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Re: open parenthesis and reduce an expression

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> 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

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