Summer sines

Using the identity $2\sin(a)\cos(b) = \sin(a - b) + \sin(a + b)$ with $a = 30^{\circ}$ and $b = 10^{\circ}$ we see that

$\sin(20^{\circ}) + \sin(40^{\circ}) = \sin(30^{\circ} - 10^{\circ}) + \sin(30^{\circ} + 10^{\circ}) = 2\sin(30^{\circ})\cos(10^{\circ}) = \cos(10^{\circ})$

since $\sin(30^{\circ}) = \dfrac{1}{2}$ . But as $\cos(\theta) = \sin(90^{\circ}- \theta)$ we know that $\cos(10^{\circ}) = \sin(80^{\circ})$ , and so the given expression is $\mathscr{E} = 1$ , giving us a desired answer of $\lfloor 1000\mathscr{E} \rfloor = \boxed{1000}$ .