Re: showing an expression equal to 0

*To*: mathgroup at smc.vnet.net*Subject*: [mg73719] Re: showing an expression equal to 0*From*: "David W. Cantrell" <DWCantrell at sigmaxi.net>*Date*: Mon, 26 Feb 2007 06:14:52 -0500 (EST)*References*: <errm70$8bp$1@smc.vnet.net>

"dimitris" <dimmechan at yahoo.com> wrote: > Consider > > In[87]:= > expr = ArcTan[y/Sqrt[R^2 - y^2]] - ArcSin[y/R]; > > The expression for R>0 and -R<y<R is equal to zero > > In[72]:= > (Plot[Chop[Evaluate[expr /. R -> #1]], {y, -#1, #1}, Axes -> False, > Frame -> True] & ) /@ Range[5] > > Based on Andrzej's resposnse on a recenet message I am able to show > this equality for given R. > > E.g. for R=4 the following demonstrates that expr is equal to zero > > In[99]:= > Simplify[expr /. R -> 4 /. y -> 2] > > Out[99]= > 0 > > In[97]:= > D[expr /. R -> 4, y] > Factor[%] > > Out[97]= > -(1/(4*Sqrt[1 - y^2/16])) + (y^2/(16 - y^2)^(3/2) + 1/Sqrt[16 - y^2])/ > (1 + y^2/(16 - y^2)) > Out[98]= > 0 > > However not specifying R (but assuming R>0&&-R<y<R) I am not able to > show that expr is equal to zero. > > Any ideas? In[4]:= Simplify[R Sin[ArcTan[y/Sqrt[R^2 - y^2]]] == y, -R<y<R] Out[4]= True Since we also know that ArcTan and ArcSin have the same range, we can conclude that ArcTan[y/Sqrt[R^2 - y^2]] == ArcSin[y/R] for -R < y < R. A similar method is In[5]:= Simplify[TrigExpand[Sin[ArcTan[y/Sqrt[R^2 - y^2]] - ArcSin[y/R]]], -R<y<R] Out[5]= 0 David