Trigonometry/The sine of 15 degrees

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

If $$\theta=15^\circ$$ then
 * $$\cos(2\theta)=\cos(30^\circ)=\frac{\sqrt3}{2}$$

so after some manipulation (left as an exercise),
 * $$\sin(15^\circ)=\frac{\sqrt6-\sqrt2}{4}=\cos(75^\circ)$$
 * $$\cos(15^\circ)=\frac{\sqrt6+\sqrt2}{4}=\sin(75^\circ)$$

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