Cotangent of 15 Degrees
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Theorem
- $\cot 15 \degrees = \cot \dfrac {\pi} {12} = 2 + \sqrt 3$
where $\cot$ denotes cotangent.
Proof
| \(\ds \cot 15^\circ\) | \(=\) | \(\ds \frac {\cos 15 \degrees} {\sin 15 \degrees}\) | Cotangent is Cosine divided by Sine | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {\frac {\sqrt 6 + \sqrt 2} 4} {\frac {\sqrt 6 - \sqrt 2} 4}\) | Cosine of $15 \degrees$ and Sine 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 $4$ |
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles