I is amazing how easily old stuff gets lost, when it is used no longer.

I bet most mathematicians don’t know how to spot the secans or the cotangens in the unit circle. Neither did I, when the question came up on the German version of geopardy (Wer wird Millionär). Of course, we can look that up in Wikipedia (or elsewhere), but the origin of the names is obvious only, if you look at the following image.

The secans is where the tangent intersects the x-axis. The cosecans is related to the y-axis. The formulas are easily derived with similarity.

\(\sec(x) = \dfrac{1}{\cos(x)}, \quad \csc{x} = \dfrac{1}{\sin(x)}\)

Here is another interesting image.

Just like the unit circle, the unit hyperbola in the image above has a simple equation.

\(x^2-y^2 = 1\)

If we set

\(x = \sinh(A) = \dfrac{e^A – e^{-A}}{2}\)

and

\(y = \cosh(A) = \dfrac{e^A + e^{-A}}{2}\)

we have with a little computation that (x,y) is on the unit hyperbola. This explains the terms „sinus hyperbolicus“ and „cosinus hyperbolicus“.

It requires a bit more computation to show that the points (x,y), (x,-y), (0,0) are indeed the corners of an area of size A. The easiest way to see this is to use a formula for the area that is swiped by the ray from 0 to g(t) in the plane between two times t1 and t2

\(\int\limits_{t_1}^{t_2} \frac{1}{2} \left| \begin{matrix} g_1(t) & g_1′(t) \\ g_2(t) & g_2′(t)\end{matrix} \right| \,dt\)

The reason for this formula is that the integral sums up the areas of the triangles between g(t) and g'(t)dt. Now, with our path along the unit hyperbola, we see that the determinant is equal to 1. Thus, we get A/2, if we integrate from 0 to any point A>0, which proves the result.

It is a nice fact that the same formula for the area is true for (cos A, sin A) on the unit circle!