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Re: Trignometric rules

  • To: mathgroup at smc.vnet.net
  • Subject: [mg118194] Re: Trignometric rules
  • From: Alexei Boulbitch <alexei.boulbitch at iee.lu>
  • Date: Sat, 16 Apr 2011 07:33:28 -0400 (EDT)

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

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