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MathGroup Archive 2007

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Re: Showing that ArcSinh[2]/ArcCsch[2] is 3?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg73528] Re: [mg73499] Showing that ArcSinh[2]/ArcCsch[2] is 3?
  • From: Bob Hanlon <hanlonr at cox.net>
  • Date: Wed, 21 Feb 2007 01:45:34 -0500 (EST)
  • Reply-to: hanlonr at cox.net

expr=ArcSinh[2]/ArcCsch[2];

Rationalize[expr,10^-100]

3

Reduce[{ArcSinh[x]/ArcCsch[x]==3,x>0},x]

x == 2

Off[N::meprec];

expr==3//FullSimplify

True

On[N::meprec];


Bob Hanlon

---- "David W.Cantrell" <DWCantrell at sigmaxi.net> wrote: 
> I hope I've just overlooked something very simple.
> I want to transform ArcSinh[2]/ArcCsch[2] to 3, using just "knowledge"
> already implemented in Mathematica. I tried FullSimplify first, and it
> doesn't help. I tried several other things. For example,
> 
> TrigToExp[ArcSinh[2]/ArcCsch[2]]  yields
> 
> Log[2 + Sqrt[5]]/Log[1/2 + Sqrt[5]/2].
> 
> But then how should we transform that to 3?
> 
> David



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