# Triple Angle Formulas/Tangent/Proof 1

$\map \tan {3 \theta} = \dfrac {3 \tan \theta - \tan^3 \theta} {1 - 3 \tan^2 \theta}$
 $\displaystyle \tan \left({3 \theta}\right)$ $=$ $\displaystyle \frac {\sin \left({3 \theta}\right)} {\cos \left({3 \theta}\right)}$ Tangent is Sine divided by Cosine $\displaystyle$ $=$ $\displaystyle \frac {3 \sin \theta - 4 \sin^3 \theta} {4 \cos^3 \theta - 3 \cos \theta}$ Triple Angle Formula for Sine and Triple Angle Formula for Cosine $\displaystyle$ $=$ $\displaystyle \frac {3 \sin \theta - 4 \sin^3 \theta} {4 \cos^3 \theta - 3 \cos \theta}\frac {\cos^3 \theta} {\cos^3 \theta}$ $\displaystyle$ $=$ $\displaystyle \frac {\frac {3 \tan \theta}{\cos^2 \theta} - 4 \tan^3 \theta} {4 - \frac {3 \cos \theta}{\cos^3 \theta} }$ Tangent is Sine divided by Cosine $\displaystyle$ $=$ $\displaystyle \frac {3 \tan \theta \sec^2 \theta - 4 \tan^3 \theta} {4 - 3 \sec^2 \theta}$ Secant is Reciprocal of Cosine $\displaystyle$ $=$ $\displaystyle \frac {3 \tan \theta \left({1 + \tan^2 \theta}\right) - 4 \tan^3 \theta} {4 - 3 \left({1 + \tan^2 \theta}\right)}$ Difference of Squares of Secant and Tangent $\displaystyle$ $=$ $\displaystyle \frac {3 \tan \theta + 3 \tan^3 \theta - 4 \tan^3 \theta} {4 - 3 - 3 \tan^2 \theta}$ multiplying out $\displaystyle$ $=$ $\displaystyle \frac {3 \tan \theta - \tan^3 \theta} {1 - 3 \tan^2 \theta}$ gathering terms
$\blacksquare$