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### Solving a System of Inverse Hyperbolic Trigonometry Equations ... with Natural Logs and the Quadratic Equation

```Date: 03/28/2010 at 01:50:57
From: Ben
Subject: solve 2 simultaneous eqns w/ 2 unknowns involving cosh

How would one solve the following set of simultaneous equations?

7 = a cosh(-x/a) - a
10 = a cosh((5 - x)/a) - a

There are 2 equations and 2 unknowns, so I figure that it is possible, but
I run into trouble when trying to tackle the simultaneous equations with
my usual methods.

cosh(x) = (e^x + e^-x) / 2; but when I attempt to solve the simultaneous
equations, my usual go-to methods fail -- namely, multiplying one equation
by a scalar, then adding together; and solving for one variable and
substituting.

It all boils down to how one cannot pull an x out of the cosh function,
with the e^x and e^-x. I'm stuck.

```

```
Date: 03/28/2010 at 20:29:09
From: Doctor Vogler
Subject: Re: solve 2 simultaneous eqns w/ 2 unknowns involving cosh

Hi Ben,

Thanks for writing to Dr Math.

That's an interesting question. In the end, I don't think you'll be able
to solve this algebraically.

But your observation about what it "boils down to" does suggest a way
forward. Again, I don't think this will lead to an algebraic solution, but
it should make calculating a numerical solution much faster.

To compute y = arccosh(z), or to solve the equation cosh(y) = z for the
variable y, you proceed as follows:

cosh(y) = z
2 cosh(y) = 2z
e^y + e^(-y) = 2z
(e^y)^2 + 1 = 2z e^y
(e^y)^2 - 2z(e^y) + 1 = 0

This is a quadratic equation in e^y, the solution of which, according to the

e^y = z +/- sqrt(z^2 - 1).

Notice that e^-y = 1/(e^y), and

1 / [z + sqrt(z^2 - 1)] = z - sqrt(z^2 - 1),

So the two solutions are ...

ln [z + sqrt(z^2 - 1)]

... and its negative,

ln [z - sqrt(z^2 - 1)].

To sum up, we have ...

y = arccosh(z) = ln [z + sqrt(z^2 - 1)],

... which is the positive inverse cosh (much like square root, this
usually refers to the *positive* square root); and the other value, the
cosh of which is z, is -y.

and show me what you have been able to do, and I will try to offer further
suggestions.

- Doctor Vogler, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Trigonometry

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