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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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