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My question is to find the roots, counted with multiplicity, of the polynomial equation

$16x^5-20x^3+5x-1=0$ using the compound angle formula $\sin\left(5\theta\right)=16\sin^5\theta-20\sin^3\theta+5\sin\theta$

So after substituting $x=\sin\theta$, I get to the equation

$\sin(5\theta)=1$

Then I get an infinitude of $\theta$ values, which when I find the sines of these values, all correspond to the distinct solutions $x=1,\sin\left(\frac{\pi}{10}\right),\sin\left(-\frac{3\pi}{10}\right)$.

What I don't understand is how I can then find which roots are repeating, since the degree of the polynomial is 5 hence there must be 5 roots when counted with multiplicity.

Also if possible, is there a way to solve this polynomial using the given compound angle formula without the need to find the distinct roots and then determine the ones which repeat?

This is because the solutions to this question say $x=1,\sin\left(\frac{\pi}{10}\right),\sin\left(\frac{9\pi}{10}\right),\sin\left(\frac{13\pi}{10}\right),\sin\left(\frac{17\pi}{10}\right)$ without any reasoning, which makes me suspect I am unaware of some related theorem.

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    $\begingroup$ Is your first equation correct? Maybe it is $16x^ 5$ instead of $x^5$. $\endgroup$
    – Pryscilla Silva
    Nov 30, 2016 at 14:11
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    $\begingroup$ Technically speaking, $x=\sin \theta $ is not a correct and proper substitution since it makes an assumption that $x \in [-1,1]$. $\endgroup$
    – SchrodingersCat
    Nov 30, 2016 at 14:49

2 Answers 2

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What I don't understand is how I can then find which roots are repeating, since the degree of the polynomial is 5 hence there must be 5 roots when counted with multiplicity.

If one knows the roots, then it is possible to find their multiplicity by finding whether or not the function's derivative(s) have the same root. See this question for a proof.

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  • $\begingroup$ Yes I am aware of that theorem, but how exactly does one figure out if any of those three distinct roots are zeros of the derivative polynomial? $\endgroup$
    – Rishi
    Nov 30, 2016 at 15:04
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    $\begingroup$ The derivative polynomial is $80x^4 - 60x^2 + 5$. So you know that $1$ definitely is not a repeated root. Also, you can see that the derivative is even. So it has $2$ positive roots and $2$ negative roots. And you have figured out that the distinct roots of the original equation are definitely $\sin(\frac{\pi}{10})$ and $\sin(\frac{-3\pi}{10})$. Do you see a complete argument? $\endgroup$
    – Shraddheya Shendre
    Nov 30, 2016 at 15:13
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As already explained, the solutions are: $$x=1,\sin\left(\frac{\pi}{10}\right),\sin\left(\frac{9\pi}{10}\right),\sin\left(\frac{13\pi}{10}\right),\sin\left(\frac{17\pi}{10}\right)$$ Because, $$\sin(5\theta)=1 \Rightarrow \theta=\frac{1}{5}\left(\frac{\pi}{2}+2k\pi\right)$$ and setting $k=0,1,2,3,4$ we get all five solutions.

But $\sin\left(\frac{\pi}{10}\right)=\sin\left(\frac{9\pi}{10}\right)$ and $\sin\left(\frac{13\pi}{10}\right)=\sin\left(\frac{17\pi}{10}\right)$.

So the five roots (with multiplicity) are $$1,\sin\left(\frac{\pi}{10}\right),\sin\left(\frac{\pi}{10}\right),\sin\left(\frac{13\pi}{10}\right),\sin\left(\frac{13\pi}{10}\right)$$.

Let's explain better why if make any other choice of $k$ we will always find an angle congruent to one of those in $S=\{\pi/10,\pi/2,9\pi/10,13\pi/10,17\pi/10\}$.

We can write $k=5n+r$ with $r \in \{0,1,2,3,4\}$ and $n \in \Bbb Z$. Replacing that at the expression of $\theta$ we get:

$$\theta=\frac{\pi}{10}+\frac{2k\pi}{5}=\frac{\pi}{10}+\frac{2(5n+r)\pi}{5}=\frac{\pi}{10}+2n\pi+\frac{2r\pi}{5} \equiv \frac{\pi}{10}+\frac{2r\pi}{5} $$

The last equivalence means that the angle $\frac{\pi}{10}+2n\pi$ is congruent to $\frac{\pi}{10}$. That means that both angle stop at the same point on the trigonometric circle and then if we apply any trigonometric function at those angles we will get the same value. The algebric value are different but geometricaly they represent the same point at the circle.

It means that if we want to find the different solution for $\sin\theta$ is enough to take

$$\frac{\pi}{10}+\frac{2r\pi}{5}$$

with $r=0,1,2,3,4$.

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  • $\begingroup$ The question seems to be more like: why the specific values of $k=0,1,2,3,4$? Why not $k=-2,-1,1,2,5$? $\endgroup$
    – Shraddheya Shendre
    Nov 30, 2016 at 14:50
  • $\begingroup$ When we choose $k=0,1,2,3,4$ we get all different values of $\theta$, wich are $S={\pi/10,\pi/2,9\pi/10,13\pi/10,17\pi/10}$. If you those any other set for $k$ your new set will be inside $S$. $\endgroup$
    – Arnaldo
    Nov 30, 2016 at 14:58
  • $\begingroup$ What do you mean by "different values of $\theta$"? If we choose different $k$, then obviously the $\theta$'s obtained will be different. And then, how can the new set of $\theta$'s be "inside" the aforementioned set $S$. $\endgroup$
    – Shraddheya Shendre
    Nov 30, 2016 at 15:05
  • $\begingroup$ @ArnaldoNascimento could you please clarify what you meant by 'your new set will be inside S'? $\endgroup$
    – Rishi
    Nov 30, 2016 at 15:05
  • $\begingroup$ Note that if $k \in \mathbb{Z}, \ \mbox{with}\ k \neq 0,1,2,3$ or $4 $ then $k = 5\cdot n + r$ with $r=0,1,2,3\ \mbox{or}\ 4, n \in \mathbb{Z}$. In this case this does not matter because $sin (\frac{1}{5}( \frac{\pi}{2} + 2kpi )) = sin (\frac{1}{5}( \frac{\pi}{2} + 2rpi ))$. $\endgroup$
    – Pryscilla Silva
    Nov 30, 2016 at 15:18

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