Re: open parenthesis and reduce an expression

*To*: mathgroup at smc.vnet.net*Subject*: [mg75151] Re: open parenthesis and reduce an expression*From*: kem <kemelmi at gmail.com>*Date*: Wed, 18 Apr 2007 05:10:04 -0400 (EDT)

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 On 4/17/07, Don Taylor <dont at rdrop.com> wrote: > > In comp.soft-sys.math.mathematica you write: > >Hi, How can I reduce the following formula for instance > >(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]) > > >to the formula > >zxx zx + zyx zx > >in mathematica? > > >I know that such a reduction is possible since I did it manually, I > >just need to test it on more complex expressions. > > (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])/. > Sin[_]->1/.Cos[_]->1/.zy[___]->zy/.zx[___]->zx/.bx->1 > > Expand[(%)/2] > > giving > > zx zxx - zx zy - zxx zy + zy^2 > > All I ask is that I be given no credit for any of this. > Thank you >