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Hyperbolic Identities and Osborne’s Rule

April 29th, 2009 |

Author: Merrin

The hyperbolic functions are defined analagously to the trigonometric functions except they follow a slightly different sort of geometry. But unlike how the trigonometric functions are built around a right triangle inscribed in a circle. The hyperbolic functions obey a different sort of rule which might be compared to an imaginary hyperbolic reference triangle. The basis for this triangle would be $latex x^2-y^2=h^2&s=1$. With this in mind all hyperbolic functions can be defined analogously to the trigonometric functions with the same combinations.

$\sinh x = (1/2)(e^x - e^{-x})\qquad \cosh x = (1/2)(e^x + e^{-x})$

$\tanh x = \cfrac{\sinh x}{\cosh x}\qquad \text{csch}\, x = \cfrac{1}{\sinh x}$

$\text{sech}\, x = \cfrac{1}{\cosh x} \qquad \coth x = \cfrac{1}{\tanh x}$

The identities regarding the hyperbolic functions will be different from the trigonometric identities because they follow a different sort of geometry. One way to approach the problem is to ignore geometry and just use the relationships between functions. You may notice that if you add the first two equations you get something similar to Euler’s formula

$\sinh x + \cosh x = e^x$

With the algebraic formulas then all the analagous hyperbolic identities to the trigonometric identities can be derived amongst relations regarding exponentials. Instead of going through all that algebra, I will show you a handy rule which gives all the answers provided you know the analagous trigonometric identity.

Osborne’s Rule: In a valid trigonometric identity anywhere you see a sine or cosine, replace either with $latex \sinh x&s=1$ or $latex \cosh x&s=1$ respectively and anytime you would multiply two $latex \sinh x&s=1$ functions together you multiply that term by a minus sign.

Pretty simple right? There should be a Corollary to Osborne’s Rule that state you can turn any hyperbolic identity into a trigonometric identity by replacing $latex \cosh x&s=1$ with $latex \cos x&s=1$ and $latex \sinh x&s=1$ and $latex \sin x&s=1$. Every time you would multiply two $latex \sin x&s=1$ functions together you must multiply that term by a minus sign.

Let’s review some of our trigonometric identities one more time and write down the analogous hyperbolic identities.

These are the Pythagorean type identities

$\cos^2 x + \sin^2 x = 1 \quad \to \quad \cosh^2 w - \sinh^2 w = 1$

$1 + \tan^2 x = \sec^2 x \quad \to \quad 1-\tanh^2 w = \text{sech}\,^2 w$

$1 + \cot^2 x = \csc^2 x \quad \to \quad 1-\coth^2 w = -\text{csch}\,^2 w$

For the double angle formula we have

$\sin(2x) = 2\sin x\cos x \quad \to \quad \sinh(2w) = 2 \sinh w \cosh w$

$\cos(2x) = 2\cos^2x-1 x\quad \to \quad \cosh(2w) = 2\cosh^2w-1$

$\tan (A+B) = \cfrac{\tan A+\tan B}{1-\tan A\tan B} \quad \to \quad \tanh(u+v)= \cfrac{\tanh u + \tanh v}{1+\tanh u \tanh v}$

From these examples you should be able to get a handle of it. Just remember the minus signs for each double power of sine multiplied together.

Example 1. You can derive with Euler’s formula and the binomial theorem that

$\sin ^{7} \theta =\cfrac{1}{64}(-\sin 7\theta +7\sin 5\theta -21\sin 3\theta +35\sin \theta )$

What is the corresponding hyperbolic identity?

Solution 1. There are three negative signs for the seventh power of $latex \sinh x&s=1$.

$(\sinh x)^{7} =\cfrac{1}{64} (\sinh 7x-7\sinh 5x+21\sinh 3x-35\sinh x)$

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One Response to “Hyperbolic Identities and Osborne’s Rule”

free style music says:

June 24, 2010 at 4:00 pm

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Hyperbolic Identities and Osborne’s Rule

One Response to “Hyperbolic Identities and Osborne’s Rule”

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