       Re: simple trigonometric expression

• To: mathgroup at smc.vnet.net
• Subject: [mg73101] Re: simple trigonometric expression
• From: "dimitris" <dimmechan at yahoo.com>
• Date: Sat, 3 Feb 2007 04:12:06 -0500 (EST)
• References: <epv3ev\$8j7\$1@smc.vnet.net>

```Dear Don,

Thanks a lot for your response!

However, something I miss.
So, can you be more specific about what do you mean?
Thanks!

Dimitris

>Don Taylor <dont at agora.rdrop.com> wrote:

>I've used Fourier transform successfully on very complicated
>trig expressions, as long as they do not have trig functions
>in the denominator in any way.

On Feb 2, 12:23 pm, "dimitris" <dimmec... at yahoo.com> wrote:
> In another post I deal with the expression
>
> tr = Cos[2*Pi/7]*Cos[4*Pi/7]*Cos[8*Pi/7]
>
> which is actually equal to 1/8.
>
> Indeed
>
> FullSimplify[tr]
> 1/8
>
> I would like to know other ways to get the same result within
> Mathematica (if possible without the use of (Full)Simplify but just
> with thre use of TrigFactor/TrigReduce and similar built-in
> functions).
>
> I try
>
> In:=
> TrigReduce[tr]
>
> Out=
> 1/4*(1 + Cos[(2*Pi)/7] + Cos[(6*Pi)/7] + Cos[(10*Pi)/7])
>
> The expression Cos[(2*Pi)/7] + Cos[(6*Pi)/7] + Cos[(10*Pi)/7] is
> actually equal to -1/2 but I can't show this without the use of
> FullSimplify.
>
> I also try
>
> In:=
> TrigFactor[tr]
> Timing[RootReduce[%]]
>
> Out=
> 1/8*(-I - (-1)^(1/7))*(I - (-1)^(1/7))*(-I + (-1)^(2/7))*(I +
> (-1)^(2/7))*(-I + (-1)^(4/7))*(I + (-1)^(4/7))
>
> Out=
> {21.203*Second, 1/8}
>
> but the needed time did not bring very enthusiasm to me; quite the
> oppposite!
>
>
> Thanks a lot,
> Dimitris
>
> P.S.
>
> All the same, we take our chances
> Laughed at by Time
> Tricked by Circumstances
> Plus ca change
> Plus c'est la meme chose
> The more that things change
> The more they stay the same
>
> Rush 1978
> Hemispheres LP

```

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