Inverse Trigonometric Functions

We need $\theta \in [0, \pi]$ with $\cos\theta = \tfrac{1}{2}$.

From the unit circle, $\cos\!\left(\tfrac{\pi}{3}\right) = \tfrac{1}{2}$.

$\tfrac{\pi}{3}$ lies in $[0,\pi]$, so $\arccos\!\left(\tfrac{1}{2}\right) = \dfrac{\pi}{3}$.

$\cos\dfrac{\pi}{2} = 0$, so $\arccos 0 = \dfrac{\pi}{2}$.

$\sin\dfrac{\pi}{6} = \dfrac{1}{2}$, and $\dfrac{\pi}{6}$ is in the principal range, so $\arcsin\dfrac{1}{2} = \dfrac{\pi}{6}$.

$\arcsin\!\left(-\dfrac{1}{2}\right) = -\arcsin\dfrac{1}{2} = -\dfrac{\pi}{6}$.

$\tan\dfrac{\pi}{4} = 1$, so $\arctan 1 = \dfrac{\pi}{4}$.

$\dfrac{2\pi}{3} \in [0, \pi]$, so $\arccos\!\left(\cos\dfrac{2\pi}{3}\right) = \dfrac{2\pi}{3}$.

$\sin\dfrac{7\pi}{6} = -\dfrac{1}{2}$. Then $\arcsin\!\left(-\dfrac{1}{2}\right) = -\dfrac{\pi}{6}$ (in the principal range).

$\tan\dfrac{\pi}{3} = \sqrt{3}$, and $\dfrac{\pi}{3} \in \!\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$, so $\arctan\sqrt{3} = \dfrac{\pi}{3}$.

$\cos^2\theta = 1 - \sin^2\theta = 1 - \dfrac{9}{25} = \dfrac{16}{25}$, and $\cos\theta \ge 0$, so $\cos\theta = \dfrac{4}{5}$.

For any $x \in [-1, 1]$, $\arcsin x + \arccos x = \dfrac{\pi}{2}$ (a standard identity, since $\arccos x = \dfrac{\pi}{2} - \arcsin x$).