Re: open parenthesis and reduce an expression

*To*: mathgroup at smc.vnet.net*Subject*: [mg75152] Re: open parenthesis and reduce an expression*From*: kem <kemelmi at gmail.com>*Date*: Wed, 18 Apr 2007 05:10:34 -0400 (EDT)

I am sorry for the confusion: my expression is actually: (zxx Cos[b[x,y]]-zyx Sin[b[x,y]]-bx Sin[b[x,y]] zx[x,y]- bx Cos[b[x,y]] zy[x,y]) (Cos[b[x,y]] zx[x,y]- Sin[b[x,y]] zy[x,y])+(Sin[b[x,y]] zx[x,y]+ Cos[b[x,y]] zy[x,y]) (zyx Cos[b[x,y]]+zxx Sin[b[x,y]]+ bx Cos[b[x,y]] zx[x,y]-bx Sin[b[x,y]] zy[x,y]) and it reduces to: zxx zx + zyx zy using Simplify[Expand[..]] so it works Thanks very much for your help. On 4/17/07, Don Taylor <dont at rdrop.com> wrote: > > > On Tue, 17 Apr 2007, kem wrote: > > Thanks for your answer. > > > > Is Sin[_]->1 mean that you assume all sin are 1? it is not really true > in > > the general case. What I meant in my question is to reduce the > expression > > to more compact form of > > zxx zx + zyx zy > > by using trigonometric identities like Cos^2+Sin^2->1 and by grouping > items > > together ... > > > > is it possible in mathematica? > > > > thanks > > kem > > You have given no information at all about what b[x,y] is, > or any of the other undefined functions that you have used. > Mathematica cannot even determine if they are Real valued > functions or perhaps simply infinity. Even FullSimplify[] > does not have enough information to use to determine how > to simplify this. > > I would like to see how you were able to make that simplification > by hand given nothing more than: > > (Cos[b[x,y]] zx[x,y]-Sin[b[x,y]] zy[x,y]) (zxx Cos[b[x,y]]+zyx > Cos[b[x,y]]+zxx Sin[b[x,y]]-zyx Sin[b[x,y]]+bx (Cos[b[x,y]]- > Sin[b[x,y]])zx[x,y]-bx (Cos[b[x,y]]+Sin[b[x,y]]) zy[x,y]) > > You cannot use anything else. You do not know the values > of any of the undefined functions or variables, or even that > they are finite or perhaps complex valued. > > Perhaps I can learn something from that. > Thank you >