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

*To*: mathgroup at smc.vnet.net*Subject*: [mg73599] Re: Showing that ArcSinh[2]/ArcCsch[2] is 3?*From*: "dimitris" <dimmechan at yahoo.com>*Date*: Fri, 23 Feb 2007 04:23:32 -0500 (EST)*References*: <erelp9$7mm$1@smc.vnet.net>

Hi again. As Andrzej Kozlowski pointed out and most of the replies you get demonstrated it is easier to prove that one thing is equal to another than to transform one of them into the other. However trying harder you can many times transform your expression in the desired way, as Dana's smart response showed you. Search in the archives for relevant to Dana's solution posts of Adam Strzebonski regarding (full)simplification. Also the Help Browser contains some relevant examples FrontEndExecute[{HelpBrowserLookup["RefGuide", "ComplexityFunction"]}] Below is some relevant material from Micheal Trott's Guidebook of Symbolics In[6]:= Simplify[Tan[x] - Cot[x]] Out[6]= -Cot[x] + Tan[x] In[7]:= Simplify[Tan[x] - Cot[x], ComplexityFunction -> (Count[#1, _Tan | _Cot, Infinity] & )] Out[7]= (-Cos[2*x])*Csc[x]*Sec[x] In[8]:= Simplify[Tan[x] - Cot[x], ComplexityFunction -> (Count[#1, _Tan | _Cot | _Sin | _Cos | _Csc | _Sec, Infinity] & ), TransformationFunctions -> {TrigToExp}] Out[8]= (I*(E^((-I)*x) - E^(I*x)))/(E^((-I)*x) + E^(I*x)) + (I*(E^((-I)*x) + E^(I*x)))/(E^((-I)*x) - E^(I*x)) Best Regards Dimitris =CF/=C7 David W.Cantrell =DD=E3=F1=E1=F8=E5: > 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