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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
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