Hyperbolic function equations. Where do they come from?

Question

My text (Stewart's Early transcendentals) is going into hyperbolic functions, but I'm a bit confused about some statements they are making. All I know about hyperbolic functions is that the distances between a point on the hyperbola and the two foci are constant. IT's the locus of all these points I think. This text I find confusing:

1. On the circle... if the angle is $\frac{\pi}{6}$, the area is $\frac{\pi}{6}$ too right? The area isn't double? Say the radius is 1, that means the area is $\pi \cdot r^2 = \pi$ and we divide the area the area by 6 because of the radian measure. But why does the text say the trig case t represents twice the area of the shaded sector?

2. What does $\cosh{t}$ represent?

3. Why does $\sinh{x}$ equal $\frac{e^x - e^{-x}}{2}$?

Answer 1

1. If you mean on the circle of radius 1, if the angle is x and $$0 < x < 2\pi,$$ and you want the area between the line joining the origin to the circle with the angle x from the positive horizontal x axis (as in your diagram above), then actually the area is not x, but half of x. To see this, note the area of the whole circle is $$\pi r^2 = \pi*1 = \pi$$ The area of half the circle is then $$\frac{\pi}{2},$$ which occurs when $$x = \pi.$$ Thus, the angle is double the area, just as in the hyperbolic definition. You can show this is true in general by using the fact that the radius cuts the circle's circumference, which is double its area.

EDIT: I saw you commented the other answer about this, so here's a full explanation. At angle x, you're cutting off a proportion of the circle's circumference. In fact, if C is the circumference of the circle, you're cutting off precisely $$\frac{x}{2\pi}*C,$$ right? Because cutting off a circle's circumference cuts off the same proportion of the area (if you think about the line from the origin to the point), then replacing that C with the area, A, works. Test this out with x being pi or half pi and so in. In the case, where our circle has radius 1, then we have: $$\frac{x}{2\pi}A = \frac{x}{2\pi}\pi = \frac{x}{2} $$ Thus our area is half the angle (or, our angle is double the area).

2. cosh(t) is the x co-ordinate of the point, and sin(t) is the y co-ordinate, just as with sin and cos, except the object whose points we're taking is a hyperbola instead of a circle.

3. See other answers

Answer 2

By definition $\cosh t$ and $\sinh t$ represent the coordinates of the point P on the hyperbola such that the shaded area between the ray OP and the hyperbola is equal to $\frac{t}2$.

For this reason the fundamental relationship for hyperbolic functions is

$$x^2-y^2=1\implies \cosh^2 x -\sinh^2 x =1$$

From integral calculation it turns out that

$\cosh x=\frac{e^x + e^{-x}}{2}$

$\sinh x=\frac{e^x - e^{-x}}{2}$

Note that also for the trigonometric circle for an angle $\theta$ the area is $\theta/2$, just note indeed that for $\theta=2\pi$ the area is $\pi$ (R=1).