Quadruple Angle Formulas/Tangent

Theorem

$\tan 4 \theta = \dfrac {4 \tan \theta - 4 \tan^3 \theta} {1 - 6 \tan^2 \theta + \tan^4 \theta}$

where $\tan$ denotes tangent.

$\ds \tan 4 \theta$

$=$

$\ds \frac {\sin 4 \theta} {\cos 4 \theta}$

$\ds $

$=$

$\ds \frac {4 \sin \theta \cos \theta - 8 \sin^3 \theta \cos \theta} {8 \cos^4 \theta - 8 \cos^2 \theta + 1}$

$\ds $

$=$

$\ds \frac {4 \frac {\tan \theta} {\cos^2 \theta} - 8 \tan^3 \theta} {8 - \frac 8 {\cos^2 \theta} + \frac 1 {\cos^4 \theta} }$

dividing top and bottom by $\cos^4 \theta$

$\ds $

$=$

$\ds \frac {4 \tan \theta \sec^2 \theta - 8 \tan^3 \theta} {8 - 8 \sec^2 \theta + \sec^4 \theta}$

$\ds $

$=$

$\ds \frac {4 \tan \theta \paren {1 + \tan^2 \theta} - 8 \tan^3 \theta} {8 - 8 \paren {1 + \tan^2 \theta} + \paren {1 + \tan^2 \theta}^2}$

$\ds $

$=$

$\ds \frac {4 \tan \theta - 4 \tan^3 \theta} {1 - 6 \tan^2 \theta + \tan^4 \theta}$

multiplying out and gathering terms

$\blacksquare$