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

*To*: mathgroup at smc.vnet.net*Subject*: [mg73617] Re: Showing that ArcSinh[2]/ArcCsch[2] is 3?*From*: "David W.Cantrell" <DWCantrell at sigmaxi.net>*Date*: Fri, 23 Feb 2007 04:33:16 -0500 (EST)*References*: <erjp6e$pv7$1@smc.vnet.net>

"Dana DeLouis" <dana.del at gmail.com> wrote: > > 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 Maybe not ideal, as you say, but, of all responses, it comes closest to doing what I wanted! Many thanks to all those who responded to my question. The title I used for the thread was very badly worded; I apologize. Nonetheless, most respondents realized from the text of my message that I wanted to get Mathematica to _transform_ ArcSinh[2]/ArcCsch[2]] to 3, rather than to get it merely to verify that the value is 3. > Dana DeLouis > Windows XP & Mathematica 5.2 > Ps. Any guesses when v. 6.0 is released? Good question. Anyone? David > "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 > >