Re: cosh

how??

by taking the natural logaritm in some way???

I don't see how.. :(

/peter

gerben47@xxxxxxxxxxx wrote:
(-Peter-) wrote: 
yes.. but this results in the following:

0.5=(e^x-e^-x)/(e^x+e^-x)

and I'm stuck :(

alainverghote@xxxxxxxx wrote: 
Peter,

just develop with exponentials
cosh  and sinh ,
put t = exp(x)  ...

good luck ,
Alain.


How about multiplying with e^x / e^x to get:

1/2 = (e^2x - 1) / (e^2x +1)

This should be solvable.

Gerben

.

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