Critical Points of cos x + cos y + cos(x+y)

Subtract the two equations before you solve either

Both partials inherit a term from $\cos(x+y)$, and since the inside derivative is 1 in each variable, that term is identical in the two equations.

Subtracting kills the shared term and leaves a relation between $x$ and $y$ alone. That is the move worth remembering whenever a coupled system shares a piece: subtract first, substitute second.

On the branch $y = x$, the first equation becomes $\sin x + \sin 2x = 0$. Replace $\sin 2x$ with $2 \sin x \cos x$ and the whole thing factors.

That gives $x = 0$, $x = \pi$ and $x = \pm\frac{2\pi}{3}$ along the diagonal. On the other branch, $y = \pi - x$ makes $\sin(x+y) = \sin\pi = 0$, so the equation reduces to $\sin x = 0$ and produces two more saddles at $(0, \pi)$ and $(\pi, 0)$.

Three different verdicts from the same formulas

Every second partial is a cosine, so the arithmetic at each point is a matter of reading three values off the unit circle.

At the origin every cosine equals 1, so $f_{xx} = f_{yy} = -2$ and $f_{xy} = -1$. Then $D = 4 - 1 = 3 > 0$ with $f_{xx} < 0$, a local maximum. The value $f(0,0) = 3$ is the global maximum as well, since no cosine can exceed 1 and all three reach it together at the origin. Every translate by $2\pi$ in either variable does the same thing, so the global maximum is attained at infinitely many points, all at height 3.

At $\left(\frac{2\pi}{3}, \frac{2\pi}{3}\right)$ we have $\cos\frac{2\pi}{3} = -\frac{1}{2}$ and $\cos\frac{4\pi}{3} = -\frac{1}{2}$, so $f_{xx} = f_{yy} = 1$ and $f_{xy} = \frac{1}{2}$.

That is a local minimum with $f = -\frac{3}{2}$. At $(\pi, \pi)$, $\cos\pi = -1$ and $\cos 2\pi = 1$, so $f_{xx} = f_{yy} = 0$ while $f_{xy} = -1$, and $D = -1$: a saddle.

The mistake students make

Seeing $f_{xx} = 0$ and $f_{yy} = 0$ at $(\pi, \pi)$, students often report the test as inconclusive. Inconclusive means $D = 0$, not $f_{xx} = 0$. When both diagonal entries vanish, $D = -f_{xy}^{2}$, which is negative unless the mixed partial is also zero. Here $f_{xy} = -1$, so $D = -1$ and the answer is a saddle with no ambiguity at all.

The other slip is stopping after the diagonal branch. Subtracting the partials gives $\sin x = \sin y$, and sine is not one to one: $y = \pi - x$ solves it just as well as $y = x$. Dropping that branch loses the saddles at $(0, \pi)$ and $(\pi, 0)$ completely. Those two are easy to confirm by substitution: at $(0, \pi)$, $f_x = -\sin 0 - \sin\pi = 0$ and $f_y = -\sin\pi - \sin\pi = 0$, so it is a genuine critical point that the diagonal branch never produces.

• Subtract the two equations to get $\sin x = \sin y$.
• Keep both branches, $y = x$ and $y = \pi - x$.
• Substitute each branch back into one of the original equations.
• Add $2\pi$ to either variable to generate every remaining copy.