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?

**Follow-Ups**:**Re: Re: Trigonometric simplification***From:*"Carl K. Woll" <carlw@wolfram.com>