Calculus: Hyperbolic Trigonometry, III

We continue where we left off on the last post about hyperbolic trigonometry.  Recall that we ended by finding an antiderivative for \sec(x) using the hyperbolic trigonometric substitution \sec(\theta)=\cosh(u).  Today, we’ll look at this substitution in more depth.

The functional relationship between \theta and u is described by the gudermannian function, defined by

\theta=\text{gd}\,u=2\arctan(e^u)-\dfrac\pi2.

This is not at all obvious, so we’ll look at the derivation of this rather surprising-looking formula.  It’s the only formula I’m aware of which involves both the arctangent and the exponential function.  We remark (as we did in the last post) that we restrict \theta to the interval (-\pi/2,\pi/2) so that this relationship is in fact invertible.

We use a technique similar to that used to derive a formula for the inverse hyperbolic cosine.  First, write

\sec\theta=\cosh u=\dfrac{e^u+e^{-u}}2,

and then multiply through by e^u to obtain the quadratic

(e^u)^2-2\sec(\theta)e^u+1=0.

This quadratic equation results in

e^u=\sec\theta\pm\tan\theta.

Which sign should we choose?  We note that \theta and u increase together, so that because e^u is an increasing function of u, then \sec\theta\pm\tan\theta must be an increasing function of \theta. It is not difficult to see that we must choose “plus,” so that e^u=\sec\theta+\tan\theta, and consequently

u=\ln(\sec\theta+\tan\theta).

We remark that no absolute values are required here; this point was discussed in the previous post.

Now to solve for \theta.  The trick is to use a lesser-known trigonometric identity:

\sec\theta+\tan\theta=\tan\left(\dfrac\pi4+\dfrac\theta2\right).

There is such a nice geometrical proof of this identity, I can’t help but include it.  Start with the usual right triangle, and extend the segment of length \tan\theta by \sec\theta in order to form an isosceles triangle.  Thus,

\tan(\theta+\alpha)=\sec\theta+\tan\theta.

Day146Figure

To find \alpha, observe that \beta is supplementary to both 2\alpha and \pi/2-\theta, so that

2\alpha=\dfrac\pi2-\theta,

which easily implies

\alpha=\dfrac\pi4-\dfrac\theta2.

Therefore

\theta+\alpha=\dfrac\pi4+\dfrac\theta2,

which is precisely what we need to prove the identity.

Now we substitute back into the previous expression for u, which results in

u=\ln\tan\left(\dfrac\pi4+\dfrac\theta2\right).

This may be solved for \theta, giving

\theta=\text{gd}\,u=2\arctan(e^u)-\dfrac\pi2.

So let’s see how to use this to relate circular and hyperbolic trigonometric functions.  We have

\sec(\text{gd}\,u)=\dfrac1{\cos(2\arctan(e^u)-\pi/2)},

which after using the usual circular trigonometric identities, becomes

\sec(\text{gd}\,u)=\dfrac{e^u+e^{-u}}2=\cosh u.

It is also an easy exercise to see that

\dfrac{d}{du}\,\text{gd}\,u=\text{sech}\, u.

So revisiting the integral

\displaystyle\int\sec\theta\,d\theta,

we may alternatively make the substitution \theta=\text{gd}\,u, giving

\displaystyle\int\sec\theta\,d\theta=\int\cosh u\,(\text{sech}\, u\,du)=\int du,

which is the same simple integral we saw in the previous post.

What about the other trigonometric functions?  Certainly we know that \cos(\text{gd}\,u)=\text{sech}\,u.  Again using the usual circular trigonometric identities, we can show that

\sin(\text{gd}\,u)=\tanh u.

Knowing these three relationships, the rest are easy to find: \tan(\text{gd}\,u)=\sinh u, \cot(\text{gd}\,u)=\text{csch}\,u, and \csc(\text{gd}\,u)=\text{coth}\,u.

I think that the gudermannian function should be more widely known.  On the face of it, circular and hyperbolic trigonometric functions are very different beasts — but they relate to each other in very interesting ways, in my opinion.

I will admit that I don’t teach students about the gudermannian function as part of a typical calculus course.  Again, there is the issue of time:  as you are well aware, students finishing one course in the calculus sequence must be adequately prepared for the next course in the sequence.

So what I do is this:  I put the exercises on the gudermannian function as extra challenge problems.  Then, if a student is already familiar with hyperbolic trigonometry, they can push a little further to learn about the gudermannian.

Not many students take on the challenge — but there are always one or two who will visit my office hours with questions.  Such a treat for a mathematics professor!  But I feel it is always necessary to give something to the very best students to chew on, so they’re not bored.  The gudermannian does the trick as far as hyperbolic trigonometry is concerned….

As a parting note, I’d like to leave you with a few more exercises which I include in my “challenge” question on the gudermannian.  I hope you enjoy working them out!

  1.  Show that \tanh\left(\dfrac x2\right)=\tan\left(\dfrac 12\text{gd}\,x\right).
  2. Show that e^x=\dfrac{1+\tan(\frac12\text{gd}\,x)}{1-\tan(\frac12\text{gd}\,x)}.
  3. Show that if h is the inverse of the gudermannian function, then h'(x)=\sec x.

Published by

Vince Matsko

Mathematician, educator, consultant, artist, puzzle designer, programmer, blogger, etc., etc. @cre8math

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