Triple Angle Formulas/Sine/Proof 1

Theorem

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

$\ds \sin 3 \theta$

$=$

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

$\ds $

$=$

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

$\ds $

$=$

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

$\ds $

$=$

$\ds 2 \sin \theta \cos^2 \theta + \cos^2 \theta \sin \theta - \sin^3 \theta$

$\ds $

$=$

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

$\ds $

$=$

$\ds 2 \sin \theta - 2 \sin^3 \theta + \sin \theta - \sin^3 \theta - \sin^3 \theta$

multiplying out

$\ds $

$=$

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

gathering terms

$\blacksquare$

1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text V$. Trigonometry: Two more useful formulae