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MathGroup Archive 2006

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Re: Trigonometric simplification

  • To: mathgroup at smc.vnet.net
  • Subject: [mg68921] Re: Trigonometric simplification
  • From: carlos at colorado.edu
  • Date: Wed, 23 Aug 2006 07:15:53 -0400 (EDT)
  • References: <ecbnnc$r29$1@smc.vnet.net><ecek3f$qpu$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Here is a correction to my second post (Thanks to Paul Abbott for
noticing the factor of 2).  The reason I need to get  2 + 4*Cos(a)^3
is to match a published solution in a homework solutions manual.

=====================================================

Thanks, that works perfectly.  Actually Sec[a]>0 as assumption
is sufficient. This is correct from the problem source, since
the angle is in the range (-Pi/2,Pi/2)

Here is a related question.  How can I get Mathematica to pass from

   d = 2 + 3*Cos[a] + Cos[3*a]     (* leaf count 10 *)

to

    2 + 4*Cos[a]^3                  (* leaf count 8 *)

TrigExpand[d] gives

   2 + 3*Cos[a] + Cos[a]^3 - 3*Cos[a]*Sin[a]^2

Applying Simplify to that yields  2 + 3*Cos[a] + Cos[3*a]  so we are
back to the beggining.

======================================================

Paul Abbott's solution, sent directly by email, is elaborate:

f[d_] := 10*Plus @@ Cases[{d}, Cos[n_ x_] | Sin[n_ x_] -> n, Infinity]
+LeafCount[d]

FullSimplify[d, ComplexityFunction -> f]

4*Cos[a]^3 + 2

Is there a simpler way?


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