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Re: open parenthesis and reduce an expression

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?


On 4/17/07, Don Taylor <dont at> 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

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