What are the inverse trigonometric functions?

They are the inverses of the trigonometric functions: arcsine, arccosine and arctangent. They answer the reverse question — not "what is the value of the function at this angle" but "which angle has this value". On a calculator they are labelled sin⁻¹, cos⁻¹ and tan⁻¹.

Why did the domain of the sine have to be restricted?

Because the sine takes every value in [−1, 1] infinitely often — sin 30° = sin 150° = sin 390° = 1/2. The question "which angle has sine 1/2" therefore has no single answer. Restricting to [−π/2, π/2], where the sine increases, makes every value occur exactly once — and only then does reversing it produce a function.

What are the ranges of arcsin, arccos and arctan?

arcsin takes values in [−π/2, π/2], arccos in [0, π] and arctan in the open interval (−π/2, π/2). The domain of arcsin and arccos is [−1, 1], because that is all the sine and cosine ever produce; the domain of arctan is every real number.

Why does arccos have a different range from arcsin?

Because on [−π/2, π/2] the cosine is not one-to-one — cos(−60°) = cos 60°. A different interval had to be chosen on which the cosine is monotone, and the natural one is [0, π], where it decreases from 1 to −1. That is why arccos(−1/2) = 2π/3 rather than −π/3.

What is arcsin(sin 150°)?

Thirty degrees, not 150°. The arcsine returns only angles between −90° and 90°, and sin 150° = sin 30° = 1/2, so arcsin(1/2) = 30°. This is the classic trap of the topic: arcsin(sin x) gives x back only when x already lies in the range of the arcus.

Is arcsin x the same as 1/sin x?

No. 1/sin x is a reciprocal (the cosecant), while arcsin x is an inverse function — the operation that undoes the assignment. The sin⁻¹ notation on a calculator is misleading, but it means the arcsine, not a power of −1.

Why does the arctangent graph have horizontal asymptotes?

Because the tangent grows without bound as the angle approaches 90°. Reflecting the graph in the line y = x turns the vertical asymptotes of the tangent into horizontal asymptotes of the arcus: arctan x approaches π/2 for large x and −π/2 for very negative x, without ever reaching either.

What are the inverse trigonometric functions used for?

Anywhere a ratio of lengths is known and the angle is wanted: a road gradient from a percentage, a lens angle of view from a focal length, the critical angle in optics, the joint angles of a robot from a gripper position. Every such problem ends with the sin⁻¹, cos⁻¹ or tan⁻¹ key.

Advanced level

Inverse trigonometric functions

Arcsine, arccosine and arctangent answer the reverse question: which angle has this value. See why the domain had to be restricted, what the ranges of the three functions are, and when arcsin(sin x) is not x.

Before you start

This topic builds on earlier ideas. Before you start, it's worth working through the lessons below — they'll make everything click:

Trigonometric graphsA sine wave is the unit circle unrolled along an axis. See where the period of 2π comes from, why the cosine is a shifted sine, and what happens to the graph of the tangent where the function does not exist.
Trigonometric equationsThe equation sin x = 1/2 has infinitely many solutions, because the sine repeats. See how to find the base solution, how to write all the others with a parameter k, and how to pick out the ones that fall inside a given interval.

Where this is used

Real situations where you count exactly the way this lesson teaches:

Road gradients
A road sign reading "8%" means the road climbs 8 m over 100 m of horizontal distance. The angle is arctan 0.08 = 4.57°. A 12% climb is arctan 0.12 = 6.84°, and the maximum permitted gradient of a wheelchair ramp (6%) corresponds to arctan 0.06 = 3.43°. Without the arctangent a percentage cannot be turned into an angle at all.
Photography
The angle of view of a lens is 2 · arctan(d / 2f), where d is the sensor diagonal and f the focal length. For full frame (d = 43.3 mm) and a 50 mm lens that is 2 · arctan 0.433 = 2 · 23.4° = 46.8°, and for a 24 mm wide angle it is 84.1°. The angle-of-view column in every manufacturer catalogue is a column of arctangents.
Diving and optics
The critical angle at a water–air boundary is arcsin(1 / 1.33) = 48.8°. A diver looking up at more than 48.8° from the vertical no longer sees the sky but a mirror image of the bottom — the whole sky fits into a cone of 97.6°, known as Snell’s window. That single number is a value of the arcsine.
Inverse kinematics
An arm has to reach the point (30 cm, 40 cm). The base rotation is arctan(40 / 30) = 53.1° and the distance to the target is √(30² + 40²) = 50 cm. A controller computes exactly those two numbers on every move; "inverse kinematics" is the name for deriving joint angles from a gripper position — that is, for a chain of arcus functions.
Satellite dishes
A dish in central Poland aimed at a geostationary satellite over 13°E needs an elevation of about arctan(0.60) = 31°. An installer reads that number off a table or computes it from a ratio of two distances — either way it is an arctangent, because the two legs are known and the angle is what is wanted.

