Trigonometry/The sine of 15 degrees

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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.