1. The problem is to find $\sin^{-1}(\cos 60^\circ)$.\n\n2. Recall that $\cos 60^\circ = \frac{1}{2}$.\n\n3. So the expression becomes $\sin^{-1}\left(\frac{1}{2}\right)$.\n\n4. The function $\sin^{-1}(x)$, or arcsine, gives the angle whose sine is $x$, with a range of $[-\frac{\pi}{2}, \frac{\pi}{2}]$ or $[-90^\circ, 90^\circ]$.\n\n5. We need to find an angle $\theta$ in $[-90^\circ, 90^\circ]$ such that $\sin \theta = \frac{1}{2}$.\n\n6. The angle that satisfies this is $\theta = 30^\circ$ or $\frac{\pi}{6}$.\n\n7. Therefore, $\sin^{-1}(\cos 60^\circ) = 30^\circ$.