Septuple Angle Formulas/Cosine

Theorem

$\cos 7 \theta = 64 \cos^7 \theta - 112 \cos^5 \theta + 56 \cos^3 \theta - 7 \cos \theta$

where $\cos$ denotes cosine.

Proof

$\ds \cos 7 \theta + i \sin 7 \theta$

$=$

$\ds \paren {\cos \theta + i \sin \theta}^7$

De Moivre's Formula

$\ds $

$=$

$\ds \paren {\cos \theta}^7 + \binom 7 1 \paren {\cos \theta}^6 \paren {i \sin \theta} + \binom 7 2 \paren {\cos \theta}^5 \paren {i \sin \theta}^2 + \binom 7 3 \paren {\cos \theta}^4 \paren {i \sin \theta}^3$

Binomial Theorem

$\ds $

$\, \ds + \, $

$\ds \binom 7 4 \paren {\cos \theta}^3 \paren {i \sin \theta}^4 + \binom 7 5 \paren {\cos \theta}^2 \paren {i \sin \theta}^5 + \binom 7 6 \paren {\cos \theta} \paren {i \sin \theta}^6 + \paren {i \sin \theta}^7$

$\ds $

$=$

$\ds \cos^7 \theta + 7 i \cos^6 \theta \sin \theta - 21 \cos^5 \sin^2 \theta - 35 i \cos^4 \theta \sin^3 \theta$

substituting for binomial coefficients

$\ds $

$\, \ds + \, $

$\ds 35 \cos^3 \theta \sin^4 \theta + 21 i \cos^2 \theta \sin^5 \theta - 7 \cos \theta \sin^6 \theta - i \sin^7 \theta$

and using $i^2 = -1$

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

$\ds $

$=$

$\ds \cos^7 \theta - 21 \cos^5 \sin^2 \theta + 35 \cos^3 \theta \sin^4 \theta - 7 \cos \theta \sin^6 \theta$

$\ds $

$\, \ds + \, $

$\ds i \paren {7 \cos^6 \theta \sin \theta - 35 \cos^4 \theta \sin^3 \theta + 21 \cos^2 \theta \sin^5 \theta - \sin^7 \theta}$

rearranging

Hence:

$\ds \cos 7 \theta$

$=$

$\ds \cos^7 \theta - 21 \cos^5 \sin^2 \theta + 35 \cos^3 \theta \sin^4 \theta - 7 \cos \theta \sin^6 \theta$

equating real parts in $(1)$

$\ds $

$=$

$\ds \cos^7 \theta - 21 \cos^5 \paren {1 - \cos^2 \theta} + 35 \cos^3 \theta \paren {1 - \cos^2 \theta}^2 - 7 \cos \theta \paren {1 - \cos^2 \theta}^3$

Sum of Squares of Sine and Cosine

$\ds $

$=$

$\ds 64 \cos^7 \theta - 112 \cos^5 \theta + 56 \cos^3 \theta - 7 \cos \theta$

multiplying out and gathering terms

$\blacksquare$