Indeed there are, but they're not usually called that. What ordinary trigonometric and hyperbolic functions have in common is that they are solutions to the differential equation $$f''(t) = af(t)$$
When $a$ is negative, the solutions are ordinary sines and cosines, scaled horizontally by a factor that depends on $a$. If you take a solution $f$ and draw the parametric plot $(x,y)=(f'(t), f(t))$, the result is an ellipse whose eccentricity depends on $a$. For $a=-1$ the ordinary sine and cosine are solutions, and you get a circle.
On the other hand, when $a$ is positive the solutions are either hyperbolic sines or hyperbolic cosines, again with a horizontal scaling factor that depends on $a$. A plot of $(x,y)=(f'(t), f(t))$ is one arm of a hyperbola with a central angle that depends on $a$. For $a=1$ the hyperbolic sine and cosine are solutions, and the hyperbola is right-angled.
Intuitively, then, since a parabola is the limiting case between an ellipse and a hyperbola, we should expect to get a "parabolic function" by setting $a=0$. Unfortunately the differential equation then becomes
$$f''(t)=0$$
whose solutions are first-degree polynomials, and it is hard to make those create a parabola. However, there's a way out (many thanks to Qiaochu Yuan for pointing this out!): Instead of $f''(t)=af(t)$ we can take the basic differential equation to be
$$f'''(t)=af'(t)$$
In the $a\ne 0$ case all this changes is to allow us to add a constant term to solutions, which just moves the conic about in the plane. But for $a=0$, the solutions are now all the polynomials of degree $\le 2$. And when we take any quadratic polynomial $f$ and plot $(x,y)=(f'(t), f(t))$, what we get is indeed a parabola centered around the $y$-axis!
If we take $f$ to be a first-degree polynomial, the parametric plot is just a straight (vertical) line, another limiting case of conic sections.
In all of the above cases, plotting $(f_1(t),f_2(t))$ for two unrelated solutions (for the same $a$) generally produces a conic of the same general kind, but perhaps moved and rotated. And the dependency on $a$ of the eccentricity/angle disappears; that was mediated through the derivative in the $x$ position.
So a "parabolic function" is simply another (redundant) term for a quadratic polynomial. It is not quite clear which should be counted as the parabolic sine and cosine, though. Cases could be made for either $\operatorname{sinp}(t) = t$ and $\operatorname{cosp}(t) = 1+\frac12 t^2$ or the other way around -- but worrying too much about that is just silly.
I'll prove $\sinh(x)$ = $\frac {e^x - e^{-x}} {2}$ and leave the proof of $\cosh(x)$ as an exercise for you. So, $\sinh(x) = x + \frac{x^3}{3!} + \frac{x^5}{5!} + ...+\infty$
And we know that, $$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ...$$
And we also know that $$e^{-x} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + ...$$
Therefore, $$e^x - e^{-x} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ... - (1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + ...)$$
Now, after we open the brackets, the negative terms will become positive and positive terms will become negative, so let's just do that
$$1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ... - 1 + x - \frac{x^2}{2!} + \frac{x^3}{3!} - ...$$
And we are left with : $$2x + 2\frac{x^3}{3!} + 2\frac{x^5}{5!} + ...$$ and we'll take the $2$ common so we get,
$$2(x + \frac{x^3}{3!} + \frac{x^5}{5!} + ...)$$
And, we saw earlier that $$\sinh(x) = x + \frac{x^3}{3!} + \frac{x^5}{5!} + ...+\infty$$.
So we get that $$e^x - e^{-x} = 2 \times \sinh(x)$$ and hence we get that $$\sinh(x) = \frac {e^x - e^{-x}}{2}$$
$Q.E.D$
Hope it helps
Best Answer
Geometrical meaning is shown in the rough hand sketch:
Hyperbolic angle magnitude is the non-dimensional yellow area of its hyperbolic sector marked $A$ divided by $a^2$. This is the argument of hyperbolic functions.
To get $x (y)$ co-ordinates we need to divide the area $A$ by square of axis $a,$ take its $\cosh (\sinh) $ and multiply by $a$.
EDIT1:
( This is continuation of my comments, posted here in the answer area for purpose of discussion only on OP's request.)
Why I propose using the dimensioned form:
$$ x = a \cosh \dfrac{A}{a^2};\;y = a \sinh \dfrac{A}{a^2}$$
in preference to (present day text-book model)
$$ x = \cosh A ;\;y = \sinh A \;?$$
First of all it is noted the "hyperbolic angle" is related to central polar euclidean angle $θ_e$ that the radius vector makes angle to x-axis. The hyperbolic angle is not included between any two lines.
$$ \theta_h= A/a^2 = \frac12\; \log (\tan(π/4+θ_e)):$$
This area relation is found by direct integration.
It is difficult for me to justify full cross-validity without going into definitions of lengths and hence the dependent hyperbolic angles in any one model of the hyperbolic geometries..
My basic reasoning about its research aspect is that:
(1) Similar relations between euclidean/hyperbolic geometry should hold good for expressing cartesian coordinates in the plane using circular Trig functions and the hyperbolic functions. Subscripts $e/h$ for euclidean/hyperbolic are used as:
$$ x=a \cos θ_e,y=a \cosθ_e ;\; x=a \cosh θ_h,y=a \sinh θ_h ; $$
in either of parametrizations for:
$$ x^2 \pm y^2 = a^2 $$
(2) The sheer physical dimensional tally. I felt quite ill at ease to see $ x = \cosh A , y= \sinh A. $
(3) We cannot directly $ \cos i \theta_e \rightarrow \cosh \theta_h $ and hope to manipulate/accommodate with area interpretation.