Quadruple Angle Formulas/Sine/Proof 3

Theorem

$\sin 4 \theta = 4 \sin \theta \cos \theta - 8 \sin^3 \theta \cos \theta$

$\ds \sin {4 \theta}$

$=$

$\ds \map \sin {2 \times 2 \theta}$

$\ds $

$=$

$\ds 2 \sin 2 \theta \cos 2 \theta$

$\ds $

$=$

$\ds 2 \paren {2 \sin \theta \cos \theta} \paren {\cos^2 \theta - \sin^2 \theta}$

$\ds $

$=$

$\ds 4 \sin \theta \cos^3 \theta - 4 \sin^3 \theta \cos \theta$

$\ds $

$=$

$\ds 4 \sin \theta \paren {1 - \sin^2 \theta} \cos \theta - 4 \sin^3 \theta \cos \theta$

$\ds $

$=$

$\ds 4 \sin \theta \cos \theta - 8 \sin^3 \theta \cos \theta$

simplification

$\blacksquare$