Secant of 15 Degrees
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Theorem
- $\sec 15 \degrees = \sec \dfrac \pi {12} = \sqrt 6 - \sqrt 2$
where $\sec$ denotes secant.
Proof
\(\ds \sec 15 \degrees\) | \(=\) | \(\ds \frac 1 {\cos 15 \degrees}\) | Secant is Reciprocal of Cosine | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 4 {\sqrt 6 + \sqrt 2}\) | Cosine of $15 \degrees$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {4 \paren {\sqrt 6 - \sqrt 2} } {\paren {\sqrt 6 + \sqrt 2} \paren {\sqrt 6 - \sqrt 2} }\) | multiplying top and bottom by $\sqrt 6 - \sqrt 2$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {4 \paren {\sqrt 6 - \sqrt 2} } {6 - 2}\) | Difference of Two Squares | |||||||||||
\(\ds \) | \(=\) | \(\ds \sqrt 6 - \sqrt 2\) | simplifying |
$\blacksquare$
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