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

• To: mathgroup at smc.vnet.net
• Subject: [mg73588] Re: Showing that ArcSinh[2]/ArcCsch[2] is 3?
• From: "Dana DeLouis" <dana.del at gmail.com>
• Date: Thu, 22 Feb 2007 04:39:37 -0500 (EST)

```> 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?

Hi.  Not ideal, but this worked with v 5.2

equ = TrigToExp[ArcSinh[2]/ArcCsch[2]];

avoid = Count[{#1}, _ArcSinh | _ArcCsch | _Log,
Infinity] & ;

FullSimplify[equ, ComplexityFunction -> avoid]

3

Dana DeLouis
Windows XP & Mathematica 5.2
Ps.  Any guesses when v. 6.0 is released?

"David W.Cantrell" <DWCantrell at sigmaxi.net> wrote in message
news:erelp9\$7mm\$1 at smc.vnet.net...
>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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