Re: Trignometric rules
- To: mathgroup at smc.vnet.net
- Subject: [mg118239] Re: Trignometric rules
- From: Bob Hanlon <hanlonr at cox.net>
- Date: Sun, 17 Apr 2011 07:55:07 -0400 (EDT)
The x_ and y_ are unnecessary in your examples so I have simplified by eliminating them. I have also eliminated the Greek letters to simplify reading the InputForm of the expressions.
2 k1^3 k2 v^2 e1^3 ec^2 Cos[d1+d3+pb+pr]/.
{A_*Cos[a_+pb]->A*pb*Sin[a]}
2*e1^3*ec^2*k1^3*k2*pb*v^2*Sin[d1 + d3 + pr]
2 k1 k2 u^2 e1 Cos[d1+d2-pb-pr]/.
A_*Cos[b_-pb]->-A*pb*Sin[b]
2*e1*k1*k2*u^2*Cos[d1 + d2 - pb - pr]
For the second example, compare the FullForm representations (which is what ReplaceAll acts on)
2 k1 k2 u^2 e1 Cos[d1+d2-pb-pr]//FullForm
Times[2,e1,k1,k2,Power[u,2],Cos[Plus[d1,d2,Times[-1,pb],Times[-1,pr]]]]
LHS of second rule
A_*Cos[b_-pb]//FullForm
Times[Cos[Plus[pb,Times[-1,Pattern[b,Blank[]]]]],Pattern[A,Blank[]]]
Note that -pb does not show up as such. The rule {A_*Cos[a_+c_.*pb] -> A*c*pb*Sin[a]} will work for both
2 k1^3 k2 v^2 e1^3 ec^2 Cos[d1+d3+pb+pr]/.
{A_*Cos[a_+c_.*pb]->A*c*pb*Sin[a]}
2*e1^3*ec^2*k1^3*k2*pb*v^2*Sin[d1 + d3 + pr]
2 k1 k2 u^2 e1 Cos[d1+d2-pb-pr]/.
A_*Cos[b_+c_.*pb]->-A*c*pb*Sin[b]
2*e1*k1*k2*pb*u^2*Sin[d1 + d2 - pr]
Bob Hanlon
---- 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
--
Alexei Boulbitch, Dr. habil.
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