unless my approach is the long way round, this involves a lot of computation and the answer is really ugly/cannot be simplified easily... just a warning! (the problem is a bash)

It is well known that $\cos(72^\circ) = \frac{\sqrt{5}-1}4 = \sin(18^\circ)$ From the pythagorean identity, $\cos(18^\circ) = \sqrt{1-\sin^2(18^\circ)} \implies \cos(18^\circ) = \frac{\sqrt{10+2\sqrt{5}}}4$ Now, $\sin(3^\circ) = \sin(18^\circ-15^\circ) = \sin(18^\circ)\cos(15^\circ)-\cos(18^\circ)\sin(15^\circ)$ Now just use half angle formula on both of $\sin(30^\circ), \cos(30^\circ)$, Im not gong to continue

Yeah, using W|A, the answer is: \[ \dfrac{ \cfrac{1}{8} \left( \sqrt{5} - 1 \right) - \cfrac{1}{4} \sqrt{ \cfrac{3}{2} \left(5 + \sqrt{5} \right) }}{ \sqrt{2} } - \dfrac{ - \cfrac{1}{8} \sqrt{3} \left( \sqrt{5} - 1\right) - \cfrac{1}{4} \sqrt{ \cfrac{1}{2} \left( 5 + \sqrt{5} \right) }}{ \sqrt{2} } \] EDIT: Fixed $\LaTeX$ Who doesn't love radical bashing? EDIT 2: Yes, I know that can be simplified, but if W|A is too lazy to do so, then why shouldn't I be too?

$\cos(15^{\circ} )= \frac{1+\sqrt{3}}{2 \sqrt{2}}$, $\sin(18^{\circ})=\frac{\sqrt{5}-1}{4} $. Using what ssilwa had and doing some simplification we can get : \[\sin(3^{\circ})=\frac{1}{16} \left(\sqrt{2} \left(1+\sqrt{3}\right) \left(\sqrt{5}-1\right)-2 \left(\sqrt{3}-1\right) \sqrt{5+\sqrt{5}}\right).\]