# Quadruple Angle Formulas/Cosine

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## Theorem

$\cos 4 \theta = 8 \cos^4 \theta - 8 \cos^2 \theta + 1$

where $\cos$ denotes cosine.

### Corollary

$\cos 4 \theta = 8 \sin^4 \theta - 8 \sin^2 \theta + 1$

### Factor Form

$\cos 4 \theta = \paren {\cos \theta - \cos \dfrac \pi 8} \paren {\cos \theta - \cos \dfrac {3 \pi} 8} \paren {\cos \theta - \cos \dfrac {5 \pi} 8} \paren {\cos \theta - \cos \dfrac {7 \pi} 8}$

## Proof 1

 $\ds \cos 4 \theta$ $=$ $\ds \cos \paren {2 \theta + 2 \theta}$ $\ds$ $=$ $\ds \cos 2 \theta \cos 2 \theta - \sin 2 \theta \sin 2 \theta$ Cosine of Sum $\ds$ $=$ $\ds \paren {\cos^2 \theta - \sin^2 \theta} \paren {\cos^2 \theta - \sin^2 \theta} - \paren {2 \sin \theta \cos \theta} \paren {2 \sin \theta \cos \theta}$ Double Angle Formulas $\ds$ $=$ $\ds \cos^4 \theta - 2 \cos^2 \theta \sin^2 \theta + \sin^4 \theta - 4 \cos^2 \theta \sin^2 \theta$ multiplying out $\ds$ $=$ $\ds \cos^4 \theta - 2 \cos^2 \theta \paren {1 - \cos^2 \theta} + \paren {1 - \cos^2 \theta}^2 - 4 \cos^2 \theta \paren {1 - \cos^2 \theta}$ Sum of Squares of Sine and Cosine $\ds$ $=$ $\ds 8 \cos^4 \theta - 8 \cos^2 \theta + 1$ multiplying out and gathering terms

$\blacksquare$

## Proof 2

We have:

 $\ds \cos 4 \theta + i \sin 4 \theta$ $=$ $\ds \paren {\cos \theta + i \sin \theta}^4$ De Moivre's Formula $\ds$ $=$ $\ds \paren {\cos \theta}^4 + \binom 4 1 \paren {\cos \theta}^3 \paren {i \sin \theta} + \binom 4 2 \paren {\cos \theta}^2 \paren {i \sin \theta}^2$ $\ds$  $\, \ds + \,$ $\ds \binom 4 3 \paren {\cos \theta} \paren {i \sin \theta}^3 + \paren {i \sin \theta}^4$ Binomial Theorem $\ds$ $=$ $\ds \cos^4 \theta + 4 i \cos^3 \theta \sin \theta - 6 \cos^2 \theta \sin^2 \theta$ substituting for binomial coefficients $\ds$  $\, \ds - \,$ $\ds 4 i \cos \theta \sin^3 \theta + \sin^4 \theta$ and using $i^2 = -1$ $\text {(1)}: \quad$ $\ds$ $=$ $\ds \cos^4 \theta - 6 \cos^2 \theta \sin^2 \theta + \sin^4 \theta$ $\ds$  $\, \ds + \,$ $\ds 4 i \cos^3 \theta \sin \theta - 4 i \cos \theta \sin^3 \theta$ rearranging

Hence:

 $\ds \cos 4 \theta$ $=$ $\ds \cos^4 \theta - 6 \cos^2 \theta \sin^2 \theta + \sin^4 \theta$ equating real parts in $(1)$ $\ds$ $=$ $\ds \cos^4 \theta - 6 \cos^2 \theta \paren {1 - \cos^2 \theta} + \paren {1 - \cos^2 \theta}^2$ Sum of Squares of Sine and Cosine $\ds$ $=$ $\ds 8 \cos^4 \theta - 8 \cos^2 \theta + 1$ multiplying out and gathering terms

$\blacksquare$