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

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

so after some manipulation (left as an exercise),

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

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

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