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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Alexei Boulbitch, Dr. habil.
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