(x)\sin(y) = 0$ (the first partial derivatives) I obtain:

$\cos(x) = 0$ or $\cos(y) = 0$
$\sin(x) = 0$ or $\sin(y) = 0$.

If $\cos(x) = 0$, then $\sin(x) = \pm 1$ and if $\sin(x) = 0$ then $\cos(x) = \pm1$. Well, here I think that we have got just two possibilities:


*

*$\cos(x) = \sin(y) = 0$;

*$\sin(x) = \cos(y) = 0$.
Now, how do I find the extrema?
I mean, with $\cos(x) = \sin(y) = 0$, which is the point $(x, y)$ and how do I substitute it in the Hessian matrix? (The solution says it must a saddle point but I don't get it, why?)
 [/INST] Find the extrema of $f(x,y) = \sin(x)\cos(y)$.
$$\begin{cases} f_x=\cos x\cos y=0 \\ f_y=-\sin x\sin y=0\end{cases} \Rightarrow 1)\begin{cases}\cos x=0 \\ \sin y=0\end{cases} \ \ \text{or} \ \ 2)\begin{cases}\cos y=0 \\ \sin x=0\end{cases}.$$
The Hessian is:
$$H=\begin{bmatrix}
-\sin x \cos y & -\cos x \sin y \\
-\cos x \sin y & -\sin x \cos y 
\end{bmatrix}\\
H_1=-\sin x\cos y; \ \ \ H_2=\sin^2 x\cos^2y-\cos^2x\sin^2y.$$
For $1)$: 
$$a) \begin{cases}\cos x=0 \\ \sin y=0 \\ \sin x=1 \\ \cos y=1\end{cases} \Rightarrow H_1=-1<0, H_2=1>0 \Rightarrow max; \\ 
b) \begin{cases}\cos x=0 \\ \sin y=0 \\ \sin x=-1 \\ \cos y=1\end{cases} \Rightarrow H_1=1>0, H_2=1>0 \Rightarrow min; \\
c) \begin{cases}\cos x=0 \\ \sin y=0 \\ \sin x=1 \\ \cos y=-1\end{cases} \Rightarrow H_1=1>0, H_2=1>0 \Rightarrow min; \\
d) \begin{cases}\cos x=0 \\ \sin y=0 \\ \sin x=-1 \\ \cos y=-1\end{cases} \Rightarrow H_1=-1<0, H_2=1>0 \Rightarrow max. \\$$
Can you handle $2)$?
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