Re: Trignometric rules
- To: mathgroup at smc.vnet.net
- Subject: [mg118243] Re: Trignometric rules
- From: DrMajorBob <btreat1 at austin.rr.com>
- Date: Sun, 17 Apr 2011 07:55:55 -0400 (EDT)
Long story short, Mathematica's pattern matching is a lousy way to do such
things, because there are myriad unexpected ways to represent expressions
you or I might consider similar.
Sums and differences are so tricky (because of the Flat and Orderless
attributes, -x represented as Times[-1,x], et cetera) that if the rule
DOES match what you'd like it to match, it's almost an accident.
The second problem can be solved with a simpler rule, however:
2 k1 k2 u^2 \[Epsilon]1 Cos[d1 + d2 - \[Phi]b - \[Phi]r] /.
Cos[x_] :> Sin[x + \[Phi]b]
2 k1 k2 u^2 \[Epsilon]1 Sin[d1 + d2 - \[Phi]r]
or (more carefully), with
2 k1 k2 u^2 \[Epsilon]1 Cos[d1 + d2 - \[Phi]b - \[Phi]r] /.
Cos[x_] /; ! FreeQ[x, \[Phi]b] :> (Sin[x /. \[Phi]b -> 0])
2 k1 k2 u^2 \[Epsilon]1 Sin[d1 + d2 - \[Phi]r]
Bobby
On Sat, 16 Apr 2011 06:33:28 -0500, Alexei Boulbitch
<alexei.boulbitch at iee.lu> wrote:
> Indeed, why this works:
>
> 2 k1^3 k2 v^2 \[Epsilon]1^3 \[Epsilon]c^2 Cos[
> d1 + d3 + \[Phi]b + \[Phi]r] /. {A_*Cos[a_ + \[Phi]b + x_] ->
> A*\[Phi]b*Sin[a + x]}
>
> 2 k1^3 k2 v^2 \[Epsilon]1^3 \[Epsilon]c^2 \[Phi]b Sin[
> d1 + d3 + \[Phi]r]
>
> while this:
>
> 2 k1 k2 u^2 \[Epsilon]1 Cos[d1 + d2 - \[Phi]b - \[Phi]r] /.
> A_*Cos[b_ - \[Phi]b + y_] -> -A*\[Phi]b*Sin[b + y]
>
> 2 k1 k2 u^2 \[Epsilon]1 Cos[d1 + d2 - \[Phi]b - \[Phi]r]
>
> does not?
>
> Best, Alexei
>
>
> I have an expression as hsown below
>
> tt = -4 k1^2 v^2 \[Epsilon]1^2 \[Epsilon]c^2 Cos[2 d1 + d2 + d3] -
> 2 k1 k2 u^2 \[Epsilon]1 Cos[d1 + d2 - \[Phi]b - \[Phi]r] +
> 2 k1^2 v^2 \[Epsilon]1^2 \[Epsilon]c^2 Cos[
> 2 d1 + d2 + d3 - \[Phi]b - \[Phi]r] +
> 2 k1^3 k2 v^2 \[Epsilon]1^3 \[Epsilon]c^2 Cos[
> d1 + d3 + \[Phi]b + \[Phi]r] +
> 2 k1^2 u^2 \[Epsilon]1^2 Cos[2 d1 + d2 + d3 + \[Phi]b + \[Phi]r] -
> 2 k1^2 v^2 \[Epsilon]1^2 \[Epsilon]c^2 Cos[
> 2 d1 + d2 + d3 + \[Phi]b + \[Phi]r];
>
> I'd like to replace any term containing Cos[+/- phib + x_] with 2 Sin
> [+/-phib] Sin[x]. How do I go about doing it.
>
> Thanks
> Chelly
>
--
DrMajorBob at yahoo.com