Is there a formula for this?

Let $\theta = \sin^{-1} (\sin 10^\circ + \sin 50^\circ)$ , then we have:

$\begin{aligned} \sin \theta & = \sin 10^\circ + \sin 50^\circ \\ & = \sin (30-20)^\circ + \sin (30+20)^\circ \\ & = \sin 30^\circ \cos 20^\circ - \cos 30^\circ \sin 20^\circ + \sin 30^\circ \cos 20^\circ + \cos 30^\circ \sin 20^\circ \\ & = 2\sin 30^\circ \cos 20^\circ \\ & = 2 \left(\frac 12 \right) \cos 20^\circ \\ & = \cos 20^\circ = \sin (90 - 20)^\circ = \sin 70^\circ \\ \implies \theta & = \boxed{70}^\circ \end{aligned}$

Using the identity $\sin(a) + \sin(b) = 2\sin\left(\dfrac{a + b}{2}\right)\cos\left(\dfrac{a - b}{2}\right)$ with $a = 50^{\circ}$ and $b = 10^{\circ}$ yields that

$\sin(50^{\circ}) + \sin(10^{\circ}) = 2\sin(30^{\circ})\cos(20^{\circ}) = \sin(70^{\circ})$ ,

since $\sin(30^{\circ}) = \dfrac{1}{2}$ and $\cos(\theta) = \sin(90^{\circ} - \theta)$ .

Taking the range of the $\sin^{-1}$ function as $[-90^{\circ}, 90^{\circ}]$ , we obtain a final answer of $\boxed{70}$ .

$\sin 10° + \sin 50° = \sin 10° + \sin(60°-10°)$

$= \sin 10° + \frac{\sqrt{3}}{2} \cos 10° - \frac{1}{2} \sin 10°$

$= \frac{\sqrt{3}}{2} \cos 10° + \frac{1}{2} \sin 10°$

$= \sin (60° + 10°) = \sin 70°$

so $\sin^{-1} (\sin 10° + \sin 50°) = 70°$