Math Core

So far you have started with an angle and asked for a ratio: given $\theta = 30^\circ$ , find $\sin\theta$ . Real problems often run the other way. You measure the sides of a ramp or the height of a ladder and need the angle. The inverse trigonometric functions answer the question "which angle has this sine, cosine or tangent?"

The problem: too many answers

To undo a function, each output must come from exactly one input. Sine fails that test badly. Many angles have a sine of $\dfrac{1}{2}$ :

\sin\frac{\pi}{6} = \frac{1}{2}, \qquad \sin\frac{5\pi}{6} = \frac{1}{2}, \qquad \sin\frac{13\pi}{6} = \frac{1}{2}, \qquad \sin\left(-\frac{7\pi}{6}\right) = \frac{1}{2}, \ \dots

In fact, there are infinitely many. On the graph of $y = \sin x$ , every horizontal line between $y = -1$ and $y = 1$ crosses the curve over and over, so sine fails the horizontal line test. If you asked a calculator "which angle has sine $\dfrac{1}{2}$ ?", it would have no single answer to give.

The fix is to restrict the domain: keep one piece of the graph that

passes the horizontal line test (it only rises or only falls),
still produces every output value exactly once, and
includes the familiar first-quadrant angles near $0$ .

For sine, the standard choice is $-\dfrac{\pi}{2} \le x \le \dfrac{\pi}{2}$ . On that interval sine rises steadily from $-1$ to $1$ , hitting each value once.

The dashed curve is all of y = sin x. The solid piece from -π/2 to π/2 passes the horizontal line test, so it can be reversed.Open in grapher →

Inverse sine

Reversing that restricted piece gives the inverse sine function.

Definition

Inverse sine

For $-1 \le x \le 1$ , $\;y = \arcsin x$ (also written $y = \sin^{-1} x$ ) means

\sin y = x \quad\text{and}\quad -\frac{\pi}{2} \le y \le \frac{\pi}{2}.

In words, $\arcsin x$ is the angle between $-\dfrac{\pi}{2}$ and $\dfrac{\pi}{2}$ whose sine is $x$ .

So $\arcsin\dfrac{1}{2} = \dfrac{\pi}{6}$ , and not $\dfrac{5\pi}{6}$ , because only $\dfrac{\pi}{6}$ lies in the allowed interval. The domain of arcsine is $[-1, 1]$ (the possible sine values) and its range is $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ (the restricted angles). The inputs and outputs have simply traded places.

The graph of an inverse is the reflection of the original graph over the line $y = x$ :

y = arcsin x is the reflection of the restricted sine curve over y = x. It runs from (-1, -π/2) to (1, π/2).Open in grapher →

Inverse cosine and inverse tangent

Cosine needs a different interval, because on $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ it rises and then falls. Instead we use $0 \le x \le \pi$ , where cosine falls steadily from $1$ to $-1$ .

Tangent already takes every real value on the open interval $-\dfrac{\pi}{2} < x < \dfrac{\pi}{2}$ , between its two asymptotes, and it rises the whole way. That piece is the one we keep.

The three inverse functions

function	means	domain (inputs)	range (angles out)
$y = \arcsin x$	$\sin y = x$	$-1 \le x \le 1$	$-\dfrac{\pi}{2} \le y \le \dfrac{\pi}{2}$
$y = \arccos x$	$\cos y = x$	$-1 \le x \le 1$	$0 \le y \le \pi$
$y = \arctan x$	$\tan y = x$	all real numbers	$-\dfrac{\pi}{2} < y < \dfrac{\pi}{2}$

A helpful way to remember the ranges is by quadrant. Arcsine and arctangent give angles in Quadrants I and IV (the right half of the unit circle). Arccosine gives angles in Quadrants I and II (the top half). A negative input to arcsine or arctangent gives a negative angle; a negative input to arccosine gives an obtuse angle between $\dfrac{\pi}{2}$ and $\pi$ .

y = arccos x falls from (-1, π) to (1, 0). y = arctan x is defined for every x and levels off toward the horizontal asymptotes y = π/2 and y = -π/2.Open in grapher →

Worked example: Reading the definition

Find each value in radians.

