# Tangent of 15 Degrees

## Theorem

$\tan 15^\circ = \tan \dfrac {\pi} {12} = 2 - \sqrt 3$

where $\tan$ denotes tangent.

## Proof 1

 $\ds \tan 15^\circ$ $=$ $\ds \tan \frac {30^\circ} 2$ $\ds$ $=$ $\ds \frac {\sin 30^\circ} {1 + \cos 30^\circ}$ Half Angle Formula for Tangent: Corollary 1 $\ds$ $=$ $\ds \frac {\frac 1 2} {1 + \frac {\sqrt 3} 2}$ Sine of $30^\circ$ and Cosine of $30^\circ$ $\ds$ $=$ $\ds \frac {\frac 1 2} {\frac {2 + \sqrt 3} 2}$ $\ds$ $=$ $\ds \frac 1 {2 + \sqrt 3}$ $\ds$ $=$ $\ds \frac {2 - \sqrt 3} {\left({2 + \sqrt 3}\right) \left({2 - \sqrt 3}\right)}$ multiplying top and bottom by $2 - \sqrt 3$ $\ds$ $=$ $\ds \frac {2 - \sqrt 3} {4 - 3}$ Difference of Two Squares $\ds$ $=$ $\ds 2 - \sqrt 3$

$\blacksquare$

## Proof 2

 $\ds \tan 15 \degrees$ $=$ $\ds \tan \frac {30 \degrees} 2$ $\ds$ $=$ $\ds \frac {1 - \cos 30 \degrees} {\sin 30 \degrees}$ Half Angle Formula for Tangent: Corollary 2 $\ds$ $=$ $\ds \frac {1 - \frac {\sqrt 3} 2} {\frac 1 2}$ Cosine of $30 \degrees$ and Sine of $30 \degrees$ $\ds$ $=$ $\ds 2 - \sqrt 3$ multiplying top and bottom by $2$

$\blacksquare$

## Proof 3

 $\ds \tan 15^\circ$ $=$ $\ds \tan \frac {30^\circ} 2$ $\ds$ $=$ $\ds \frac {\sin 15^\circ} {\cos 15^\circ}$ Tangent is Sine divided by Cosine $\ds$ $=$ $\ds \frac {\frac {\sqrt 6 - \sqrt 2} 4} {\frac {\sqrt 6 + \sqrt 2} 4}$ Sine of $15^\circ$ and Cosine of $15^\circ$ $\ds$ $=$ $\ds \frac {\sqrt 6 - \sqrt 2} {\sqrt 6 + \sqrt 2}$ simplifying $\ds$ $=$ $\ds \frac {\left({\sqrt 6 - \sqrt 2}\right)^2} {\left({\sqrt 6 + \sqrt 2}\right) \left({\sqrt 6 - \sqrt 2}\right)}$ multiplying top and bottom by $\sqrt 6 - \sqrt 2$ $\ds$ $=$ $\ds \frac {6 - 2 \sqrt 6 \sqrt 2 + 2 } {6 - 2}$ multiplying out, and Difference of Two Squares $\ds$ $=$ $\ds \frac {8 - 4 \sqrt 3} 4$ simplifying $\ds$ $=$ $\ds 2 - \sqrt 3$ dividing top and bottom by $4$

$\blacksquare$