# Double Angle Formulas/Tangent

## Theorem

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

where $\tan$ denotes tangent.

### Corollary

Let $u = \tan \dfrac \theta 2$.

Then:

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

## Proof 1

 $\ds \tan 2 \theta$ $=$ $\ds \frac {\sin 2 \theta} {\cos 2 \theta}$ Tangent is Sine divided by Cosine $\ds$ $=$ $\ds \frac {2 \cos \theta \sin \theta} {\cos^2 \theta - \sin^2 \theta}$ Double Angle Formula for Sine and Double Angle Formula for Cosine $\ds$ $=$ $\ds \frac {\frac {2 \cos \theta \sin \theta} {\cos^2 \theta} } {\frac {\cos^2 \theta - \sin^2 \theta} {\cos^2 \theta} }$ dividing numerator and denominator by $\cos^2 \theta$ $\ds$ $=$ $\ds \frac {2 \tan \theta} {1 - \tan^2 \theta}$ simplifying: Tangent is Sine divided by Cosine

$\blacksquare$

## Proof 2

 $\ds \tan 2 \theta$ $=$ $\ds \map \tan {\theta + \theta}$ $\ds$ $=$ $\ds \frac {\tan \theta + \tan \theta} {1 - \tan \theta \tan \theta}$ Tangent of Sum $\ds$ $=$ $\ds \frac {2 \tan \theta} {1 - \tan^2 \theta}$

$\blacksquare$

## Proof 3

 $\ds \frac {2 \tan \theta} {1 - \tan^2 \theta}$ $=$ $\ds \dfrac {2 i \dfrac {1 - e^{2 i \theta} } {1 + e^{2 i \theta} } } {1 - \paren {i \dfrac {1 - e^{2 i \theta} } {1 + e^{2 i \theta} } }^2}$ Tangent Exponential Formulation $\ds$ $=$ $\ds \dfrac {2 i \paren {1 - e^{2 i \theta} } \paren {1 + e^{2 i \theta} } } {\paren {1 + e^{2 i \theta} }^2 + \paren {1 - e^{2 i \theta} }^2}$ multiplying both numerator and denominator by $\paren {1 + e^{2 i \theta} }^2$; $i$ is the imaginary unit $\ds$ $=$ $\ds \dfrac {2 i \paren {1 - e^{4 i \theta} } } {1 + 2 e^{2 i \theta} + e^{4 i \theta} + 1 - 2 e^{2 i \theta} + e^{4 i \theta} }$ Difference of Two Squares, Square of Sum, Square of Difference $\ds$ $=$ $\ds \dfrac {2 i \paren {1 - e^{4 i \theta} } } {2 + 2 e^{4 i \theta} }$ simplifying $\ds$ $=$ $\ds i \dfrac {1 - e^{4 i \theta} } {1 + e^{4 i \theta} }$ simplifying $\ds$ $=$ $\ds \tan 2 \theta$ Tangent Exponential Formulation

$\blacksquare$