Inverse trig domain and range. Game on.

$\cos 60^{\circ} = \dfrac{1}{2}$ and $60^{\circ}$ lies in $[0^{\circ}, 180^{\circ}]$, so $\arccos\left(\dfrac{1}{2}\right) = 60^{\circ}$.

$\sin 30^{\circ} = \dfrac{1}{2}$ and $30^{\circ}$ lies in $[-90^{\circ}, 90^{\circ}]$, so $\arcsin\left(\dfrac{1}{2}\right) = 30^{\circ}$.

$\tan 45^{\circ} = 1$ and $45^{\circ}$ lies in arctangent's principal range, so $\arctan(1) = 45^{\circ}$.

The domain of $\arcsin(x)$ is $[-1,\, 1]$.

$\tan\!\left(\dfrac{\pi}{4}\right) = 1$, so $\arctan(1) = \dfrac{\pi}{4}$.

$\cos 90^{\circ} = 0$ and $90^{\circ}$ lies in $[0^{\circ}, 180^{\circ}]$, so $\arccos(0) = 90^{\circ}$.

$\sin 90^{\circ} = 1$, the top of arcsine's range $[-90^{\circ}, 90^{\circ}]$, so $\arcsin(1) = 90^{\circ}$.

$\tan 0^{\circ} = 0$, so $\arctan(0) = 0^{\circ}$.