Quadruple Angle Formulas/Sine/Proof 1

Theorem

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

$\ds \map \sin {4 \theta}$

$=$

$\ds \map \sin {3 \theta + \theta}$

$\ds $

$=$

$\ds \sin 3 \theta \cos \theta + \cos 3 \theta \sin \theta$

$\ds $

$=$

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

$\ds $

$=$

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

$\ds $

$=$

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

$\ds $

$=$

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

multiplying out and gathering terms

$\blacksquare$