Prove that the equation $x^4-4x^3-14x^2-4x+1=0$ has a root $x = \tan 9^\circ$

Question

Prove that the equation $x^4-4x^3-14x^2-4x+1=0$ has a root $x = \tan 9^\circ$. I have tried calculating $\tan 9^\circ$ and substituting it in the equation but it is pretty tough to calculate that. Can anyone help me with a hint that leads to a better approach to this problem?

Answer 1

Let $x=\tan(9^\circ)$, $y=\tan(18^\circ)$, and $z=\tan(36^\circ)$. Then$$y=\frac{2x}{1-x^2}\quad\text{and}\quad z=\frac{2y}{1-y^2}=\frac{4x-4x^3}{x^4-6 x^2+1}.$$But then\begin{align}1&=\tan(45^\circ)\\&=\frac{x+z}{1-xz}\\&=\frac{x^5-10x^3+5x}{5x^4-10x^2+1},\end{align}and therefore\begin{align}0&=\frac{x^5-10x^3+5x}{5x^4-10x^2+1}-1\\&=\frac{x^5-5x^4-10x^3+10x^2+5 x-1}{5 x^4-10 x^2+1}.\end{align}So, $x^5-5x^4-10x^3+10x^2+5x-1=0$. But$$x^5-5x^4-10x^3+10x^2+5x-1=(x-1)\left(x^4-4x^3-14x^2-4x+1\right).$$Since $x\ne1$, $x^4-4x^3-14x^2-4x+1=0$.

Answer 2

Alternative approach

$\underline{\text{Preliminary Result}}$

PR-1
$\displaystyle \sin(18^\circ) = \cos(72^\circ) = \frac{\sqrt{5} - 1}{4}.$
Proof:
See this article.

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The original equation, $~x^4-4x^3-14x^2-4x+1=0, ~$ is a symmetrical quartic equation. This means that the coefficients of the $~x^4~$ and $~x^0~$ terms are equal and that the absolute value of the coefficients of the $~x^3~$ and $~x^1~$ terms are equal.

Further, since the coefficient of the $~x^0~$ term is not equal to $~0,~$ you know that $~(x=0)~$ is not a root of the equation.

The standard approach to such a problem is to divide the original equation by $~x^2,~$ and then either use the change of variable $~u = x + \dfrac{1}{x},~$ or use the change of variable $~u = x - \dfrac{1}{x}.$

Dividing the original equation by $~x^2~$ gives

$$~\left( ~x^2 + \frac{1}{x^2} ~\right) - 4 \left( ~x + \frac{1}{x} ~\right) - 14 = 0. \tag1 $$

Note that
$~u = \left( ~x + \dfrac{1}{x} ~\right) \implies ~$ $u^2 = \left( ~x^2 + \dfrac{1}{x^2} ~\right) + 2.$

Therefore, with the change of variable, $u = \left( ~x + \dfrac{1}{x} ~\right),~$ the equation in (1) above becomes

$$u^2 - 4u - 16 = 0. \tag2 $$

So, the entire problem reduces to proving that $~\left[ ~\tan(9^\circ) + \dfrac{1}{\tan(9^\circ)} ~\right] ~$ is one of the roots of the equation in (2) above.

This will be done by using PR-1.

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The roots of (2) above are $~\displaystyle u = 2 \pm 2\sqrt{5}.$

Since $~\tan(9^\circ) > 0,~$ the problem has been reduced to using PR-1 to prove that $~\displaystyle \left[ ~\tan(9^\circ) + \dfrac{1}{\tan(9^\circ)} ~\right] = 2 + 2\sqrt{5}.$

$$2\sin(9^\circ)\cos(9^\circ) = \sin(18^\circ) = \frac{\sqrt{5} - 1}{4} = \frac{1}{\sqrt{5} + 1} \implies $$

$$\sin(9^\circ)\cos(9^\circ) = \frac{1}{2 \left( ~\sqrt{5} + 1 ~\right)} \implies $$

$$\frac{1}{\cos(9^\circ)} = 2 \left( ~\sqrt{5} + 1 ~\right) \sin(9^\circ) \implies $$

$$\sec^2(9^\circ) = \frac{1}{\cos^2(9^\circ)} = 2 \left( ~\sqrt{5} + 1 ~\right) \frac{\sin(9^\circ)}{\cos(9^\circ)} \implies $$

$$\tan^2(9^\circ) + 1 = 2 \left( ~\sqrt{5} + 1 ~\right) \tan(9^\circ) \implies $$

$$\left[ ~\tan(9^\circ) + \dfrac{1}{\tan(9^\circ)} ~\right] = 2 + 2\sqrt{5}.$$

Answer 3

Hint: $\;\tan 5\theta =\dfrac{5\tan\theta-10\tan^3\theta+\tan^5\theta}{1-10\tan^2\theta+5\tan^4\theta}\,$. For $\theta = 9^\circ$ and $\tan \theta = x\,$:

$$ 1 = \dfrac{5x-10x^3+x^5}{1-10x^2+5x^4} \;\;\iff\;\; x^5 - 5 x^4 - 10 x^3 + 10 x^2 + 5 x - 1 = 0 $$

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[ EDIT ] $\;$ References:

• Multiple angle $\tan$ formula stated here and proved here:

$$\tan(n\theta )={\frac {\sum _{k{\text{ odd}}}(-1)^{\frac {k-1}{2}}{n \choose k}\tan ^{k}\theta }{\sum _{k{\text{ even}}}(-1)^{\frac {k}{2}}{n \choose k}\tan ^{k}\theta }} $$

• General sum of angles $\tan$ formula stated here and proved here, which reduces to the sum of angles formula when $\theta_1=\theta_2=\dots=\theta_n=\theta$:

$$ {\begin{aligned}\tan \left(\sum _{i}\theta _{i}\right)&={\frac {\displaystyle \sum _{{\text{odd}}\ k\geq 1}(-1)^{\frac {k-1}{2}}\sum _{\begin{smallmatrix}A\subseteq \{\,1,2,3,\dots \,\}\\\left|A\right|=k\end{smallmatrix}}\prod _{i\in A}\tan \theta _{i}}{\displaystyle \sum _{{\text{even}}\ k\geq 0}~(-1)^{\frac {k}{2}}~~\sum _{\begin{smallmatrix}A\subseteq \{\,1,2,3,\dots \,\}\\\left|A\right|=k\end{smallmatrix}}\prod _{i\in A}\tan \theta _{i}}}={\frac {e_{1}-e_{3}+e_{5}-\cdots }{e_{0}-e_{2}+e_{4}-\cdots }}\end{aligned}} $$

Answer 4

We have $x=\tan(45^\circ-36^\circ)=\frac{1-\tan 36^{\circ}}{1+\tan 36^{\circ}}$ and thus $\tan 36^\circ=\frac{1-x}{1+x}$.

On the other hand, a well-known identity is that $\cos 36^\circ=\frac\phi 2$ where $\phi=\frac{\sqrt5 +1}{2}$ is the golden-ratio. So we have $\tan 36^{\circ}=\frac{\sqrt{4-\phi^2}}{\phi}$.

Hence, $$\frac{1-x}{1+x}=\frac{\sqrt{4-\phi^2}}{\phi}.$$ We can get the result after some algebraic manuplations: $x^4-4x^3-14x^2-4x+1=0.$