Cosecant of 15 Degrees

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Theorem

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

where $\csc$ denotes cosecant.


Proof

\(\ds \csc 15 \degrees\) \(=\) \(\ds \frac 1 {\sin 15 \degrees}\) Cosecant is Reciprocal of Sine
\(\ds \) \(=\) \(\ds \frac 1 {\frac {\sqrt 6 - \sqrt 2} 4}\) Sine of $15 \degrees$
\(\ds \) \(=\) \(\ds \frac 4 {\sqrt 6 - \sqrt 2}\) multiplying top and bottom by $4$
\(\ds \) \(=\) \(\ds \frac {4 \paren {\sqrt 6 + \sqrt 2} } {\paren {\sqrt 6 - \sqrt 2} \paren {\sqrt 6 + \sqrt 2} }\) 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