Re: Trigonometric simplification

*To*: mathgroup at smc.vnet.net*Subject*: [mg69267] Re: Trigonometric simplification*From*: Paul Abbott <paul at physics.uwa.edu.au>*Date*: Mon, 4 Sep 2006 04:47:08 -0400 (EDT)*Organization*: The University of Western Australia*References*: <ecbnnc$r29$1@smc.vnet.net> 1115.HAA25039@sssssssssssss <ecmgm2$99o$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

In article <ecmgm2$99o$1 at smc.vnet.net>, "Carl K. Woll" <carlw at wolfram.com> wrote: > carlos at colorado.edu wrote: > > 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. > > > > A while back Maxim gave a nice method to convert Cos[n a] into Cos[a]^_.: > > d = 2 + 3*Cos[a] + Cos[3*a] > > First, replace a with ArcCos[x] and use TrigExpand: > > In[5]:= TrigExpand[d /. a -> ArcCos[x]] > > Out[5]= 4 x^3 + 2 > > Then, replace x with Cos[a]. All together we have: > > In[6]:= TrigExpand[d /. a -> ArcCos[x]] /. x -> Cos[a] > > Out[6]= 4 (Cos[a])^3 + 2 Personally, I prefer to use Cos[n a] == ChebyshevT[n, Cos[a]]. So 2 + 3 Cos[a] + Cos[3 a] /. Cos[n_ a] :> ChebyshevT[n, Cos[a]] yields 2 + 4 Cos[a]^3 Cheers, Paul _______________________________________________________________________ Paul Abbott Phone: 61 8 6488 2734 School of Physics, M013 Fax: +61 8 6488 1014 The University of Western Australia (CRICOS Provider No 00126G) AUSTRALIA http://physics.uwa.edu.au/~paul

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