$\arcsin 1$
$\arccos 0$
$\arctan 0$
$\arccos(-1)$

Solutions. For each one, ask "which angle in the allowed range has this ratio?"

The angle in $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ with sine $1$ is $\dfrac{\pi}{2}$ .
The angle in $[0, \pi]$ with cosine $0$ is $\dfrac{\pi}{2}$ .
The angle in $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$ with tangent $0$ is $0$ .
The angle in $[0, \pi]$ with cosine $-1$ is $\pi$ .

Common mistake

The $-1$ in $\sin^{-1} x$ is not an exponent. It means "inverse function," not "reciprocal":

\sin^{-1} x = \arcsin x \qquad\text{but}\qquad (\sin x)^{-1} = \frac{1}{\sin x} = \csc x.

For example, $\sin^{-1}\dfrac{1}{2} = \dfrac{\pi}{6}$ , while $\left(\sin\dfrac{\pi}{6}\right)^{-1} = \left(\dfrac{1}{2}\right)^{-1} = 2$ . When in doubt, write $\arcsin$ .

Inputs that don't work

Since sine and cosine only produce values from $-1$ to $1$ , arcsine and arccosine only accept inputs in that interval. An expression like $\arcsin 2$ is undefined: no angle has a sine of $2$ . A calculator will report an error. Arctangent has no such limit, because tangent takes every real value.

Worked example: Which are defined?

Decide whether each expression is defined: $\arccos 1.2$ , $\;\arctan 50$ , $\;\arcsin(-0.7)$ .

Solution.

$\arccos 1.2$ is undefined, since $1.2 > 1$ and no cosine is larger than $1$ .
$\arctan 50$ is defined. Arctangent accepts every real number; the answer is an angle just under $\dfrac{\pi}{2}$ .
$\arcsin(-0.7)$ is defined, since $-1 \le -0.7 \le 1$ . It is a negative angle in Quadrant IV.

Finding angles in right triangles

This is where inverse functions earn their keep. In a right triangle, pick the ratio that uses the two sides you know, then apply the matching inverse function. Set your calculator to degree mode if you want degrees (the keys are usually labeled $\sin^{-1}$ , $\cos^{-1}$ , $\tan^{-1}$ ).

Worked example: The angle of a ramp

A wheelchair ramp rises $2$ feet over a horizontal distance of $24$ feet. What angle does the ramp make with the ground, to the nearest tenth of a degree?

Solution. The rise is opposite the angle $\theta$ and the horizontal run is adjacent, so use tangent:

\tan\theta = \frac{2}{24} = \frac{1}{12}.

Since $\theta$ is an acute angle, $\theta = \arctan\dfrac{1}{12} \approx 4.763^\circ$ , or about $4.8^\circ$ .

Worked example: Using the hypotenuse

In right triangle $ABC$ with right angle $C$ , the hypotenuse is $AB = 13$ and side $BC = 5$ . Find $\angle A$ to the nearest tenth of a degree.

Solution. Side $BC$ is opposite $\angle A$ , and $AB$ is the hypotenuse, so

\sin A = \frac{5}{13} \quad\Longrightarrow\quad A = \arcsin\frac{5}{13} \approx 22.62^\circ \approx 22.6^\circ.

Check: the other acute angle is $\arccos\dfrac{5}{13} \approx 67.38^\circ$ , and $22.62^\circ + 67.38^\circ = 90^\circ$ , as it should be.

Tip

In a right triangle every acute angle is between $0^\circ$ and $90^\circ$ , which lies inside the range of all three inverse functions. So the calculator's answer is always the angle you want there. The range restrictions only start to matter for negative inputs and angles outside the first quadrant, which the next lesson covers.

Practice

Practice 1

What is the range of $y = \arcsin x$ ?

Practice 2

Find $\arccos(-1)$ in radians.