Cosecant of 15 Degrees

From ProofWiki
Jump to navigation Jump to search

Theorem

$\csc 15^\circ = \csc \dfrac \pi {12} = \sqrt 6 + \sqrt 2$

where $\csc$ denotes cosecant.


Proof

\(\displaystyle \csc 15^\circ\) \(=\) \(\displaystyle \frac 1 {\sin 15^\circ}\) Cosecant is Reciprocal of Sine
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {\frac {\sqrt 6 - \sqrt 2} 4}\) Sine of 15 Degrees
\(\displaystyle \) \(=\) \(\displaystyle \frac 4 {\sqrt 6 - \sqrt 2}\) multiplying top and bottom by $4$
\(\displaystyle \) \(=\) \(\displaystyle \frac {4 \left({\sqrt 6 + \sqrt 2}\right)} {\left({\sqrt 6 - \sqrt 2}\right) \left({\sqrt 6 + \sqrt 2}\right)}\) multiplying top and bottom by $\sqrt 6 + \sqrt 2$
\(\displaystyle \) \(=\) \(\displaystyle \frac {4 \left({\sqrt 6 + \sqrt 2}\right)} {6 - 2}\) Difference of Two Squares
\(\displaystyle \) \(=\) \(\displaystyle \sqrt 6 + \sqrt 2\) simplifying

$\blacksquare$


Sources