Smooth parametrization of curve $\cosh x+\cosh y = \operatorname{constant}$ I have a curve (image below)
$$
\cosh x+\cosh y = C,\qquad C>2.
$$
I would like to get its smooth parametrization of form
$$
x = f(t),\qquad y=g(t),\qquad t\in[a,b],
$$
so for every point on the curve there is a corresponding parameter $t$. (In the same manner, as for curve $x^2+y^2=1$, there is a smooth parametrization $x=\cos t$, $y = \sin t$).
I would appreciate any help. Thanks!

 A: Write your equation in the form $(\cosh x-1) +(\cosh y-1)=C-2$, hence
$$\sinh^2{x\over2}+\sinh^2 {y\over2}={C-2\over2}=:\rho^2\ .$$
It follows that a smoth parametrization of your curve is given by
$$\left.\eqalign{x(t)&=2\,{\rm arsinh}(\rho\cos t)\cr y(t)&=2\,{\rm arsinh}(\rho\sin t)\cr}\right\}\qquad(0\leq t\leq 2\pi)\ .$$
A: This solution is less elementary than those in the other answers, but it points out that this question is connected to some rich, classical mathematics in a perhaps nonobvious way and that this theory leads to a reasonably natural parameterization.
Changing coordinates to $x = \log s, y = \log t$ and clearing denominators, the curve is equivalent to
$$r^2 s + r s^2 - 2 C r s + r + s = 0 .$$
If we parameterize this curve, we can always parameterize the original curve with the inverse coordinate change, $x = \log r, y = \log s$.
This curve and cubic and has discriminant a nonzero multiple of $(C + 2) C^2 (C - 2)$, which is nonvanishing in the cases $C > 2$ under consideration. So, this is an elliptic curve, and we can put it in Weierstrass normal form [pdf warning]: In the coordinates $(u, v)$ defined by
$$r = \frac{3(-6 C u + 3 v + 2 C^3 - 4 C)}{2 (3 u - C^2 - 1) (3 u - C^2 + 2)}, \qquad s = \frac{3(-6 C u - 3 v + 2 C^3 - 4 C)}{2 (3 u - C^2 - 1) (3 u - C^2 + 2)} ,$$
the curve is
$$v^2 = 4 u^3 - \color{red}{\frac{4}{3}(C^4 - 4 C^2 + 1)} u - \color{blue}{\left[-\frac{4}{27}(2 C^4 - 8 C^2 + 1)(C^2 - 2)\right]}.$$
On the other hand, a Weierstrass elliptic function $\wp(z)$ satisfies the differential equation
$$\wp'(z)^2 = 4 \wp(z)^3 - \color{red}{g_2} \wp(z) - \color{blue}{g_3},$$ for some $g_2, g_3$, and we can choose an elliptic function $\wp(z)$ such that $g_2, g_3$ respectively match the corresponding coefficients of the previous display equation, so for that $\wp(z)$ we can parameterize the equation in $u, v$ by
$$(u, v) = (\wp(t), \wp'(t)) .$$
Conversely, we can ask which real elliptic curves arise this way: The $j$-invariant of this curve is
$$\frac{1728 \color{red}{g}_{\color{red}{2}}^3}{\color{red}{g}_{\color{red}{2}}^3 - 27 \color{blue}{g}_{\color{blue}{3}}^2} = \frac{256 (C^4 - 4 C^2 + 1)^3}{(C + 2) C^2 (C - 2)},$$
which achieves its minimum of $j = 1$ on $(2, \infty)$ at $C = \frac{1}{2} \sqrt{8+6 \sqrt{2}} = 2.0301\!\ldots$ and is unbounded as $C \to \infty$. So, up to complexification and isomorphism every real elliptic curve with $j$-invariant $j \geq 1$ arises this way.
A: Here's something more symmetric. I'll write the steps if anyone wants to modify the idea to get something else.
First, we can remove the pesky hyperbolic functions simply by letting $x = \ln t,\ y = \ln s$ so $$\cosh x + \cosh y = c \iff t + \frac 1t + s + \frac 1s = 2c.$$
Now, we'd like to "invert" $z\mapsto z + 1/z$. More precisely, we want $t + 1/t = 2u^2+2$ and $s + 1/s = 2v^2 + 2$, so the whole thing becomes $ u^2 + v^2 = c - 2$ which we know how to parametrize.
Solving the appropriate quadratics, we get $$t = 1 + u^2\pm \sqrt{2u^2+u^4},\ s = 1+v^2 \pm \sqrt{2v^2 + v^4}.$$ Noting that we want $ u^2 + v^2 = c - 2$, we further choose $u = \sqrt {c-2} \cos\varphi$ and $v = \sqrt {c-2}\sin\varphi$.
All in all, $$x = \ln \left( 1 + (c-2)\cos^2\varphi \pm \sqrt{2(c-2)\cos^2\varphi + (c-2)^2\cos^4\varphi} \right),\\
y = \ln \left( 1 + (c-2)\sin^2\varphi \pm \sqrt{2(c-2)\sin^2\varphi + (c-2)^2\sin^4\varphi} \right).
$$
Now you must be wondering what to do with these $\pm$ signs. Well, if you choose one of the four possibilities, you will get the part of the curve in the corresponding quadrant. So, to deal with that:
$$x = \ln \left( 1 + (c-2)\cos^2\varphi + \cos\varphi \sqrt{2(c-2) + (c-2)^2\cos^2\varphi} \right),\\
y = \ln \left( 1 + (c-2)\sin^2\varphi + \sin\varphi \sqrt{2(c-2) + (c-2)^2\sin^2\varphi} \right),
$$ and we are finally done.
A: I think this is rather terrible, but it works:
Define
$$
\alpha=\cosh^{-1}(C-1)
$$
and use this as follows:
$$
x(t)=\alpha\cdot\cos(t),\qquad t\in[0,2\pi]
$$
and then
$$
y(t)=\operatorname{sign}(\sin(t))\cdot \cosh^{-1}(C-\cosh(x(t)))
$$
