# Triple Angle Formulas/Tangent

## Theorem

$\map \tan {3 \theta} = \dfrac {3 \tan \theta - \tan^3 \theta} {1 - 3 \tan^2 \theta}$

where $\tan$ denotes tangent.

## Proof 1

 $\displaystyle \tan \left({3 \theta}\right)$ $=$ $\displaystyle \frac {\sin \left({3 \theta}\right)} {\cos \left({3 \theta}\right)}$ 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

$\blacksquare$

## Proof 2

 $\displaystyle \tan \left({3 \theta}\right)$ $=$ $\displaystyle \dfrac {\tan \theta + \tan \left({2 \theta}\right)} {1 - \tan \theta \tan \left({2 \theta}\right)}$ Tangent of Sum $\displaystyle$ $=$ $\displaystyle \dfrac {\tan \theta + \dfrac {2\tan \theta} {1 - \tan^2 \theta} } {1 - \tan \theta \dfrac {2 \tan \theta} {1 - \tan^2 \theta} }$ Double Angle Formula for Tangent $\displaystyle$ $=$ $\displaystyle \dfrac {\tan \theta \left({1 - \tan^2 \theta}\right) + 2 \tan \theta} {\left({1 - \tan^2 \theta}\right) - 2 \tan^2 \theta}$ simplifying $\displaystyle$ $=$ $\displaystyle \dfrac {\tan \theta - \tan^3 \theta + 2 \tan \theta} {1 - \tan^2 \theta - 2 \tan^2 \theta}$ simplifying $\displaystyle$ $=$ $\displaystyle \dfrac {3 \tan \theta - \tan^3 \theta} {1 - 3 \tan^2 \theta}$ simplifying

$\blacksquare$