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I was finding the roots of the polynomial $y=x^3-x^2-9x+1$. And I got one of the roots of the polynomial to be $$\dfrac{4\sqrt{7}}{3}\cos{\left(\dfrac{1}{3}\arccos{\dfrac{1}{\sqrt{28}}}\right)}+\dfrac{1}{3}\tag{1}$$ which can be simplified into $$2\left(\cos\frac {\pi}{7}+\cos\frac {2\pi}{7}+\cos\frac {3\pi}{7}\right)\tag{2}$$ And according to this problem, it can be further simplified into $$4\cos\frac {2\pi}{7}+1\tag{3}$$

My question is: How do you get from $(1)$ to $(2)$ to $(3)$?

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1 Answers1

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The number in (1) comes from Viète's solution of a cubic with three real roots by using the cosine triplication formula, see The solution of cubic equations on Wikibooks for a very short account. The given number should be the largest root ($\approx3.494$).

Let's see number (3). Set $x=t+1$; then the polynomial becomes $$ t^3+3t^2+3t+1-t^2-2t-1-9t-9+1=t^3+2t^2-8t-8 $$ Now set $t=2(u+u^{-1})$, to get $$ 8u^3+24u+24u^{-1}+8u^{-3}+8u^2+16+8u^{-2}-16u-16u^{-1}-8=\\ 8u^3+8u^2+8u+8+8u^{-1}+8u^{-2}+8u^{-3} $$ and equalling to $0$ gives $$ u^6+u^5+u^4+u^3+u^2+u+1=0 $$ so the roots are the seventh roots of $1$ (excluding $1$). The root with the smallest argument is $e^{2i\pi/7}$, which gives $$ t=2(u+u^{-1})=4\cos\frac{2\pi}{7} $$ so $$ x=1+4\cos\frac{2\pi}{7} $$ Check that this is the largest root ($\approx3.494$)

egreg
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  • Hm.. then how would you simplify a random solution, for example, $\frac {2\sqrt{15}}{3}\cdot\cos\left(\frac {1}{3}\arccos\left(-\frac {9\sqrt{15}}{50}\right)\right)-1$? – Frank Jul 07 '16 at 00:57
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    @Frank I wouldn't. Expression (1) can always be given explicitly for a cubic with three distinct roots, but (3) depends on the very particular equation. – egreg Jul 07 '16 at 08:39