Re: Trignometric rules
- To: mathgroup at smc.vnet.net
- Subject: [mg118238] Re: Trignometric rules
- From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
- Date: Sun, 17 Apr 2011 07:54:55 -0400 (EDT)
Because you are allowing Cos[b_ - \[Phi]b + y_] to evaluate and when it does so there is no longer a match. The simplest way to avoid this problem is to use HoldPattern:
2 k1 k2 u^2 \[Epsilon]1 Cos[d1+d2-\[Phi]b-\[Phi]r]/.HoldPattern[A_*Cos[b_-\[Phi]b+y_]]->-A*\[Phi]b*Sin[b+y]
-2 k1 k2 u^2 \[Epsilon]1 \[Phi]b sin(d1+d2-\[Phi]r)
Andrzej Kozlowski
On 16 Apr 2011, at 13:33, Alexei Boulbitch 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
>
> --
> Alexei Boulbitch, Dr. habil.
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