Sine of 15 Degrees

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Theorem

$\sin 15 \degrees = \sin \dfrac \pi {12} = \dfrac {\sqrt 6 - \sqrt 2} 4$

where $\sin$ denotes the sine function.


Proof 1

\(\ds \sin 15 \degrees\) \(=\) \(\ds \sin \frac {30 \degrees} 2\)
\(\ds \) \(=\) \(\ds \sqrt {\frac {1 - \cos 30 \degrees} 2}\) Half Angle Formula for Sine: $\theta$ is in Quadrant I
\(\ds \) \(=\) \(\ds \sqrt {\frac {1 - \frac {\sqrt 3} 2} 2}\) Cosine of $30 \degrees$
\(\ds \) \(=\) \(\ds \sqrt {\frac {2 - \sqrt 3} 4}\)
\(\ds \) \(=\) \(\ds \sqrt {\frac {8 - 4 \sqrt 3} {16} }\)
\(\ds \) \(=\) \(\ds \sqrt {\frac {6 + 2 - 2 \sqrt 2 \sqrt 6} {16} }\)
\(\ds \) \(=\) \(\ds \sqrt {\frac {\paren {\sqrt 6 - \sqrt 2}^2} {16} }\)
\(\ds \) \(=\) \(\ds \frac {\sqrt 6 - \sqrt 2} 4\) positive because $\theta$ is in the first quadrant

$\blacksquare$


Proof 2

\(\ds \sin 15 \degrees\) \(=\) \(\ds \map \sin {45 \degrees - 30 \degrees}\)
\(\ds \) \(=\) \(\ds \sin 45 \degrees \cos 30 \degrees - \cos 45 \degrees \sin 30 \degrees\) Sine of Difference
\(\ds \) \(=\) \(\ds \paren {\frac {\sqrt 2} 2} \paren {\frac {\sqrt 3} 2} - \paren {\frac {\sqrt 2} 2} \paren {\dfrac 1 2}\) Sine of $45 \degrees$, Cosine of $30 \degrees$, Cosine of $45 \degrees$, Sine of $30 \degrees$
\(\ds \) \(=\) \(\ds \frac {\sqrt 6} 4 - \frac {\sqrt 2} 4\)
\(\ds \) \(=\) \(\ds \frac {\sqrt 6 - \sqrt 2} 4\)

$\blacksquare$


Sources

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