RE: Transformation Problem
- To: mathgroup at smc.vnet.net
- Subject: [mg43504] RE: [mg43454] Transformation Problem
- From: "David Park" <djmp at earthlink.net>
- Date: Thu, 18 Sep 2003 05:39:35 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
Geir,
Mathematica is going to make the simplification. If it is really important
to show the exact form then you will have to use a HoldForm.
f = Sin[x] + Cos[x]
HoldForm @@ {f}
% /. Sin[x] :> Cos[x + Pi/2]
Cos[x] + Cos[x + Pi/2]
The delayed rule was necessary here.
David Park
djmp at earthlink.net
http://home.earthlink.net/~djmp/
From: Geir Ove [mailto:geirAoves at online.no]
To: mathgroup at smc.vnet.net
Hello,
I am trying to transform an expression that involves Sin[x] and Cos[x] into
an expression that only contains Cos[x] terms. However, I do not succeed.
The very problem seems to be that the built in rules that Mathematica knows,
will always rewrite e.g. terms like -Cos[x + Pi/2] to Sin[x].
E.g. assume:
f = Sin[x] + Cos[x]
Now, for the sake of the example, I want mathematica to start represening f
as Cos[x] - Cos[x + Pi/2]. So I try:
In : f /. Sin[x] -> -Cos[x + Pi/2]
Out: Sin[x] + Cos[x]
It seems that mathematica does nothing, but it does: It does the
replacement, then uses its internal definitions to simplify back -Cos[x +
Pi/2] to Sin[x]. You can convince yourself about this by doing:
In : f /. Sin[x] -> Cos[x + Pi/2]
Out: Sin[x] - Cos[x]
Mathematica changed the sign since you gave the wring rule.
QUESTION:
How can I achieve what I want, namely the form: Cos[x] - Cos[x + Pi/2]. ???
The expression I wan to simplify is of course more complex, but the same
principle applies.
In advance, thanks for any help.
Geir Ove
Norway