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MathGroup Archive 2006

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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


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