Hyperbolic Functions

The functions hyperbolic and hyperbolic cosine, written, respectively as sinh and cosh, are well known functions defined by the formulae

ex − e−x ex + e−x sinh (x) := , and cosh (x) := , 2 2 were first studied by Riccati in the mid-18th century. He applied them to the solution of general quadratic equations with real coefficients and he found a number of the standard identities that are similar to those familiar from the study of sine and consine functions. The functions have a number of applica- tions. For one, they were crucial in the development of navigational charts using Mercator Projections. The architects Eero Saarinen and Hannskari Bandal de- signed the St. Louis Gateway Arch by using the hyperbolc cosine function. Indeed the inside shape of the Arch is a slightly flattened catenary. You have probably observed catenary curves. That is the shape of a perfictly flexible ca- ble hanging between two posts, and the chains that often hang on the borders of gardens to prevent people from walking on grass are approximate catenaries. More importantly, engineers use these functions in various design problems, for example in the design of high voltage lines.

It is interesting to note that, from Euler’s relation we get the two complex exponential formulae

ei x − e−i x ei x + e−i x sin (x) = , and cos(x) = . 2i 2

This, together with the definitions of the hyperbolic sine and cosine suggests that there are formulae that relate the trigonometric to the hyyperbolic functions and this is indeed the case.

Exercise: Use Euler’s relation to show the following: Recalling that (1/i) = (1/i)(i/i) = (i/ − 1) = −i,

(a) sin (x), = −i sinh (i x) , and cos (x) = cosh (i x); (b) sinh (x), = −i sin (i x) , and cosh (x) = cos (i x).

The use of is made easier by a number of identities that are regularly used. We will list some of them here. It is a matter of straight- forward computation to check that

cosh2 (x) − sinh2 (x) = 1

1 Taking the sum of the sinh and cosh funcitons of course yields an exponential as the following computation shows:

ex + e−x ex − e−x ex + e−x + ex − e−x 2 ex cosh (x) + sinh (x) = + = 2) = 2 2 ( 2

= ex .

Exercise: Show in the same fashion that cosh (x) − sinh (x) = e−x.

Like their counterparts, sinh (x) is an odd function.and cosh (x) is even. These follow directly from the definitions as can be easily checked. This fact can be useful in proving addition formulae. To establish addition formulae for the hyperbolic functions is usually a matter of mind-numbing, but elementary cal- culation. Here is one example:

sinh (x + y) = sinh (x) cosh (y) + cosh (x) sinh (y) . We establish this , using the two preceeding representations for ex and e−x, as follows:

ex ey − e−x e−y ex ey − e−x ey + e−x ey − e−xe−y sinh (x + y) = = 2 2

ex − e−x  ey − e−y)  = ey + e−x = sinh (x) [cosh (y) + sinh (y) + cosh (x) − sinh (y)] sinh (y) 2 2

= sinh (x) cosh (y) + sinh2 (y) + cosh (x) sinh (y) − sinh2 (y) = sinh (x) cosh (y) + cosh (x) sinh (y) .

In order to get the corresponding identity for sinh (x − y) we can use the even and odd properties of these functions. Thus sinh (x − y) = sinh (x) cosh −(y)+cosh (x) sinh (−y) = sinh (x) cos (y)−cosh (x) sinh (y) .

We can make a list of the most important identities:

cosh (x ± y) = cosh (x cosh (y) ± sinh (x) sinh (y) .

1 − tanh2(x) = sech2(x) and ctnh2(x) − 1 = csch2(x) .

2 Here, in analogy to the trigonometric functions we define

sinh (x) ex − e−x tanh (x) = = , cosh (x) ex + e−x

1 2 sech(x) = = , cosh (x) ex + e−x

1 2 csch(x) = = , sinh (x) ex − e−x

1 ex + e−x ctnh(x) = = . tanh (x) ex − e−x

Again, referring to the definitions of sinh and cosh, we can compute the deriva- tives:

d ex − e−x  ex + e−x (sinh (x))0 = = = cosh (x) , dx 2 2 and a similar comutation shows that cosh (x)0 = sinh (x). This fact, alone, makes these functions easy to work with in various situations involving differ- ential equations.

d 2 Exercise: Show that dx (tanh(x)) = sech (x).

Since the derivative of the hyperbolic sine is the hyperbolic cosine which is always positive, the sinh function is strictly increasing and, in particular, in- vertible. Moewocwe rhw dunxrion ia xonrinuoua ang we have

lim sinh (x) = +∞ and lim sinh (x) = −∞ , x→+∞ x→−∞ the range of the hyperbolic sine is the entire line. The function is invertible and it is easy to calculate its inverse. Indeed, if

ex − e−x y = sinh (x) =, 2 then

2 y ex = e2x − 1 , or (ex)2 − 2 y ex − 1 = 0 . This is a quadratic equation in eIx which we can solve ! Indeed we have

ex = y ± p(y2 + 1) .

3 Since ex is always strictly positive, only the plus sign is appropriate. Hence

ex = yp(y2 + 1) . From this it follows that

sinh−1 (y) = x = ln (y + p(y2 + 1) .

The derivative of the inverse hyperbolic sine function is obtained by a straight- forward computation: for p y = sinh−1 x = ln (x + x2 + 1) , we have dy 1 d p 1  1  = (x + x2 + 1) = √ 1 + √ dx x + p(x2 + 1) dx x + x2 + 1) x + x2 + 1)

1 = . p(x2 + 1)

4