Triple Angle Formulas/Cosine/2 cos 3 theta + 1/Mistake

Source Work

 * Chapter $3$: Roots of Unity:
 * Example $6$
 * Example $6$

This mistake can be seen in the $1960$ edition as published by Routledge & Kegan Paul.

Mistake

 * Show that
 * $z^6 + z^3 + 1 = \paren {z^2 - 2 z \cos \dfrac {2 \pi} 9 + 1} \paren {z^2 - 2 z \cos \dfrac {4 \pi} 9 + 1} \paren {z^2 - 2 z \cos \dfrac {8 \pi} 9 + 1} \qquad (3.14)$


 * and deduce that
 * $2 \cos 3 \theta + 1 = \paren {\cos \theta - \cos \dfrac {2 \pi} 9} \paren {\cos \theta - \cos \dfrac {4 \pi} 9} \paren {\cos \theta - \cos \dfrac {8 \pi} 9} \qquad (3.15)$


 * Use $(3.13)$ when $n = 9$ and observe that the factor that corresponds to $r = 3$, is $z^2 - 2 z \cos \dfrac {2 \pi} 3 + 1 = z^2 + z + 1$, the remaining three quadratic factors being as on the right-hand side of $(3.14)$. This expression is therefore equal to $\paren {z^9 - 1} / \paren {z - 1} \paren {z^2 + z + 1} = \paren {z^9 - 1} / \paren {z^3 - 1} = z^6 + z^3 + 1$, which proves $(3.14)$. Next, divide $(3.14)$ throughout by $z^3$ and then put $z = e^i$. With this value of $z$, $z + z^{-1} = 2 \cos \theta$, $z^3 + z^{-3} = 2 \cos 3 \theta$, and $(3.15)$ is an immediate consequence.

Correction
The substitution $z = e^i$ should instead be $z = e^{i \theta}$.

The full derivation of this exercise is given in Complex Algebra Examples: $z^6 + z^3 + 1$ and Triple Angle Formula for Cosine: $2 \cos 3 \theta + 1$.