Tangent of 15 Degrees/Proof 1

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Theorem

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


Proof

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

$\blacksquare$