Tangent of 15 Degrees/Proof 1
Jump to navigation
Jump to search
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 | |||||||||||
\(\ds \) | \(=\) | \(\ds 2 - \sqrt 3\) |
$\blacksquare$