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 Dipl.-Math. Adalbert Hanszen