The Quintic in Trig

Note: I provide this tutorial with no guarentee of its accurracy. It is here for your own enjoyment. If you would like to use it in your own papers or theories, the work should be checked and a refrence to the author would be nice.

This is based off of the triginomic expansion of Sin[5*Theta].

This is a quintic equation of which we know one of the solutions, at least in trigomic form. If we have the Sin[5*x], then the solution is Sin[x]. But what if we have a non-triginomic value? Let's consider the equation in the form of where a == Sin[5*Theta]. We know that the ArcSin[x] == Theta and that ArcSin[a] == 5*Theta. Therefore Theta == 1/5*ArcSin[a]. Hence ArcSin[x]==1/5*ArcSin[a] or x == Sin[1/5*ArcSin[a]]. However, we have -Sin[5*Theta] in the original equation and positive a in our psuedo equation. Therefore x == -Sin[1/5*ArcSin[a]].

Alright, now we know one of the solutions! But where does that get us? This is where that annoying, yet useful idea of synthetic substitution comes in. We plug in a known solution to the coefficents and out pops an equation of smaller degree with the same roots. In our case this is a polynomial of degree 4 of which we can use the quartic solution to get the exact roots. I'll spare you the algebra since it gets quite large. In the end, the simplified solution looks like this:

Cool, huh?

As an example, the exact roots of 16*x^5-20*x^3+5*x+2 == 0 are:

As far as I can tell, this works for all values of a, however, I have no way of proving this so keep that in mind.

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