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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
>
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