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Hyperbolic "trig" functions such as $\sinh$, $\cosh$, have close analogies with regular trig functions such as $\sin$ and $\cos$. Yet the hyperbolic versions seem to be encountered relatively rarely. (My frame of reference is that of someone with college freshman/sophomore, but not advanced math.)

Why is that? Is it because the hyperbolic versions of these functions are less common/useful than the circular versions?

Can you do the "usual" applications (Taylor series, Fourier series) with hyperbolic functions as you can with trigonometric?

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Hyperbolic trig functions are essentially the usual trig functions rotated as graphs in $\Bbb C^2$. It's conics every way you turn, but parametrizations of circles I guess my be more important or ubiquitous than hyperbolas. –  anon Aug 30 '12 at 19:38
If you say hyperbolic functions are encountered rarely, then I say you have not studied complex analysis, physics, or engineering. –  GEdgar Aug 31 '12 at 0:00
Like GEdgar, I disagree with the premise of the question. Hyperbolic trigonometric functions naturally appear in special relativity, so if you do special relativity, you'll encounter them quite frequently! –  Qiaochu Yuan Aug 31 '12 at 0:18
@GEdgar:I'm not a professional mathematician. I've had three semesters of calculus and one of linear algebra/differential equations, and "barely" know about hyperbolic functions. The question is with that frame of reference. –  Tom Au Aug 31 '12 at 0:42
I can tell you hyperbolic functions often occur in complex analysis. Did you know that $$-i \sin (i x) = \sinh x$$ –  Argon Aug 31 '12 at 0:50

5 Answers 5

up vote 2 down vote accepted

$\sinh$ and $\coth$ seem to appear less then their circular counterparts in real analysis as is explained well in some other answers. However, hyperbolic functions appear quite commonly in complex analysis. From Euler's formula and its subsequent trigonometric definitions, one finds that

$$\cos (ix) = \cosh x$$ $$\sinh (ix)=i\sin x$$

which is a clear connection between hyperbolic and circular trigonometric functions.

One use of hyperbolic functions, that I have personally used, is in integration. Using the identity

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

we can often be used for substitutions to evaluate integrals with $x^2+1$ and $x^2-1$ (instead of $\sec$ and $\tan$), just as the identity $\cos^2 x+\sin^2 x=1$ can be used for substitutions to evaluate integrals with $1-x^2$.

For example, to evaluate

$$\int \frac{dx}{\sqrt{x^2+1}}$$

we may substitute $x = \sinh u \implies dx = \cosh u \, du$ so the integral becomes

$$\int \frac{\cosh u}{\sqrt{\sinh^2 u+1}}\, du = \int 1 \, du= u +C = \operatorname{arsinh} x + C$$

And, of course, the hyperbolic functions provide a parametrization of the standard hyperbola.

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I can think of two reasons.

  1. When we do geometry, we usually work in Euclidean space, where the intrinsic property of a line segment between two points is its length, given (in two dimensions) by $\ell^2 = \Delta x^2 + \Delta y^2$. We are allowed to change our reference frame as long as we preserve lengths, which means that the transformation of $(\Delta x,\Delta y)$ is a rotation, and the transformed vector must lie on the circle $\Delta x^2 + \Delta y^2 = \operatorname{const}$, which is naturally parametrized by $\sin$ and $\cos$. The hyperbolic variants don't have much to do with circles or with rotation, so are not relevant here (unless you supply them with imaginary arguments, at which point you're really working just with $\sin$ and $\cos$ in disguise).

    On the other hand, the natural setting for special relativity is Minkowski space, where the invariant property of the interval between two points in spacetime is of the form $s^2=\Delta x^2-\Delta t^2$, with a minus sign. Here the allowed changes of reference frame are given by Lorentz transformations, the transformed interval lies on $\Delta x^2-\Delta t^2=\operatorname{const}$ which is a hyperbola, and indeed one finds hyperbolic trigonometric functions to be quite useful in special relativity.

