Double Angle Formulas/Cosine/Proof 1

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Theorem

$\cos 2 \theta = \cos^2 \theta - \sin^2 \theta$


Proof

\(\ds \cos 2 \theta + i \sin 2 \theta\) \(=\) \(\ds \paren {\cos \theta + i \sin \theta}^2\) De Moivre's Formula
\(\ds \) \(=\) \(\ds \cos^2 \theta + i^2 \sin^2 \theta + 2 i \cos \theta \sin \theta\)
\(\ds \) \(=\) \(\ds \cos^2 \theta - \sin^2 \theta + 2 i \cos \theta \sin \theta\)
\(\ds \leadsto \ \ \) \(\ds \cos 2 \theta\) \(=\) \(\ds \cos^2 \theta - \sin^2 \theta\) equating real parts

$\blacksquare$