       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:=
> 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:=
> (Plot[Chop[Evaluate[expr /. R -> #1]], {y, -#1, #1}, Axes -> False,
> Frame -> True] & ) /@ Range
>
> 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:=
> Simplify[expr /. R -> 4 /. y -> 2]
>
> Out=
> 0
>
> In:=
> D[expr /. R -> 4, y]
> Factor[%]
>
> Out=
> -(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=
> 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:= Simplify[R Sin[ArcTan[y/Sqrt[R^2 - y^2]]] == y, -R<y<R]

Out= 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:=
Simplify[TrigExpand[Sin[ArcTan[y/Sqrt[R^2 - y^2]] -  ArcSin[y/R]]], -R<y<R]

Out= 0

David

```

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