Sine takes each value between $-1$ and $1$ infinitely many times, so "the angle whose sine is $\frac{1}{2}$" is not one angle: $\frac{\pi}{6}$, $\frac{5\pi}{6}$, $\frac{13\pi}{6}$ and more all work. As with any function that is not one-to-one, we restrict the domain to one piece that takes every value once, and invert that piece.
Key idea
$\arcsin x$ (also written $\sin^{-1}x$) is the angle in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ whose sine is $x$, for $-1 \le x \le 1$.
$\arccos x$ ($\cos^{-1}x$) is the angle in $[0, \pi]$ whose cosine is $x$, for $-1 \le x \le 1$.
$\arctan x$ ($\tan^{-1}x$) is the angle in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ whose tangent is $x$, for any $x$.
$\arccos x$ ($\cos^{-1}x$) is the angle in $[0, \pi]$ whose cosine is $x$, for $-1 \le x \le 1$.
$\arctan x$ ($\tan^{-1}x$) is the angle in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ whose tangent is $x$, for any $x$.
Why it works
On the unit circle: the right half (Quadrants I and IV, from $-\frac{\pi}{2}$ up to $\frac{\pi}{2}$) visits every height from $-1$ to $1$ exactly once, so it is the natural piece for sine and tangent. The top half ($0$ to $\pi$) visits every $x$-coordinate from $1$ down to $-1$ exactly once, so it is the piece for cosine. Each piece includes the familiar first-quadrant angles.
Worked examples
Worked example: A first-quadrant answer
Find $\arcsin\frac{1}{2}$.
- We need an angle in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ with sine $\frac{1}{2}$.
- $\sin\frac{\pi}{6} = \frac{1}{2}$, and $\frac{\pi}{6}$ is in the range.
Answer: $\frac{\pi}{6}$
Worked example: A negative input to arccos
Find $\arccos\left(-\frac{\sqrt{2}}{2}\right)$.
- The answer must be in $[0, \pi]$, the top half of the circle. Cosine is negative there only in Quadrant II.
- The reference angle is $\frac{\pi}{4}$, so the angle is $\pi - \frac{\pi}{4} = \frac{3\pi}{4}$.
Answer: $\frac{3\pi}{4}$
Worked example: A negative input to arctan
Find $\tan^{-1}\left(-\sqrt{3}\right)$.
- The answer is in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$; a negative tangent puts it in Quadrant IV.
- $\tan\frac{\pi}{3} = \sqrt{3}$, so the angle is $-\frac{\pi}{3}$.
Answer: $-\frac{\pi}{3}$
Common mistake
Two traps. First, $\sin^{-1}x$ is not $\frac{1}{\sin x}$; that reciprocal is $\csc x$. The $-1$ here means "inverse function," just like $f^{-1}$. Second, giving a correct angle outside the range, such as $\arcsin\left(-\frac{1}{2}\right) = \frac{7\pi}{6}$. Its sine is right, but arcsine only gives angles between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$, so the answer is $-\frac{\pi}{6}$.