# 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 θ = 15º 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 3º and its multiples.

Last modified on 21 August 2013, at 21:40