Associated Topics || Dr. Math Home || Search Dr. Math

### Logarithmic Equivalent of the Inverse Hybolic Cosine Function

```
Date: 10/28/97 at 06:58:22
From: Stephen Gurnell
Subject: Logarithmic equivalent of the inverse hybolic cosine function

Dr. Math,

First, thank you very much for your recent help and swift replies.

I have recently attempted a question which I think I have done wrong.
I was wondering if there was a set formula for this problem:

key
---

__
/   = square root

Q. Use the logarithmic equivalent of the inverse hyperbolic cosine
function, to deduce that:

__
-1 (70)     (35+/69)
cosh  (--) = ln(------)
(68)     (  34  )

A?.
-1
cosh   1.029411765 = ln(1.273724231)

0.241945074 = ln(1.273724231)

0.241945074 = 0.241945074

0.242 = 0.242

As you can see from the above I feel that this method of solving the
problem must be wrong. as it appears too simple.

Many thanks.
Stephen Gurnell.
```

```
Date: 10/28/97 at 10:07:58
From: Doctor Anthony
Subject: Re: Logarithmic equivalent of the inverse hybolic cosine
function

If cosh^(-1)(x) = y, then   x = cosh(y)

e^y + e^(-y)
x = --------------
2

2x = e^y + e^-y     and   e^y - 2x + e^(-y) = 0

Multiply by e^y             e^(2y) - 2x.e^y + 1 = 0

2x +- sqrt(4x^2 - 4)
e^y = ----------------------
2

e^y = x +- sqrt(x^2-1)

y  = ln(x +-sqrt(x^2-1))

taking + and - sign this gives the two values

y = +- ln(x^2+sqrt(x^2-1))

because x-sqrt(x^2-1) = 1/(x+sqrt(x^2-1))

and so cosh^(-1)(x) = +-ln(x+sqrt(x^2-1))

cosh(-1)(35/34) = +-ln(35/34 + sqrt((35/34)^2 - 1))

(35 + sqrt(69)
= ln[--------------]
34

-Doctor Anthony,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```

```
Date: 10/28/97 at 10:09:26
From: Doctor Mitteldorf
Subject: Re: Logarithmic equivalent of the inverse hybolic cosine
function

Dear Stephen,

This is a fine method of showing that the identity is true, but it's
not a mathematician's way. The mathematician might say, "How do you
know it isn't just a coincidence that the two sides come out
approximately equal? Maybe if you evaluated just one more decimal
place than your hand calculator is capable of, they would come out
different."

Not very likely, but there's also a pedagogical reason: if you go back
to your definitions, apply some algebra, and find that the two are
exactly equivalent, you've practiced using some skills that may help
you next time when the numbers are different - or maybe when there are
variables in the problem, and not just numbers.

So here's a start on the other way:

Take cosh of both sides. Cosh of cosh^-1 is just the argument, 70/68.
On the right, use the definition of cosh:

cosh(x) = (e^x + e^(-x))/2

e^x here is (35+/69)
(------)
(  34  )

since e^x is just the inverse of the ln function.  What is e^(-x)?
It's just the reciprocal of e^x.

You take it from here...

-Doctor Mitteldorf,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Exponents
High School Trigonometry

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search