#hyperbolic functions

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Hyperbolic functions

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Is the stuff in purple valid? If not, could someone please help me?

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A blog about mathematics.

Updated Section 3.6: Hyperbolic Functions!

Added the inverse hyperbolic functions of sech, csch, and coth.

Questions, comments, corrections? Message me!

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An Unusual Relationship

While it is common to see the trigonometric functions used on a day to day basis, most students seem to forget (or ignore) the existence of the hyperbolic functions. These functions have multiple analogies to the trigonometric functions, and yet they rarely see use in areas of mathematics outside of differential equations and complex analysis. The hyperbolic functions have very basic definitions, as will be shown below. \( \sinh (u) \), pronounced as either "sinch" or "shine" is the hyperbolic sine function. \( \cosh (u) \), pronounced "kosh" is the hyperbolic cosine function. The four other hyperbolic functions, hyperbolic tangent \( \tanh (u) \), hyperbolic cotangent \( \coth (u) \), hyperbolic secant \( \operatorname{sech}(u) \), and hyperbolic cosecant \( \operatorname{csch}(u) \), are all defined in relation to these first two hyperbolic functions.

$$ \sinh (u) = \frac{e^u - e^{-u}}{2} $$

$$ \cosh (u) = \frac{e^u + e^{-u}}{2} $$

$$ \tanh (u) = \frac{\sinh (u)}{\cosh (u)} = \frac{e^u - e^{-u}}{e^u + e^{-u}} $$

$$ \coth (u) = \frac{\cosh (u)}{\sinh (u)} = \frac{e^u + e^{-u}}{e^u - e^{-u}} $$

$$ \operatorname{sech}(u) = \frac1{\cosh(u)} = \frac{2}{e^u + e^{-u}} $$

$$ \operatorname{csch}(u) = \frac1{\sinh(u)} = \frac{2}{e^u - e^{-u}} $$

Additional definitions are given at the Wikipedia page on Hyperbolic Functions.

[Images taken from Wikipedia entry on Hyperbolic function]

These images above show the hyperbolic sine and cosine functions, and how their definitions are related to the exponential function \( e^x \): the hyperbolic cosine function is the average of \( e^x \) and \( e^{-x} \), while the hyperbolic sine function is half the difference of \( e^x \) and \( e^{-x} \).

Initially, despite having similar names, it would seem as if these functions have no relation to the trigonometric functions; the trigonometric functions are used for finding angles and calculating components of vectors, and it seems like there isn't any relationship between the trigonometric functions and the exponential function...so what's the catch?

Well, it turns out the hyperbolic functions are just as good as the trigonometric functions, but they have their own uses. While the trigonometric functions can tell you a circular angle at a given point, the hyperbolic functions can give you a hyperbolic angle at that point. This hyperbolic angle also happens to be the area of a hyperbolic sector up to that point, as seen in the image below.

[Image taken from Wikipedia entry on Hyperbolic function]

The image depicts the unit hyperbola, where the hyperbolic angle \( a\) reaches up to the point \( (\cosh (a), \sinh (a) ) \). This is similar to how a circular angle, \( \theta \), will produce a point \( (\cos(\theta), \sin(\theta)) \) on the unit circle. Clearly, the trigonometric functions and the hyperbolic functions have more in common than we originally thought. But wait, there's more!

Discovered around 1740 and published to the mathematical world in 1748, a famous equation was figured out by Leonhard Euler that provided an astonishing amount of groundwork for complex analysis. This equation is known today as Euler's formula. No, not the one relating the vertices, sides, and edges of a polyhedron.

$$ e^{i \theta} = \cos(\theta) + i \sin (\theta) $$

You'll probably recognize it for a certain value of \( \theta = \pi \):

$$ e^{i \pi} = \cos(\pi) + i \sin(\pi) = -1 + i (0) = -1 $$

Yes, this formula has found its way across the internet, scaring teenagers with its daunting, yet elegant, complexity. Most stumble over the \(i = \sqrt{-1}\) and the \(\pi \) being multiplied together and then exponentiated, and they say something along the lines of "How does this even produce a number? Exponentiating an imaginary number has no actual meaning!" Euler's formula says it does have a meaning: a rotation in the complex plane by an angle of \( \theta = \pi \), or 180 degrees (a half-turn). This is what makes this formula so amazing and eloquent, despite its simplicity.

The uses for Euler's formula doesn't stop there, however! It shows us a relationship between the exponential function and the trigonometric functions! Using Euler's formula, we can define the trigonometric functions in terms of the exponential function in the same way as the definitions of the hyperbolic functions. First, we realize that

$$ e^{-i \theta} = \cos(-\theta) + i \sin(-\theta) = \cos(\theta) - i \sin(\theta) $$

since the cosine function is even and the sine function is odd. Finding the sum of this result with Euler's formula, we can remove the imaginary part of these equations to obtain a formula for cosine:

$$ e^{i \theta} + e^{-i \theta} = 2\cos(\theta) $$

$$ \cos(\theta) = \frac{e^{i \theta} + e^{-i \theta}}{2} $$

We repeat the same process in order to obtain a formula for sine, but instead compute \( e^{i \theta} - e^{-i \theta} \) in order to remove the real part of the equations:

$$ e^{i \theta} - e^{-i \theta} = 2i \sin(\theta) $$

$$ \sin(\theta) = \frac{e^{i \theta} - e^{-i \theta}}{2i} $$

These formulas definitely look very similar to the formulas for \( \cosh(u) \) and \( \sinh(u) \). In fact, we can define each function in terms of the other to truly show how closely they're related:

$$ \cos(x) = \cosh(i x) \qquad \cosh(x) = \cos(i x) $$

$$ \sin(x) = -i \sinh(i x) \qquad \sinh(x) = -i \sin(i x) $$

This astounding result only makes us question the one remaining question about this surprising relation between trigonometric functions and hyperbolic functions: Why aren't the hyperbolic functions known or used as often as the trigonometric functions?

Well, that's beyond the scope of this post. ;)

[mathwork]

[maths-help]

Do you do hyperbolic functions? If so, please help with this: √(3)cosh + √(2)sinhx = √3 (solve for x). completely stuck on it, have tried a myriad of different things but I can't get the answer that's in the back of my textbook. thanks

turns into

Wow, so let’s collect similar terms on the LHS to get:

Um, this is getting kind of complicated. At this point, it’s worthwhile making the substitution e^x = y, because then we can write e^(-x) as 1/y and get:

which we can see is a quadratic equation as soon as we multiply across by y (and move terms to one side)!

Admittedly it’s a really nasty looking quadratic, but a quadratic nonetheless. So we…solve, using whatever method of solving quadratics you’d like (quadratic formula, personally), and get…oh:

Now remember that y = e^x, so then taking logs, we end up with:

(Which I hope is the answer in the back of your textbook, because it’s late.)

Hope that helped!

- H2