Compound or half.

From half angle formula we have that $\sin\dfrac{\phi}{2}=\sqrt{\dfrac{1-\cos \phi}{2}}$ now set $\phi = 6^{\circ}$ . Giving us $\sin3^{\circ} =\sqrt{\dfrac{1-\cos 6^{\circ}}{2}}= \sqrt{\dfrac{1-\cos(36^{\circ} -30^{\circ})}{2}}$ Now using compound angle formula we can get $\begin{aligned} \cos(36^{\circ} -30^{\circ}) & = \cos36^{\circ}\cos30^{\circ}+\sin36^{\circ}\sin30^{\circ} \\&= \left(\dfrac{\sqrt 5 +1}{4} \right) \dfrac{\sqrt 3}{2} +\left(\dfrac{\sqrt{10-2\sqrt5}}{4}\right)\dfrac{1}{2} \\&=\dfrac{\sqrt{10-2\sqrt 5}+\sqrt 3 +\sqrt {15}}{8} \end{aligned}$ putting in original equation $\begin{aligned}\sin3^{\circ} & =\dfrac{1}{\sqrt 2} \left( \sqrt{1- \dfrac{\sqrt{10-2\sqrt 5}+\sqrt 3 +\sqrt {15}}{8} }\right)\\ & = \dfrac{1}{4}\left(\sqrt{8-\sqrt{10-2\sqrt 5}-\sqrt 3 -\sqrt {15} }\right)\end{aligned}$ Thus $A= 4, B= 8 ,C = 8 ,D = 2 ,E=5 ,F= 3, G=15$ and sum is $47$ .

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Note we have $\sin18^{\circ} =\dfrac{\sqrt 5-1}{ 4}$ then $\cos36^{\circ} =\dfrac{\sqrt 5+1}{ 4}$ from half angle formula (in the begining) and square it and solving a bit we can get $\sin36^{\circ} =\dfrac{\sqrt {10- 2\sqrt 5}}{ 4}$ .

The second way to solve is using compound angle formula for $\sin3^{\circ} = \sin(18^{\circ} -15^{\circ})$ .

Other solutions are most welcomed