Re: Abs[a] Sin[Abs[a]]

• To: mathgroup at smc.vnet.net
• Subject: [mg21025] Re: [mg20923] Abs[a] Sin[Abs[a]]
• From: BobHanlon at aol.com
• Date: Thu, 2 Dec 1999 21:41:36 -0500 (EST)
• Sender: owner-wri-mathgroup at wolfram.com

```Edward,

Since both

Simplify[Abs[a]*Sin[Abs[a]], a >= 0]

a Sin[a]

and

FullSimplify[Abs[a]*Sin[Abs[a]], a < 0]

a Sin[a]

then

Abs[a]*Sin[Abs[a]] /. Abs[x_]*Sin[Abs[x_]] :> x*Sin[x]

a Sin[a]

Or, since

FullSimplify[Abs[a] == a*Sign[a], Element[a, Reals]]

True

then

FullSimplify[Abs[a]*Sin[Abs[a]] /. Abs[x_] :> x*Sign[x], Element[a, Reals]]

a Sin[a]

Bob Hanlon

In a message dated 12/1/1999 3:19:20 AM, gold at isitel1.isi.kfa-juelich.de
writes:

>How to compell Mathematica to simplify
>  Abs[a] Sin[Abs[a]]
>to
>  a Sin[a]
>
>?
>Why it does not do it automatically? I also tried
>assumptions like a :enum: Reals.
>

```

• Prev by Date: Re: A question of simplification
• Next by Date: RE: ExcludedForms
• Previous by thread: Abs[a] Sin[Abs[a]]
• Next by thread: Re: Abs[a] Sin[Abs[a]]