Secant of 15 Degrees

From ProofWiki
Jump to navigation Jump to search

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