Definitions and Domains

Undoing a Sine

We know $\sin 30° = \frac{1}{2}$ . Now run it backwards: which angle has a sine of $\frac{1}{2}$ ?

You might say 30°. But $150°$ works too. So does $390°$ , and $510°$ , and infinitely many others.

The yellow line is one sine value. It crosses the wave again and again, so a single value comes from many angles. To build an inverse, we have to pick just one answer.

Why Sine Has No Inverse (As Is)

An inverse function must send each output back to one input. Sine fails this badly: it repeats forever, so countless angles map to the same value.

A function can only be inverted if it is one-to-one: no two inputs share an output.

Sine, cosine, and tangent all repeat, so none of them are one-to-one. We cannot invert them until we fix that.

The Fix: Restrict the Domain

The trick is to keep only one piece of the curve, chosen so it climbs (or falls) through every possible output exactly once.

For sine, we keep the piece from $-90°$ to $90°$ , the green band in the animation. On that stretch sine rises steadily from $-1$ to $1$ , hitting each value just once.

$\arcsin x$ takes a value in $[-1, 1]$ and returns the single angle in $[-90°, 90°]$ whose sine is $x$ .

That chosen output range is called the principal branch, and its answers are principal values.

The Notation Trap

The inverse of sine is written two ways:

$\arcsin x = \sin^{-1} x$

That little $-1$ is not a power.

$\sin^{-1} x$ means the inverse function, not $\dfrac{1}{\sin x}$ .

If you want $\frac{1}{\sin x}$ , that is the cosecant, $\csc x$ . Keep them separate.

The Three Inverses

Each function gets its own restricted range, chosen so it stays one-to-one while still reaching every output.

Inverse	Input (domain)	Output (range)
$\arcsin x$	$[-1, 1]$	$[-90°, 90°]$
$\arccos x$	$[-1, 1]$	$[0°, 180°]$
$\arctan x$	all real numbers	$(-90°, 90°)$

A few things worth noticing:

$\arcsin$ and $\arccos$ only accept inputs from $-1$ to $1$ , because sine and cosine never leave that range.
$\arctan$ accepts any number, because tangent shoots off to every value.
$\arccos$ uses $0°$ to $180°$ , a different band, because cosine (not sine) is one-to-one there.

Reading Off Values

$\arcsin\frac{1}{2}$ : which angle in $[-90°, 90°]$ has sine $\frac{1}{2}$ ? That is $30°$ .

$\arccos 0$ : which angle in $[0°, 180°]$ has cosine $0$ ? That is $90°$ .

$\arctan 1$ : which angle in $(-90°, 90°)$ has tangent $1$ ? That is $45°$ .

Watch the Signs

Negative inputs land in different places for each inverse.

Expression	Answer	Where it lives
$\arcsin(-\tfrac{1}{2})$	$-30°$	below zero (sine’s range dips negative)
$\arctan(-1)$	$-45°$	below zero (tangent’s range dips negative)
$\arccos(-\tfrac{1}{2})$	$120°$	still positive, now obtuse

$\arcsin$ and $\arctan$ answer with negative angles for negative inputs. $\arccos$ never goes negative; it swings up toward $180°$ instead.