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

$\blacksquare$


Sources

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