  2. The usual trigonometric functions $\sin$ and $\cos$ are two real solutions to the differential equation $y'' = -y$, which describes a simple harmonic oscillator (a conservative system in a parabolic potential well) which is a central example in much of classical physics. The hyperbolic versions $\sinh$ and $\cosh$ are solutions to $y'' = y$ instead. This equation is rarely useful to model physical systems in real life because all its solutions are unbounded and gain infinite amounts of kinetic energy. Even taken locally, the equation describes an unstable equilibrium, so any real system will not spend most of its time there without additional forcing.

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Putting 2. a bit more generally: we are more often interested in DEs whose solutions are bounded than those with unbounded solutions. –  J. M. Aug 30 '12 at 23:14
Regarding 2: The equation $y''=y$ is important too. But its solutions can be written in terms of exponentials. –  timur Aug 31 '12 at 3:52
@timur: Can you say a little more about what contexts it is important in? –  Rahul Aug 31 '12 at 3:55

I'd say because hyperbolic functions can be written pretty easily in terms of exponential functions. By that I mean you don't need $i$ like when you express $\sin$ and $\cos$ using $\exp$. That means in the "real" world, it's not necessary to use $\sinh$ and $\cosh$ because you can always resort to $\exp$, but the same is not true for $\sin$ and $\cos$. However, I wouldn't say hyperbolic functions are rarely encountered.

Very often I find myself choosing hyperbolic functions over exponential functions in the context of ODEs and PDEs with boundary/initial conditions at $0$. (I believe sophomores should have taken/be taking these classes, no?) The fact that $\sinh(0) = \cosh'(0) = 0$ and $\cosh(0) = \sinh'(0) = 1$ makes your expression much cleaner. Also, you get $\sinh(ax)$ and $\cosh(ax)$ as two solutions to the ODE $y'' - a^2y = 0$ immediately, similar to the way you get $\sin(ax)$ and $\cos(ax)$ from $y'' + a^2y = 0$. Variation of parameters also gives $\sinh$ as a kernel when you solve the non-homogeneous ODE with Dirichlet boundary conditions. There are just so many things you can express cleanly in terms of hyperbolic functions.

Oh, and don't forget that $\tanh$ is a really nice bijection from $\mathbb R$ to $(-1, 1)$. Kind of funny that $\tanh$ looks like $\frac 2\pi \arctan$.

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Anon's answer is a good one, with one small excepton (or perhaps one wide misconception). What people commonly refer to as the complex plane is an unfortunate misnomer. It should be called the "complex number plane". By calling it the complex plane (which, by the way, is a quite widespread phenomenon), one is led to think of it as being C^2. It is not. It is actually the real plane R^2 used as a device to represent the complex numbers C (= C^1). Thus I would maintain (in support of Anon) that the best way to understand the hyperbolic sine and cosine functions is via restriction of the analytic continuation of the real sin and cosine functions to the imaginary number line of the complex number plane. In a sense. the analytic continuation of the real exponential function does both for the price of one.

One more thing, regarding your (over)reaction to Edgar's comment. Being a trained mathematician is vastly beyond taking courses in which you would first encounter hyperbolic trig functions. If you are a major in the hard sciences or in engineering, you will be exposed to them in the near future. I think that was Edgar's intended message, nothing more.

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I took my last math course 30 years ago, in 1983. There is no "near future" in the hard sciences or engineering for me, unfortunately. Which is why Edgar's statement has less meaning than for me than it deserves. –  Tom Au Oct 13 '13 at 20:19

Well, yes, I would say so. After all, sine and cosine describe everything from lengths of sides of right-angled triangle to angular momentum. Rotation is so fundamental that we're bound to see sine and cosine very often. Of course $\sin$ and $\cos$ are just the opposite side of the exponential coin from $\sinh$ and $\cosh$ since

$$ \cosh x = \frac{1}{2}(e^{x} + e^{-x}) \, , $$ $$ \sinh x = \frac{1}{2}(e^{x} - e^{-x}) \, , $$ $$ \cos x = \frac{1}{2}(e^{ix} + e^{-ix}) \, , $$ $$ \sin x = \frac{1}{2i}(e^{ix} - e^{-ix}) \, . $$

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It's BECAUSE they are so easily rotated that I would expect a lot more "back and forth" between the two. –  Tom Au Aug 30 '12 at 20:14

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