Why do hyperbolic "trig" functions seem to be encountered rarely?
I can think of two reasons.
When we do geometry, we usually work in Euclidean space, where the intrinsic property of a line segment between two points is its length, given (in two dimensions) by $\ell^2 = \Delta x^2 + \Delta y^2$. We are allowed to change our reference frame as long as we preserve lengths, which means that the transformation of $(\Delta x,\Delta y)$ is a rotation, and the transformed vector must lie on the circle $\Delta x^2 + \Delta y^2 = \operatorname{const}$, which is naturally parametrized by $\sin$ and $\cos$. The hyperbolic variants don't have much to do with circles or with rotation, so are not relevant here (unless you supply them with imaginary arguments, at which point you're really working just with $\sin$ and $\cos$ in disguise).
On the other hand, the natural setting for special relativity is Minkowski space, where the invariant property of the interval between two points in spacetime is of the form $s^2=\Delta x^2-\Delta t^2$, with a minus sign. Here the allowed changes of reference frame are given by Lorentz transformations, the transformed interval lies on $\Delta x^2-\Delta t^2=\operatorname{const}$ which is a hyperbola, and indeed one finds hyperbolic trigonometric functions to be quite useful in special relativity.
The usual trigonometric functions $\sin$ and $\cos$ are two real solutions to the differential equation $y'' = -y$, which describes a simple harmonic oscillator (a conservative system in a parabolic potential well) which is a central example in much of classical physics. The hyperbolic versions $\sinh$ and $\cosh$ are solutions to $y'' = y$ instead. This equation is rarely useful to model physical systems in real life because all its solutions are unbounded and gain infinite amounts of kinetic energy. Even taken locally, the equation describes an unstable equilibrium, so any real system will not spend most of its time there without additional forcing.
I'd say because hyperbolic functions can be written pretty easily in terms of exponential functions. By that I mean you don't need $i$ like when you express $\sin$ and $\cos$ using $\exp$. That means in the "real" world, it's not necessary to use $\sinh$ and $\cosh$ because you can always resort to $\exp$, but the same is not true for $\sin$ and $\cos$. However, I wouldn't say hyperbolic functions are rarely encountered.
Very often I find myself choosing hyperbolic functions over exponential functions in the context of ODEs and PDEs with boundary/initial conditions at $0$. (I believe sophomores should have taken/be taking these classes, no?) The fact that $\sinh(0) = \cosh'(0) = 0$ and $\cosh(0) = \sinh'(0) = 1$ makes your expression much cleaner. Also, you get $\sinh(ax)$ and $\cosh(ax)$ as two solutions to the ODE $y'' - a^2y = 0$ immediately, similar to the way you get $\sin(ax)$ and $\cos(ax)$ from $y'' + a^2y = 0$. Variation of parameters also gives $\sinh$ as a kernel when you solve the non-homogeneous ODE with Dirichlet boundary conditions. There are just so many things you can express cleanly in terms of hyperbolic functions.
Oh, and don't forget that $\tanh$ is a really nice bijection from $\mathbb R$ to $(-1, 1)$. Kind of funny that $\tanh$ looks like $\frac 2\pi \arctan$.
$\sinh$ and $\cosh$ seem to appear less then their circular counterparts in real analysis as is explained well in some other answers. However, hyperbolic functions appear quite commonly in complex analysis. From Euler's formula and its subsequent trigonometric definitions, one finds that
$$\cos (ix) = \cosh x$$ $$\sinh (ix)=i\sin x$$
which is a clear connection between hyperbolic and circular trigonometric functions.
One use of hyperbolic functions, that I have personally used, is in integration. $$\cosh^2 x - \sinh^2 x=1$$ This identity can often be used for substitutions to evaluate integrals with $x^2+1$ and $x^2-1$ (instead of $\sec$ and $\tan$), just as the identity $\cos^2 x+\sin^2 x=1$ can be used for substitutions to evaluate integrals with $1-x^2$.
For example, to evaluate
$$\int \frac{dx}{\sqrt{x^2+1}}$$
we may substitute $x = \sinh u \implies dx = \cosh u \, du$ so the integral becomes
$$\int \frac{\cosh u}{\sqrt{\sinh^2 u+1}}\, du = \int 1 \, du= u +C = \operatorname{arsinh} x + C$$
And, of course, the hyperbolic functions provide a parametrization of the standard hyperbola.