       Re: Trig Substitution

• To: mathgroup at smc.vnet.net
• Subject: [mg93454] Re: [mg93434] Trig Substitution
• From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
• Date: Sun, 9 Nov 2008 05:24:40 -0500 (EST)
• References: <200811080857.DAA14180@smc.vnet.net>

```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

```

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