# Double Angle Formulas/Cosine/Proof 1

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## 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

$\blacksquare$