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
>