All formulas

Arcsine
$y = \arcsin x \iff \sin y = x \ \text{ and } \ y \in \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$
domain [−1, 1], range [−π/2, π/2]
Arccosine
$y = \arccos x \iff \cos y = x \ \text{ and } \ y \in \left[0, \pi\right]$
domain [−1, 1], range [0, π]
Arctangent
$y = \arctan x \iff \tan y = x \ \text{ and } \ y \in \left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)$
domain: every real number
Composition with the arcus inside
$\sin(\arcsin x) = x \quad \text{for } x \in [-1, 1]$
always holds within the domain of the arcus
Composition with the arcus outside
$\arcsin(\sin x) = x \quad \text{only for } x \in \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$
outside that interval the result is a DIFFERENT angle
Arcsine and arccosine together
$\arcsin x + \arccos x = \frac{\pi}{2}$
straight from the reduction formula for the complement

All the trigonometry so far answered the question: given an angle, what is the value? Practice usually asks the reverse: given a ratio of sides, what is the angle? A road sign says "8%", and the driver wants to know how many degrees that is. That is what the inverse trigonometric functions are for.

Reversing the assignment

A function assigns exactly one value to each argument. Reading it backwards — from a value to an argument — is unambiguous only when no value occurs twice.

And here is the trouble. The sine takes the value $\frac{1}{2}$ infinitely often:

\sin 30^\circ = \sin 150^\circ = \sin 390^\circ = \sin(-210^\circ) = \frac{1}{2}

The question "which angle has sine $\frac{1}{2}$ " therefore has no single answer — which means that reversing the sine is not a function. This is exactly the problem seen in trigonometric equations, where the answer was a whole infinite set.

There is one way out: restrict the domain so that every value occurs exactly once.

arcsin: the sine on [−π/2, π/2]

On the interval from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$ the sine increases — from $-1$ to $1$ , with no repeats. On that stretch the reversal is unambiguous and defines a function:

y = \arcsin x \iff \sin y = x \ \text{ and } \ y \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]

The graph of an inverse is obtained by reflecting in the line $y = x$ — because that reflection swaps the coordinates, which is exactly swapping argument and value.

The sine on its restricted interval and its reflection in the line y = x. The domain of one function is the range of the other, and the other way round.

The domain and range follow at once: the arguments of the arcsine are numbers in $[-1, 1]$ (all the sine ever produces), and its values are angles in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ .

arccos and arctan: different intervals, same idea

For the cosine the interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ will not do: the cosine is even there, so $\cos(-60^\circ) = \cos 60^\circ$ and the value $\frac{1}{2}$ occurs twice. We pick instead an interval on which the cosine decreases from $1$ to $-1$ :

y = \arccos x \iff \cos y = x \ \text{ and } \ y \in [0, \pi]

For the tangent one asymptote-free period is enough — an open interval, since at its ends the tangent does not exist:

y = \arctan x \iff \tan y = x \ \text{ and } \ y \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)

The tangent takes every real value, so the arctangent is defined on the whole line — the only one of the three with no restriction on its domain.

Reflected, the vertical asymptotes of the tangent become horizontal asymptotes of the arctangent. The graph approaches ±π/2 and never reaches them.

function	domain	range
$\arcsin x$	$[-1, 1]$	$\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
$\arccos x$	$[-1, 1]$	$[0, \pi]$
$\arctan x$	$\mathbb{R}$	$\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$

Exact values

The table of special angles is now read from right to left.

$x$	$-1$	$-\frac{\sqrt{3}}{2}$	$-\frac{1}{2}$	$0$	$\frac{1}{2}$	$\frac{\sqrt{2}}{2}$	$\frac{\sqrt{3}}{2}$	$1$
$\arcsin x$	$-\frac{\pi}{2}$	$-\frac{\pi}{3}$	$-\frac{\pi}{6}$	$0$	$\frac{\pi}{6}$	$\frac{\pi}{4}$	$\frac{\pi}{3}$	$\frac{\pi}{2}$
$\arccos x$	$\pi$	$\frac{5\pi}{6}$	$\frac{2\pi}{3}$	$\frac{\pi}{2}$	$\frac{\pi}{3}$	$\frac{\pi}{4}$	$\frac{\pi}{6}$	$0$

The two rows do not differ by accident. The reduction formula $\cos\left(\frac{\pi}{2} - \alpha\right) = \sin \alpha$ gives

\arcsin x + \arccos x = \frac{\pi}{2}

— check it against any column of the table.

