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According to Wolfram MathWorld, $\cos(\frac{\pi}{7})$ is a root of $8x^3-4x^2-4x+1=0$. Similarly, $\cos(\frac{2\pi}{7})$ is a root of $8x^3+4x^2-4x-1=0$. What's the procedure to generate these polynomials? I understand you can solve the cubic equations and check the cosines are indeed roots. My question is how does one arrive at those polynomials in the first place?

bjorn93
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2 Answers2

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$z_1:=\cos(2\pi/7)$ is the real part of $z_2:=\exp(i2\pi/7)$. Of course $z_2$ is a zero of $z^7-1$. Factoring this, and checking which factor to use, $z_2$ is a zero of $$z^6+z^5+z^4+z^2+z+1. \tag{2}$$ and $z_2\ne 0$, so $z_2$ is a zero of $$ z^3+z^2+z+1+z^{-1}+z^{-2}+z^{-3} \tag{3}$$ Now $\overline{z_2}$ is $1/z_2$. Then from $(3)$ compute: $z_3=z_2+1/z_2$ is a zero of $$ z^3+z^2-2z-1 $$ Finally, $z_1 = z_3/2$ is a zero of $$ 8z^3+4z^2-4z-1 $$

Can you do the other one using the same method?

GEdgar
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  • For the other one, $z_1=\cos(\pi/7)$ is the real part of $z_2=\exp(i\pi/7)$. But $z_2$ is a not a zero of $z^7-1$. How do we go forward? – bjorn93 Oct 28 '19 at 18:46
  • In the other one, $z_2$ is a zero of $z^7+1$. – GEdgar Oct 28 '19 at 19:37
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    @bjorn93 yes, Reuschle first does the primes up to 100 in order, the original book is numbered page 6, while the program showing the image calls it page X, like Roman numerals. Wonderful book, with some fascinating uncertainties later https://books.google.com/books?id=wt7lgfeYqMQC&pg=PR1&lpg=PR1&dq=reuschle++tafeln+complexer+primzahlen&source=bl&ots=VGZFPrfUBn&sig=MlQ667PqXaQ9rAvLWkG3_F1rwsk&hl=en&sa=X&ved=0ahUKEwiIwtSvm9TQAhUJ-2MKHXJIA_kQ6AEIODAE#v=onepage&q=reuschle%20%20tafeln%20complexer%20primzahlen&f=false – Will Jagy Oct 28 '19 at 21:55
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By DeMoivre's theorem,

$$(\cos x+i\sin x)^7=\cos7x+i\sin7x\quad\quad(*)$$

With $x=\frac\pi7$, the right hand side of $(*)$ reduces to $-1$. Expand the binomial on the left hand side and match up the real and imaginary parts (the latter of which is $0$), leaving us with

$$\cos^7x-21\cos^5x\sin^2x+35\cos^3x\sin^4x-7\cos x\sin^6x=-1$$

Rewrite each instance of $\sin^2x$ as $1-\cos^2x$ and simplify the result; you should end up with

$$64\cos^7x-112\cos^5x+56\cos^3x-7\cos x+1=0$$

the left hand side of which can be factorized as

$$(1+\cos x)(8\cos^3x-4\cos^2x-4\cos x+1)^2=0$$

but $\cos\frac\pi7\neq-1$,

$$8\cos^3\frac\pi7-4\cos^2\frac\pi7-4\cos\frac\pi7+1=0$$

Similarly, with $x=\frac{2\pi}7$, the right hand side of $(*)$ reduces to $1$, and so on.

user170231
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