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

From ProofWiki
Jump to navigation Jump to search

Theorem

$2 \cos 3 \theta + 1 = \paren {\cos \theta - \cos \dfrac {2 \pi} 9} \paren {\cos \theta - \cos \dfrac {4 \pi} 9} \paren {\cos \theta - \cos \dfrac {8 \pi} 9}$


Proof

\(\displaystyle z^6 + z^3 + 1\) \(=\) \(\displaystyle \paren {z^2 - 2 z \cos \dfrac {2 \pi} 9 + 1} \paren {z^2 - 2 z \cos \dfrac {4 \pi} 9 + 1} \paren {z^2 - 2 z \cos \dfrac {8 \pi} 9 + 1}\) Complex Algebra Examples: $z^6 + z^3 + 1$
\(\displaystyle \leadsto \ \ \) \(\displaystyle z^3 + z^0 + z^{-3}\) \(=\) \(\displaystyle \paren {z - 2 \cos \dfrac {2 \pi} 9 + z^{-1} } \paren {z - 2 \cos \dfrac {4 \pi} 9 + z^{-1} } \paren {z - 2 \cos \dfrac {8 \pi} 9 + z^{-1} }\)


Setting $z = e^{i \theta}$:

\(\displaystyle e^{3 i \theta} + 1 + e^{-3 i \theta}\) \(=\) \(\displaystyle \paren {e^{i \theta} - 2 \cos \dfrac {2 \pi} 9 + e^{-i \theta} } \paren {e^{i \theta} - 2 \cos \dfrac {4 \pi} 9 + e^{-i \theta} } \paren {e^{i \theta} - 2 \cos \dfrac {8 \pi} 9 + e^{-i \theta} }\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle 2 \dfrac {e^{3 i \theta} + e^{-3 i \theta} } 2 + 1\) \(=\) \(\displaystyle \paren {2 \dfrac {e^{i \theta} + e^{-i \theta} } 2 - 2 \cos \dfrac {2 \pi} 9} \paren {2 \dfrac {e^{i \theta} + e^{-i \theta} } 2 - 2 \cos \dfrac {4 \pi} 9} \paren {2 \dfrac {e^{i \theta} + e^{-i \theta} } 2 - 2 \cos \dfrac {8 \pi} 9}\) rearranging
\(\displaystyle \leadsto \ \ \) \(\displaystyle 2 \cos 3 \theta + 1\) \(=\) \(\displaystyle \paren {2 \cos \theta - 2 \cos \dfrac {2 \pi} 9} \paren {2 \cos \theta - 2 \cos \dfrac {4 \pi} 9} \paren {2 \cos \theta - 2 \cos \dfrac {8 \pi} 9}\) Cosine Exponential Formulation

$\blacksquare$


Sources