MathGroup Archive 2004

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Re: Hyperbolic function identity


This works correctly:

FullSimplify[TrigToExp[ArcCosh[1+(z^2)/2]-2*ArcSinh[z/2]], z > 0]

It looks like that mathematica works better with logarithms and 
exponents then with trigonometric functions.

Carlos Felippa wrote:

>Why 
>
>   FullSimplify[ArcCosh[1+z^2/2]-2*ArcSinh[z/2],z>0];
>
>does not evaluate to 0?
>
>  
>



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