Why does $\cos(x) + \cos(y) - \cos(x + y) = 0$ look like an ellipse?

The solution set of $\cos(x) + \cos(y) - \cos(x + y) = 0$ looks like an ellipse. Is it actually an ellipse, and if so, is there a way of writing down its equation (without any trig functions)?

What motivates this is the following example. The solution set of $\cos(x) - \cos(3x + 2y) = 0$ looks like two straight lines, and indeed we can determine the equations of those lines.

\begin{align} \cos(x) &= \cos(3x + 2y) \\ \implies x &= \pm (3x + 2y) \\ \implies x + y &= 0 \text{ or } 2x + y = 0 \end{align}

Can we do a similar thing for the first equation?

• Completely unrelated, but that is one of the coolest graphs I've ever seen – imulsion Jun 23 '15 at 19:32
• $cos(x)+cos(y)$ may be transormed into multiplication, using the identity. – hyperkahler Jun 23 '15 at 19:33
• An obvious approach would be to expand $\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y)$, then rewrite $\sin$ in terms of $\cos$ using $\cos^2+\sin^2=1$ and substituting $t=\cos x, s=\cos y$. This should transform this into a degree $4$ polynomial in $t, s$, which might turn out to be an ellipse equation squared with some luck.. just a guess. – Damian Reding Jun 23 '15 at 19:37