# Sine of 75 Degrees

## Theorem

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

where $\sin$ denotes the sine function.

## Proof 1

 $\displaystyle \sin 75^\circ$ $=$ $\displaystyle \sin \left({60^\circ + 15^\circ}\right)$ $\displaystyle$ $=$ $\displaystyle \sin 60^\circ \cos 15^\circ + \cos 60^\circ \sin 15^\circ$ Sine of Sum $\displaystyle$ $=$ $\displaystyle \left({\frac {\sqrt 3} 2}\right) \left({\frac {\sqrt 6 + \sqrt 2} 4}\right) + \left({\frac 1 2}\right) \left({\dfrac {\sqrt 6 - \sqrt 2} 4}\right)$ Sine of $60^\circ$, Cosine of $15^\circ$, Cosine of $60^\circ$, Sine of $15^\circ$ $\displaystyle$ $=$ $\displaystyle \frac 1 8 \left({\sqrt 3 \left({\sqrt 6 + \sqrt 2}\right) + \sqrt 6 - \sqrt 2}\right)$ $\displaystyle$ $=$ $\displaystyle \frac 1 8 \left({\sqrt 3 \sqrt 3 \sqrt 2 + \sqrt 3 \sqrt 2 + \sqrt 3 \sqrt 2 - \sqrt 2}\right)$ $\displaystyle$ $=$ $\displaystyle \frac 1 8 \left({3 \sqrt 2 + 2 \sqrt 3 \sqrt 2 - \sqrt 2}\right)$ $\displaystyle$ $=$ $\displaystyle \frac 1 8 \left({2 \sqrt 2 + 2 \sqrt 6}\right)$ $\displaystyle$ $=$ $\displaystyle \frac {\sqrt 6 + \sqrt 2} 4$

$\blacksquare$

## Proof 2

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

$\blacksquare$