# Quintuple Angle Formulas/Tangent

## Theorem

$\tan 5 \theta = \dfrac {\tan^5 \theta - 10 \tan^3 \theta + 5 \tan \theta} {1 - 10 \tan^2 \theta + 5 \tan^4 \theta}$

where $\tan$ denotes tangent.

## Proof

 $\displaystyle \tan 5 \theta$ $=$ $\displaystyle \frac {\sin 5 \theta} {\cos 5 \theta}$ Tangent is Sine divided by Cosine $\displaystyle$ $=$ $\displaystyle \frac {5 \sin \theta - 20 \sin^3 \theta + 16 \sin^5 \theta} {16 \cos^5 \theta - 20 \cos^3 \theta + 5 \cos \theta}$ Quintuple Angle Formulas $\displaystyle$ $=$ $\displaystyle \frac {5 \frac {\tan \theta} {\cos^4 \theta} - 20 \frac {\tan^3 \theta} {\cos^2 \theta} + 16 \tan^5 \theta} {16 \cos \theta - \frac {20} {\cos^2 \theta} + \frac 5 {\cos^4 \theta} }$ dividing top and bottom by $\cos^5 \theta$ $\displaystyle$ $=$ $\displaystyle \frac {5 \tan \theta \sec^4 \theta - 20 \tan^3 \theta \sec^2 \theta + 16 \tan^5 \theta} {16 - 20 \sec^2 \theta + 5 \sec^4 \theta}$ Secant is Reciprocal of Cosine $\displaystyle$ $=$ $\displaystyle \frac {5 \tan \theta \paren {1 + \tan^2 \theta}^2 - 20 \tan^3 \theta \paren {1 + \tan^2 \theta} + 16 \tan^5 \theta} {16 - 20 \paren {1 + \tan^2 \theta} + 5 \paren {1 + \tan^2 \theta}^2}$ Difference of Squares of Secant and Tangent $\displaystyle$ $=$ $\displaystyle \frac {\tan^5 \theta - 10 \tan^3 \theta + 5 \tan \theta} {1 - 10 \tan^2 \theta + 5 \tan^4 \theta}$ multiplying out and gathering terms

$\blacksquare$