What is the use of hyperbolic trigonometric functions if they are easily expressible algebraically? I get that there are uses for $\sin(x)$ and $\cos(x)$ because they are defined with imaginary exponents which aren't as easily worked with but the hyperbolic functions are simply $\frac12(e^x\pm e^{-x})$. I don't see why it would be necessary to define a function to represent this. It is almost like defining a function to represent the value of $2$.
The only other time I can think of where there is a "useless" function is for pedagogical purposes such as the identity function.
 A: In the same way that the point $(\cos\theta,\sin\theta)$ is on the circle, the point $(\cosh\theta, \sinh\theta)$ is on a hyperbola (hence the hyperbolic part). They show up all the time in solutions to the heat equation, and a hanging rope is actually a hyperbolic cosine function. Many many reasons why we would want to have a notation for it.
A: $\cos(\theta)$ and $\sin(\theta)$ are essential for understanding geometry on the sphere, as ocean navigators have known for a couple of millennia. $\cosh(t)$ and $\sinh(t)$ are similarly essential for understanding geometry on the hyperbolic plane.
A: consider $y''+y=0$ with initial conditions $y(0)=0,y'(0)=1$ and $y(0)=1, y'(0)=1$.  the solutions are $\sin x$ and $\cos x$.
similarly, $y''-y=0$ with initial conditions $y(0)=0,y'(0)=1$ and $y(0)=1, y'(0)=1$ give $\sinh x$ and $\cosh x$.
(transcendental solutions of differential equations often get names when they are used often: $e^x$, bessel functions, weierstrass $\wp$, etc.)
just as $x=\cos t, y=\sin t$ gives a unit speed parameterization of the unit circle, $x=\cosh t, y=\sinh t$ gives a 'unit speed' parameterization of the unit hyperbola.
A: But $\sin x = \frac{\mathrm{e}^{\mathrm{i}x} - \mathrm{e}^{-\mathrm{i}x}}{2\mathrm{i}}$ and $\cos x = \frac{\mathrm{e}^{\mathrm{i}x} + \mathrm{e}^{-\mathrm{i}x}}{2}$.  So we don't need sine or cosine either.
The hyperbolic functions are handy for computing sine and cosine with complex arguments... \begin{align*}
\sin(x+\mathrm{i}y) &= \sin x \cosh y + \mathrm{i} \cos x \sinh y \\
\cos(x+\mathrm{i}y) &= \cos x \cosh y - \mathrm{i} \sin x \sinh y
\end{align*}
