Re: TrigExpand Bug?
- To: mathgroup at smc.vnet.net
- Subject: [mg39084] Re: TrigExpand Bug?
- From: "News Admin" <news at news.demon.net>
- Date: Wed, 29 Jan 2003 03:35:25 -0500 (EST)
- References: <b15o8t$erh$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
David,
I don't know what the answer is but I notice the following:
expr = (1 + Cos[t] + Cos[t]^2 + Sin[t]^2)/
(2 + Cos[t] + Cos[t]^2 + Sin[t]^2);
TrigExpand[expr]
(2 + Cos[t])/(2 + Cos[t] + Cos[t]^2 + Sin[t]^2)
TrigExpand //@ expr
(2 + Cos[t])/(3 + Cos[t])
--
Allan
---------------------
Allan Hayes
Mathematica Training and Consulting
Leicester UK
hay at haystack.demon.co.uk
Voice: +44 (0)116 271 4198
Fax: +44 (0)870 164 0565
"David Park" <djmp at earthlink.net> wrote in message
news:b15o8t$erh$1 at smc.vnet.net...
> Is there a reason for this strange behavior of TrigExpand?
>
> All of these expressions simplify both the numerator and denominator.
> (Cos[t]^2 + Sin[t]^2 is replaced by 1.)
>
> expr = (1 + Cos[t]^2 + Sin[t]^2)/(2 + Cos[t]^2 + Sin[t]^2);
> TrigExpand[expr]
> 2/3
>
> expr = (1 + f[t] + Cos[t]^2 + Sin[t]^2)/(2 + f[t] + Cos[t]^2 +
> Sin[t]^2);
> TrigExpand[expr]
> (2 + f[t])/(3 + f[t])
>
> expr = (1 + Cos[t] + Cos[t]^2 + Sin[t]^2)/(2 + f[t] + Cos[t]^2 +
> Sin[t]^2);
> TrigExpand[expr]
> (2 + Cos[t])/(3 + f[t])
>
> But the following leaves the denominator untouched.
>
> expr = (1 + Cos[t] + Cos[t]^2 + Sin[t]^2)/(2 + Cos[t] + Cos[t]^2 +
> Sin[t]^2);
> TrigExpand[expr]
> (2 + Cos[t])/(2 + Cos[t] + Cos[t]^2 + Sin[t]^2)
>
> On the other hand, Simplify, which uses the trig identities works.
>
> expr = (1 + Cos[t] + Cos[t]^2 + Sin[t]^2)/(2 + Cos[t] + Cos[t]^2 +
> Sin[t]^2);
> Simplify[expr]
> (2 + Cos[t])/(3 + Cos[t])
>
> David Park
> djmp at earthlink.net
> http://home.earthlink.net/~djmp/
>
>