The value of $\sin^{-1}\left(\dfrac{1}{2}\right)$ is:

$\sin\theta = \dfrac{1}{2}$ has the principal solution $\theta = \dfrac{\pi}{6}$ (30°), which lies in the standard range $[-\pi/2, \pi/2]$. Memorise the three classic values: $\sin^{-1}\dfrac{1}{2} = \dfrac{\pi}{6}$, $\sin^{-1}\dfrac{1}{\sqrt 2} = \dfrac{\pi}{4}$, $\sin^{-1}\dfrac{\sqrt 3}{2} = \dfrac{\pi}{3}$.