Answer
The critical points are: $\left( {x,y} \right) = \left( {m\pi ,\frac{{\left( {2n + 1} \right)\pi }}{2}} \right)$, where $m,n = 0, \pm 1, \pm 2,...$
${\rm{type{\ } of{\ } critical{\ } point : }}\left\{ {\begin{array}{*{20}{c}}
{n{\ } {\rm{is{\ } odd,}}}&{\left\{ {\begin{array}{*{20}{c}}
{m{\ } {\rm{is{\ } odd,}}}&{\rm{local}{\ } {\rm{maximum}}}\\
{m{\ } {\rm{is{\ } even,}}}&{\rm{local}{\ } {\rm{minimum}}}
\end{array}} \right.}\\
{n{\ } {\rm{is{\ } even,}}}&{{\rm{saddle{\ } point}}}
\end{array}} \right.$
We list here some of the critical points and their nature:
$\begin{array}{*{20}{c}}
{}&{}\\
n&m\\
{ - 2}&{ - 1}\\
0&0\\
2&1\\
4&2\\
{ - 3}&1\\
{ - 1}&2\\
1&3\\
3&4
\end{array}\begin{array}{*{20}{c}}
{{\rm{Critical{\ } Point}}}\\
{\left( {m\pi ,{{\left( {2n + 1} \right)\pi }}/{2}} \right)}&{{\rm{Type}}}\\
{\left( { - \pi , - {{3\pi }}/{2}} \right)}&{{\rm{saddle{\ } point}}}\\
{\left( {0,{\pi }/{2}} \right)}&{{\rm{saddle{\ } point}}}\\
{\left( {\pi ,{{5\pi }}/{2}} \right)}&{{\rm{saddle{\ } point}}}\\
{\left( {2\pi ,{{9\pi }}/{2}} \right)}&{{\rm{saddle{\ } point}}}\\
{\left( {\pi , - {{5\pi }}/{2}} \right)}&{{\rm{local{\ } maximum}}}\\
{\left( {2\pi , - {\pi }/{2}} \right)}&{{\rm{local{\ } minimum}}}\\
{\left( {3\pi ,{{3\pi }}/{2}} \right)}&{{\rm{local{\ } maximum}}}\\
{\left( {4\pi ,{{7\pi }}/{2}} \right)}&{{\rm{local{\ } minimum}}}
\end{array}$
Work Step by Step
We have $f\left( {x,y} \right) = \sin \left( {x + y} \right) - \cos x$.
The partial derivatives are
${f_x} = \cos \left( {x + y} \right) + \sin x$, ${\ \ }$ ${f_y} = \cos \left( {x + y} \right)$
${f_{xx}} = - \sin \left( {x + y} \right) + \cos x$
${f_{yy}} = - \sin \left( {x + y} \right)$
${f_{xy}} = - \sin \left( {x + y} \right)$
To find the critical points we solve the equations ${f_x} = 0$ and ${f_y} = 0$:
${f_x} = \cos \left( {x + y} \right) + \sin x = 0$, ${\ \ }$ ${f_y} = \cos \left( {x + y} \right) = 0$
The two equations imply that $\sin x = 0$. The solutions are $x = m\pi $, where $m = 0, \pm 1, \pm 2,...$
The solutions for the second equation are $x + y = \frac{{\left( {2n + 1} \right)\pi }}{2}$, where $n = 0, \pm 1, \pm 2,...$
Since $x = m\pi $, so $y = \frac{{\left( {2n + 1} \right)\pi }}{2} - m\pi $.
Thus, the critical points are: $\left( {x,y} \right) = \left( {m\pi ,\frac{{\left( {2n + 1} \right)\pi }}{2}} \right)$, where $m,n = 0, \pm 1, \pm 2,...$
Recall from previous results:
${f_{xx}} = - \sin \left( {x + y} \right) + \cos x$
${f_{yy}} = - \sin \left( {x + y} \right)$
${f_{xy}} = - \sin \left( {x + y} \right)$
So, the discriminant $D$ is
$D = {f_{xx}}{f_{yy}} - {f_{xy}}^2$
$D = - \sin \left( {x + y} \right)\left( { - \sin \left( {x + y} \right) + \cos x} \right) - {\sin ^2}\left( {x + y} \right)$
$D = - \cos x\sin \left( {x + y} \right)$
Since $\left( {x,y} \right) = \left( {m\pi ,\frac{{\left( {2n + 1} \right)\pi }}{2}} \right)$, for $m,n = 0, \pm 1, \pm 2,...$
$D = - \cos m\pi \sin \left( {m\pi + \frac{{\left( {2n + 1} \right)\pi }}{2}} \right)$
$ = - \cos m\pi \sin m\pi \cos \frac{{\left( {2n + 1} \right)\pi }}{2} - {\cos ^2}m\pi \sin \frac{{\left( {2n + 1} \right)\pi }}{2}$
Since $\sin m\pi = 0$, we get
$D = - {\cos ^2}m\pi \sin \frac{{\left( {2n + 1} \right)\pi }}{2}$
$D = \left\{ {\begin{array}{*{20}{c}}
{ + 1,}&{{\rm{n{\ } is{\ } odd}}}\\
{ - 1,}&{{\rm{n{\ } is{\ } even}}}
\end{array}} \right.$
Case 1. $n$ is even
Since $D < 0$, the critical point is a saddle point.
