RE: Simplify for ca^2+sa^2==1

• To: mathgroup at smc.vnet.net
• Subject: [mg26350] RE: [mg26339] Simplify for ca^2+sa^2==1
• From: "Ingolf Dahl" <f9aid at fy.chalmers.se>
• Date: Wed, 13 Dec 2000 02:41:13 -0500 (EST)
• Sender: owner-wri-mathgroup at wolfram.com

```Hi!
I would substitute all even powers of ca by corresponding expression in sa,
in the following way:

Expand[(your expression) //. {ca^2 -> 1 - sa^2, ca^3 -> ca*(1 - sa^2),
ca^4 -> (1 - sa^2)^2}]

If you have higher powers in ca, you could just add more terms in the
substitution list, or try to generalize the substitution rule to arbitrary
power. Is there any better way? I am curious, because I have made a lot of
similar simplifications. At the end no powers of ca (higher than one) will
appear in your expression, which will make it easier to handle the
substitution ca=(+/-)Sqrt[1-sa^2], and choose the right branch.

Kind regards

Ingolf Dahl
Chalmers University

> Adalbert Hanssen [hanssen at Zeiss.de] wrote in [mg26339] Simplify for
ca^2+sa^2==1:

Hi, MathGroup,

in a lengthy expression, I know, a lot
of simplification can be done, if Simplify
and the like would take into account that
for varaibles ca and sa

ca^2+sa^2==1

I know, that I can set ca=Sqrt[1-sa^2] and
deal with the branch cut by hand.

The bad thing is, that these ca^2 and sa^2
are expanded out in lenghty subexpressions
involving lots of other symbols. So far, I
have found no way (but would be glad, if
someone could advise me one), to factor out
(ca^2+sa^2).

Unfortunately, there are also terms, where
(1+ca^2+sa^2) or (2*ca^2+2*sa^2) can be
factored out, also others with (-ca^2-sa^2)
and so on.

Any general tip, how to best cope with such
algebraic manipulations?

kind regards