Sine of 75 Degrees

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Theorem

$\sin 75 \degrees = \sin \dfrac {5 \pi} {12} = \dfrac {\sqrt 6 + \sqrt 2} 4$

where $\sin$ denotes the sine function.


Proof 1

\(\ds \sin 75 \degrees\) \(=\) \(\ds \map \sin {60 \degrees + 15 \degrees}\)
\(\ds \) \(=\) \(\ds \sin 60 \degrees \cos 15 \degrees + \cos 60 \degrees \sin 15 \degrees\) Sine of Sum
\(\ds \) \(=\) \(\ds \paren {\frac {\sqrt 3} 2} \paren {\frac {\sqrt 6 + \sqrt 2} 4} + \paren {\frac 1 2} \paren {\dfrac {\sqrt 6 - \sqrt 2} 4}\) Sine of $60 \degrees$, Cosine of $15 \degrees$, Cosine of $60 \degrees$, Sine of $15 \degrees$
\(\ds \) \(=\) \(\ds \frac 1 8 \paren {\sqrt 3 \paren {\sqrt 6 + \sqrt 2} + \sqrt 6 - \sqrt 2}\)
\(\ds \) \(=\) \(\ds \frac 1 8 \paren {\sqrt 3 \sqrt 3 \sqrt 2 + \sqrt 3 \sqrt 2 + \sqrt 3 \sqrt 2 - \sqrt 2}\)
\(\ds \) \(=\) \(\ds \frac 1 8 \paren {3 \sqrt 2 + 2 \sqrt 3 \sqrt 2 - \sqrt 2}\)
\(\ds \) \(=\) \(\ds \frac 1 8 \paren {2 \sqrt 2 + 2 \sqrt 6}\)
\(\ds \) \(=\) \(\ds \frac {\sqrt 6 + \sqrt 2} 4\)

$\blacksquare$


Proof 2

\(\ds \sin 75 \degrees\) \(=\) \(\ds \map \cos {90 \degrees - 75 \degrees}\) Cosine of Complement equals Sine
\(\ds \) \(=\) \(\ds \cos 15^\circ\)
\(\ds \) \(=\) \(\ds \frac {\sqrt 6 + \sqrt 2} 4\) Cosine of $15 \degrees$

$\blacksquare$


Sources