Double Angle Formulas/Cosine/Proof 1

From ProofWiki
Jump to navigation Jump to search

Theorem

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


Proof

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

$\blacksquare$