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I was experimenting with trig functions and their connections to roots of polynomials when I derived a sort of quadratic formula:

$$ax^2+bx+c=0\implies x=n\cos\left(\frac12\arccos\left(\frac{b^2-4ac-2a^2n^2}{2a^2n^2}\right)\right)-\frac b{2a}\tag{for all $n$}$$

And here is the little proof I composed:

$$\cos(2\arccos(x))=2x^2-1\tag{trig identity}$$

$$ax^2+bx+c=0$$

$$ay^2-\frac{b^2}{4a}+c=0\tag{$x=y-\frac b{2a}$}$$

$$y^2=\frac{b^2-4ac}{4a^2}$$

$$n^2t^2=\frac{b^2-4ac}{4a^2}\tag{$y=nt$}$$

$$2t^2=\frac{b^2-4ac}{2a^2n^2}$$

$$2t^2-1=\frac{b^2-4ac}{2a^2n^2}-1=\frac{b^2-4ac-2a^2n^2}{2a^2n^2}$$

$$\cos(2\arccos(t))=\frac{b^2-4ac-2a^2n^2}{2a^2n^2}\tag{apply trig identity}$$

$$t=\cos\left(\frac12\arccos\left(\frac{b^2-4ac-2a^2n^2}{2a^2n^2}\right)\right)$$

$$x=n\cos\left(\frac12\arccos\left(\frac{b^2-4ac-2a^2n^2}{2a^2n^2}\right)\right)-\frac b{2a}$$

Since $n$ can be anything, choose it so that we don't deal with complex numbers, if possible. Feel free to comment on the above formula.


Anyways, there exists similar derivations for cubic equations and quartic equations, but I was wondering on, more specifically, the solution to the quintic polynomial with trigonometric functions. Is it possible? And can general formulas be made to find the roots of even higher degree polynomials using trig functions?

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  • $\begingroup$ While I did answer the question to a number of degrees, I'n still a bit wary of the last link, as there's no explicit proof (or even blatant statement) I encountered in my quick read-througg last night. I'll try to find time when possible to review the article and ones like it, and post a proof in my answer should I find time (though I'm quite busy the next few days) $\endgroup$ – Brevan Ellefsen Oct 6 '16 at 12:44
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For a warm up here, I recommend you browse the section Wikipedia has on solving the cubic equation using trigonometry. A direct link can be found here. As you mention, the quartic equation can be solved the same way. Further note that one the page for the Quintic Function we have the following quote:

In 1858 Charles Hermite showed that the Bring radical could be characterized in terms of the Jacobi theta functions and their associated elliptic modular functions, using an approach similar to the more familiar approach of solving cubic equations by means of trigonometric functions.

I think the main problem you are having here is that the Abel–Ruffini theorem is often stated as such: "There exists no algebraic solution to the general quintic". This definition leaves a little bit up to the imagination, as the definition of algebraic solution does not mention trigonometric equations. As far as I can tell, the question of whether or not any solution expressible in terms of elementary functions can be expressed in terms of roots is an open question. Take for example the following quote from Dave L. Renfro, taken from a post on Math.SE concerning "polynomials with degree 5 solvable in elementary functions" (which you might find an interesting read):

As of 1999 this was not known -- see Conjecture 2 at the bottom of p. 442 of What is a closed-form number? by Timothy Y. Chow [American Mathematical Monthly 106 #5 (May 1999), 440-448]. I suspect it's still not known, since I believe such a result would have become relatively well known given its intrinsic interest and the fact that understanding the issue at hand doesn't require a lot of advanced mathematical training.

Moreover, here we find a post linking to a couple others, explaining how we can solve the general quintic in terms of the Jacobi Theta Function (which is, in some essence, a trigonometric function) by transforming the general quintic into Brioschi form.

While this is an interesting step, this is not quite what we want. On Wikipedia's page on Closed-form expressions we find the following quote:

Similarly solutions of cubic and quartic (third and fourth degree) equations can be expressed using arithmetic, square roots, and cube roots, or alternatively using arithmetic and trigonometric functions. However, there are quintic equations without closed-form solutions using elementary functions, such as $x^5 − x + 1 = 0$.

This quote makes the claim you desire. Now, how do we prove this? The obvious way is to show that the solutions must be in terms of an antiderivative, as we have known algorithms for determining whether or not an antiderivative can be expressed in terms of elementary functions. In my searchings, I found the following paper by R. Bruce King entitled "Beyond the Quartic Equation", which seems to disprove your conjecture that any polynomials of degree $\geq 5$ can be solved in terms of trigonometric functions. The link is primarily focused on disproving your conjecture for the quintic, but mentions generalizations in a later section (although these pages appear to not be in the free preview. I'll let you know if I find a better link. I would likewise appreciate it if someone else could find such a link).

I hope this provides a comprehensive analysis of your question! If you see any glaring flaws in my answer please share and I will do my best to amend them. Hopefully this should answer your question in as simple a manner as possible!

Note: Admittedly, it's a little hard to get a hold of a good Galois Theory course while in high school, so I don't have a strong enough background to answer with that approach. The only result I use from the field seems to concern the Risch Algorithm, which is used implicitly in the final paper I share.

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  • $\begingroup$ Thank you for the very explanatory answer. I'll do my best to grasp it all. $\endgroup$ – Simply Beautiful Art Oct 6 '16 at 13:08
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    $\begingroup$ @SimpleArt sure thing, glad to help out :D let me know if you have any questions... Or any suggestions on getting a better link to replace that last one. I really want a constructive and explicit proof of this... Such an interesting question! I've actually wondered about this as well in the past. I strongly assumed that, even with all elementary functions, the problem would be impossible. Never considered proving it before yesterday though. $\endgroup$ – Brevan Ellefsen Oct 6 '16 at 14:16
  • $\begingroup$ @SimplyBeautifulArt I'm considering going back to this answer. I was never satisfied by the final paper I linked - I am fairly confident in my results, but I don't like the conclusion. $\endgroup$ – Brevan Ellefsen Feb 14 '17 at 1:40
  • $\begingroup$ Lol, thanks for all the efforts! :D $\endgroup$ – Simply Beautiful Art Feb 14 '17 at 1:42
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I would say no in the general case. Your calculations above are highly dependent on solving the quadratic equation for $x$ (completing the square, lines 2 to 4) and would already be significantly more complicated for a cubic equation.

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The formulas for solving the cubic by radicals often involves taking the cube-root of a complex number. The best way to do that is with trigonometry.

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  • $\begingroup$ Yes, I know this... found out the hard way that the radicals are outrageously complicated. :( $\endgroup$ – Simply Beautiful Art Sep 25 '16 at 14:08

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