hyperbolic-functions
I know that the same way circular trigonometry is defined over the circle $ x^2 + y^2 = 1 $, hyperbolic trigonometry is defined over the hyperbola $ x^2 – y^2 = 1 $.
What I don't know is how deduced the formulas
$$ \sinh x = \frac {e^x – e^{-x}} {2} \quad \text{and} \quad \cosh x = \frac {e^x + e^{-x}} {2} $$
are deduced.
My question is: How are the formulas for $ \sinh x $ and $ \cosh x $ deduced from the equation $ x^2 – y^2 = 1 $ of the unit hyperbola?
The easy arrangement around the circle, as you showed it, makes use of the fact that every radius of the circle is a unit length, so triangles with that unit length as one edge and a right angle at one corner are easy to find. Furthermore, right angles can be used to transfer the angle argument $\theta$ between these triangles.
In the hyperbola situation, things are a bit more complicated. The point $(1,0)$ is easily obtained as the location of one vertex, which has distance $1$ from the origin. Using e.g. a line parallel to one asymptote, you can also obtain a point $(0,1)$. The area cannot easly be transferred, so one has to make do with the single occurence of that area, and find everything else in relation to it.
Using these two unit points together with the coordinate axes and various right angles, you can find all four hyperbolic functions you mentioned like this:
The meaning of the parameter $a$ in $\sinh(a)$ and in the case $\cosh(a)$ is the same as in the case of $\sin(a)$ and $\cos(a)$. Take a look at the figure below.
In both cases the parameter equals the half area of the red region.
Best Answer
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