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MathGroup Archive 2006

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Re: Re: Trigonometric simplification

  • To: mathgroup at smc.vnet.net
  • Subject: [mg68939] Re: [mg68921] Re: Trigonometric simplification
  • From: "Carl K. Woll" <carlw at wolfram.com>
  • Date: Fri, 25 Aug 2006 05:34:51 -0400 (EDT)
  • References: <ecbnnc$r29$1@smc.vnet.net><ecek3f$qpu$1@smc.vnet.net> <200608231115.HAA25039@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

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

Carl Woll
Wolfram Research

> =====================================================
> 
> 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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