# Trigonometry/The sine of 15 degrees

We have

${\displaystyle \sin(\theta )={\sqrt {\frac {1-\cos(2\theta )}{2}}}}$
${\displaystyle \cos(\theta )={\sqrt {\frac {1+\cos(2\theta )}{2}}}}$

If ${\displaystyle \theta =15^{\circ }}$ then

${\displaystyle \cos(2\theta )=\cos(30^{\circ })={\frac {\sqrt {3}}{2}}}$

so after some manipulation (left as an exercise),

${\displaystyle \sin(15^{\circ })={\frac {{\sqrt {6}}-{\sqrt {2}}}{4}}=\cos(75^{\circ })}$
${\displaystyle \cos(15^{\circ })={\frac {{\sqrt {6}}+{\sqrt {2}}}{4}}=\sin(75^{\circ })}$

These results may be combined with those from the previous section to find the sines and cosines of ${\displaystyle =3^{\circ }}$ and its multiples.