Tangent of 15 Degrees/Proof 3
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Theorem
- $\tan 15^\circ = \tan \dfrac {\pi} {12} = 2 - \sqrt 3$
Proof
| \(\ds \tan 15 \degrees\) | \(=\) | \(\ds \frac {\sin 15 \degrees} {\cos 15 \degrees}\) | Tangent is Sine divided by Cosine | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {\frac {\sqrt 6 - \sqrt 2} 4} {\frac {\sqrt 6 + \sqrt 2} 4}\) | Sine of $15 \degrees$ and Cosine of $15 \degrees$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {\sqrt 6 - \sqrt 2} {\sqrt 6 + \sqrt 2}\) | simplifying | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {\paren {\sqrt 6 - \sqrt 2}^2} {\paren {\sqrt 6 + \sqrt 2} \paren {\sqrt 6 - \sqrt 2} }\) | 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 $4$ |
$\blacksquare$