Re: Re: Trigonometric simplification

• To: mathgroup at smc.vnet.net
• Subject: [mg68925] Re: [mg68881] Re: Trigonometric simplification
• From: Bob Hanlon <hanlonr at cox.net>
• Date: Wed, 23 Aug 2006 07:15:58 -0400 (EDT)
• Sender: owner-wri-mathgroup at wolfram.com

```They are not equal.

d = 2 + 3*Cos[a] + Cos[3*a];

Plot[{d,1 + 2*Cos[a]^3},{a,-Pi,Pi},PlotStyle->{Blue,Red}];

LeafCount[d]

10

LeafCount[2(1 + 2*Cos[a]^3)]

10

FullSimplify[d//TrigExpand,ExcludedForms->{Cos[_]^_}]

4*Cos[a]^3 + 2

Bob Hanlon

> > Hi Carlos,
> >
> > Using TrigReduce before Simplify will do it:
> >
> > r = Tan[a]^2/(Sec[a]^2)^(3/2);
> > Simplify[TrigReduce[r], Assumptions -> {a > 0, Sec[a] > 0}]
> >
> > --> Cos[a]*Sin[a]^2
> >
> > Best regards,
> > Jean-Marc
>
> 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
>
>     1 + 2*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.
>

```

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