Case 2. $n$ is odd
For $n$ is odd, $D > 0$. However, we still need to examine ${f_{xx}}$.
${f_{xx}} = - \sin \left( {x + y} \right) + \cos x$
${f_{xx}} = - \sin \left( {m\pi + \frac{{\left( {2n + 1} \right)\pi }}{2}} \right) + \cos m\pi $
$ = - \sin m\pi \cos \frac{{\left( {2n + 1} \right)\pi }}{2} - \cos m\pi \sin \frac{{\left( {2n + 1} \right)\pi }}{2} + \cos m\pi $
Since $\sin m\pi = 0$, we get
${f_{xx}} = \left( {1 - \sin \frac{{\left( {2n + 1} \right)\pi }}{2}} \right)\cos m\pi $
Since $n$ is odd, $\sin \frac{{\left( {2n + 1} \right)\pi }}{2} = - 1$, thus ${f_{xx}} = 2\cos m\pi $.
${f_{xx}} = \left\{ {\begin{array}{*{20}{c}}
{ - 2,}&{{\rm{m{\ } is{\ } odd}}}\\
{2,}&{{\rm{m{\ } is{\ } even}}}
\end{array}} \right.$
From these results we conclude that
${\rm{type{\ } of{\ } critical{\ } point : }}\left\{ {\begin{array}{*{20}{c}}
{n{\ } {\rm{is{\ } odd,}}}&{\left\{ {\begin{array}{*{20}{c}}
{m{\ } {\rm{is{\ } odd,}}}&{\rm{local}{\ } {\rm{maximum}}}\\
{m{\ } {\rm{is{\ } even,}}}&{\rm{local}{\ } {\rm{minimum}}}
\end{array}} \right.}\\
{n{\ } {\rm{is{\ } even,}}}&{{\rm{saddle{\ } point}}}
\end{array}} \right.$
Using the Second Derivative Test we list some critical points in the following table:
$\begin{array}{*{20}{c}}
{}&{}&{{\rm{Critical{\ } Point}}}\\
n&m&{\left( {m\pi ,{{\left( {2n + 1} \right)\pi }}/{2}} \right)}\\
{ - 2}&{ - 1}&{\left( { - \pi , - {{3\pi }}/{2}} \right)}\\
0&0&{\left( {0,{\pi }/{2}} \right)}\\
2&1&{\left( {\pi ,{{5\pi }}/{2}} \right)}\\
4&2&{\left( {2\pi ,{{9\pi }}/{2}} \right)}\\
{ - 3}&1&{\left( {\pi , - {{5\pi }}/{2}} \right)}\\
{ - 1}&2&{\left( {2\pi , - {\pi }/{2}} \right)}\\
1&3&{\left( {3\pi ,{{3\pi }}/{2}} \right)}\\
3&4&{\left( {4\pi ,{{7\pi }}/{2}} \right)}
\end{array}\begin{array}{*{20}{c}}
{}&{}&{}\\
{{f_{xx}}}&{{f_{yy}}}&{{f_{xy}}}\\
0&1&1\\
0&{ - 1}&{ - 1}\\
0&1&1\\
0&{ - 1}&{ - 1}\\
{ - 2}&{ - 1}&{ - 1}\\
2&1&1\\
{ - 2}&{ - 1}&{ - 1}\\
2&1&1
\end{array}\begin{array}{*{20}{c}}
{{\rm{Discriminant}}}&{}\\
{D = {f_{xx}}{f_{yy}} - {f_{xy}}^2}&{{\rm{Type}}}\\
{ - 1}&{{\rm{saddle{\ } point}}}\\
{ - 1}&{{\rm{saddle{\ } point}}}\\
{ - 1}&{{\rm{saddle{\ } point}}}\\
{ - 1}&{{\rm{saddle{\ } point}}}\\
1&{{\rm{local{\ } maximum}}}\\
1&{{\rm{local{\ } minimum}}}\\
1&{{\rm{local{\ } maximum}}}\\
1&{{\rm{local{\ } minimum}}}
\end{array}$
