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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:


 /   = square root

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

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

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

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

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

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

This is a quadratic in e^y,  and applying the quadratic formula

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

                    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[--------------]

-Doctor Anthony,  The Math Forum
 Check out our web site!   

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

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 

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!   
Associated Topics:
High School Exponents
High School Trigonometry

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