Sextuple Angle Formulas/Cosine

Theorem

$\cos 6 \theta = 32 \cos^6 \theta - 48 \cos^4 \theta + 18 \cos^2 \theta - 1$

where $\cos$ denotes cosine.

Proof

\(\ds \cos 6 \theta + i \sin 6 \theta\)

\(=\)

\(\ds \paren {\cos \theta + i \sin \theta}^6\)

De Moivre's Formula

\(\ds \)

\(=\)

\(\ds \paren {\cos \theta}^6 + \binom 6 1 \paren {\cos \theta}^5 \paren {i \sin \theta} + \binom 6 2 \paren {\cos \theta}^4 \paren {i \sin \theta}^2\)

Binomial Theorem

\(\ds \)

\(\)

\(\, \ds + \, \)

\(\ds \binom 6 3 \paren {\cos \theta}^3 \paren {i \sin \theta}^3 + \binom 6 4 \paren {\cos \theta}^2 \paren {i \sin \theta}^4 + \binom 6 5 \paren {\cos \theta} \paren {i \sin \theta}^5 + \paren {i \sin \theta}^6\)

\(\ds \)

\(=\)

\(\ds \cos^6 \theta + 6 i \cos^5 \theta \sin \theta - 15 \cos^4 \sin^2 \theta\)

substituting for binomial coefficients

\(\ds \)

\(\)

\(\, \ds - \, \)

\(\ds 20 i \cos^3 \theta \sin^3 \theta + 15 \cos^2 \theta \sin^4 \theta + 6 i \cos \theta \sin^5 \theta - \sin^6 \theta\)

and using $i^2 = -1$

\(\text {(1)}: \quad\)

\(\ds \)

\(=\)

\(\ds \cos^6 \theta - 15 \cos^4 \sin^2 \theta + 15 \cos^2 \theta \sin^4 \theta - \sin^6 \theta\)

\(\ds \)

\(\)

\(\, \ds + \, \)

\(\ds i \paren {6 \cos^5 \theta \sin \theta - 20 \cos^3 \theta \sin^3 \theta + 6 \cos \theta \sin^5 \theta}\)

rearranging

Hence:

\(\ds \cos 6 \theta\)

\(=\)

\(\ds \cos^6 \theta - 15 \cos^4 \sin^2 \theta + 15 \cos^2 \theta \sin^4 \theta - \sin^6 \theta\)

equating real parts in $(1)$

\(\ds \)

\(=\)

\(\ds \cos^6 \theta - 15 \cos^4 \paren {1 - \cos^2 \theta} + 15 \cos^2 \theta \paren {1 - \cos^2 \theta}^2 \theta - \paren {1 - \cos^2 \theta}^3\)

Sum of Squares of Sine and Cosine

\(\ds \)

\(=\)

\(\ds 32 \cos^6 \theta - 48 \cos^4 \theta + 18 \cos^2 \theta - 1\)

multiplying out and gathering terms

$\blacksquare$

Sources

• 1960: Walter Ledermann: Complex Numbers ... (previous) ... (next): $\S 2$. Geometrical Representations: Exercise $9$