Why is $\sinh(45°)$ not infinity? How does it ever intersect with the hyperbola, seeing as it goes along the asymptote? From what I know, the hyperbolic trigonometric functions are almost the same as the circular trigonometric functions ($\sin, \cos, \tan$, et cetera without the $h$ suffix), except they output when a line coming from the centre at the given angle hits the surface of a hyperbola, rather than a circle.
From what I've seen, the relevant hyperbola has asymptotes at $45^\circ$, $-135^\circ$, $225^\circ$, and $315^\circ$, so basically the lines $y = 0$ and $x = 0$, but rotated $45^\circ$. This means that a line told to go along any of these asymptotes — defined by the fact that the hyperbola will never meet them — surely will never meet the hyperbola because it basically is the asymptote!?
And anywhere that goes through, judging by the diagrams I've looked at on-line, the empty space above and below the origin (excluding where the hyperbola really is,) should, as well, surely be infinity, since it never gets to the hyperbola?
The summarised question: How come giving a hyperbolic function that treads along the asymptotes does not give infinity or some other undefined answer?
 A: Because the argument to the hyperbolic trigonometric functions is not the angle that your line makes with the $x$-axis, but rather the area between your line, the $x$-axis and the hyperbola (although to be more exact, with the $a/2$ in the picture, it's the area between your line, its reflection across the $x$-axis, and the hyperbola, with a sign to differentiate between the two sides of the $x$-axis). See the image below:

Note that you could say that the regular sine and cosine functions do the same thing, taking in the area of a sector of the unit circle, so it is actually the same, in a way.
A: The hyperbolic trig functions aren't defined by the angle turned off the $x$-axis, but by the area between the hyperbola and the line at angle $\theta.$  There is infinite area between the asymptote and the curve, so as $\theta$ goes from $0$ to $45$ degrees, the area goes from $0$ to infinity.  It's the area that's plugged into the function, not the angle.  While typing I see that Arthur has answered with a diagram.  That's a good diagram.  As the red area goes to infinity, the angle goes to $45$ degrees, but $\sinh( \mbox{area} )$ is the $y$-coordinate of the intersection point.  It goes to infinity with the area.
