Double Angle Formulas/Sine/Proof 1

Theorem

$\sin 2 \theta = 2 \sin \theta \cos \theta$

Proof

\(\ds \cos 2 \theta + i \sin 2 \theta\)

\(=\)

\(\ds \paren {\cos \theta + i \sin \theta}^2\)

De Moivre's Formula

\(\ds \)

\(=\)

\(\ds \cos^2 \theta + i^2 \sin^2 \theta + 2 i \cos \theta \sin \theta\)

\(\ds \)

\(=\)

\(\ds \cos^2 \theta - \sin^2 \theta + 2 i \cos \theta \sin \theta\)

\(\ds \leadsto \ \ \)

\(\ds \sin 2 \theta\)

\(=\)

\(\ds 2 \cos \theta \sin \theta\)

equating imaginary parts

$\blacksquare$