Tangent of 15 Degrees

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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 $4$

$\blacksquare$


Sources