Re: Trig Substitution

• To: mathgroup at smc.vnet.net
• Subject: [mg93458] Re: [mg93434] Trig Substitution
• From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
• Date: Sun, 9 Nov 2008 05:25:26 -0500 (EST)
• References: <200811080857.DAA14180@smc.vnet.net> <9517FF96-AEDF-42F5-8F88-84C70505D5EC@mimuw.edu.pl>

```On 8 Nov 2008, at 21:45, Andrzej Kozlowski wrote:

>
> On 8 Nov 2008, at 17:57, Bruce Colletti wrote:
>
>> Re Mathematica 6.0.3 under WinXP.
>>
>>
>>
>> Given a sum of (positive integer) powers of sin(t), I wish to cast
>> (as much
>> as possible) the sum in terms of cos t, chiefly by substituting 1 -
>> (cos
>> t)^2 for (sin t)^2 throughout.  The code below does this.
>>
>>
>>
>> X=Sin[t]+Sin[t]^2+Sin[t]^3;
>>
>>
>>
>> Y=Apply[List,X]/.{Sin[t]^n_?EvenQ:>Expand[(1-Cos[t]^2)^(n/
>> 2)],Sin[t]^n_:>Exp
>> and[(1-Cos[t]^2)^Quotient[n,2] Sin[t]]};
>>
>>
>>
>> Apply[Plus,Y]
>>
>>
>>
>> What is an easier way?  Thankx.
>>
>>
>>
>> Bruce
>>
>
> Obviously this depends on the definition of "easier". If by "easier"
> you mean a method that is shorter then yes:
>
> gs = GroebnerBasis[{x - (Sin[t] + Sin[t]^2 + Sin[t]^3),
>       Sin[t]^2 + Cos[t]^2 - 1}, {x, Sin[t], Cos[t]}]
> {Cos[t]^2 + Sin[t]^2 - 1, Sin[t]*Cos[t]^2 + Cos[t]^2 + x - 2*Sin[t]
> - 1}
>
> Solve[Last[gs] == 0, x]
> {{x -> (-Sin[t])*Cos[t]^2 - Cos[t]^2 + 2*Sin[t] + 1}}
>
> If however, by "easier" means that it also should be easy to see why
> it works, then this method probably does not quality.
>
> Andrzej Kozlowski

Actually, from the point of view of purely "mathematical elegance" it
might be better to do it without using Solve:

gb = GroebnerBasis[{x - (Sin[t] + Sin[t]^2 + Sin[t]^3),
Sin[t]^2 + Cos[t]^2 - 1}, {x, Sin[t], Cos[t]}];

Last[PolynomialReduce[x, gb, {x, Sin[t], Cos[t]}]]
(-Sin[t])*Cos[t]^2 - Cos[t]^2 + 2*Sin[t] + 1

But I suspect this does not make really any "easier" in the latter
sense above.

Andrzej Kozlowski

```

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