Composition: two directions and one trap

The composition with the arcus inside always works, as long as the argument is in the domain:

\sin(\arcsin x) = x \quad \text{for } x \in [-1, 1]

Because $\arcsin x$ is by definition the angle whose sine is $x$ — the sine merely undoes what the arcus did.

The composition with the arcus outside does not always work:

\arcsin(\sin x) = x \quad \text{only for } x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]

For the same reason the domain had to be restricted: the arcus must return an angle from its own range, so if $x$ is not there, the result is a different angle with the same sine.

Find arcsin(sin 150°) and arccos(cos 300°).

The calculator and the sin⁻¹ key

The key labelled sin⁻¹ (or asin) is exactly the arcsine, not a power of $-1$ . That is why, asked for the angle whose sine is $\frac{1}{2}$ , a calculator answers "30" and says nothing about $150^\circ$ — not because that angle fails to fit, but because it lies outside the range of the function being computed.

This is the quiet reason why the second solution has to be added by hand in the law of sines: the machine returns one angle, because an inverse function returns one by definition.

A road has an 8% gradient. How many degrees is that?

Exercises

Where a value of an arcus is asked for, the answer is a multiple of π (e.g. π/6, 2π/3, −π/4). Where a composition is asked for, give the angle in degrees (e.g. 30) — or, when the outer function is a sine or a cosine, the fraction that comes out (e.g. −3/5).

Practice

Work through a set of exercises — they get harder as you go. At the end you'll see your score and the mistakes worth reviewing.

Exercise 1 of 8Score: 0

arcsin 0 =

Common mistakes

Reading $\arcsin x$ as $\frac{1}{\sin x}$ — the notation $\sin^{-1}$ means the inverse function, not a reciprocal. The reciprocal of the sine is the cosecant.
$\arcsin(\sin 150^\circ) = 150^\circ$ — the arcsine returns only angles in $[-90^\circ, 90^\circ]$ ; the correct answer is $30^\circ$ .
$\arccos(-\tfrac{1}{2}) = -60^\circ$ — the range of the arccosine is $[0^\circ, 180^\circ]$ , so the answer is $120^\circ$ .
Computing $\arcsin 2$ — the domain of the arcsine is $[-1, 1]$ ; outside it the expression is meaningless and a calculator returns an error.
Assuming the calculator lost a solution — an inverse function returns one value by definition; the second solution of an equation is added from a reduction formula.
Mixing degrees and radians — in RAD mode a calculator answers $\arctan 0.08$ with $0.0798$ rather than $4.57$ ; it is the same value in a different measure.

Formula card

Topic: Inverse trigonometric functions

Arcsine
$y = \arcsin x \iff \sin y = x \ \text{ and } \ y \in \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$
domain [−1, 1], range [−π/2, π/2]
Arccosine
$y = \arccos x \iff \cos y = x \ \text{ and } \ y \in \left[0, \pi\right]$
domain [−1, 1], range [0, π]
Arctangent
$y = \arctan x \iff \tan y = x \ \text{ and } \ y \in \left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)$
domain: every real number
Composition with the arcus inside
$\sin(\arcsin x) = x \quad \text{for } x \in [-1, 1]$
always holds within the domain of the arcus
Composition with the arcus outside
$\arcsin(\sin x) = x \quad \text{only for } x \in \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$
outside that interval the result is a DIFFERENT angle
Arcsine and arccosine together
$\arcsin x + \arccos x = \frac{\pi}{2}$
straight from the reduction formula for the complement

The graph of the arcsine is the graph of the sine restricted to [−π/2, π/2], reflected in the line y = x. The reflection swaps the domain with the range.

The arctangent is defined for every real number, but its values never leave the interval from −π/2 to π/2. The graph approaches two horizontal lines and never touches them.

Frequently asked questions

A luminous unit circle with a radius sweeping out an angle, a sine wave unrolling from it as a streak of light, beside a right triangle drawn in stars, against a cosmic nebula in violet and magenta

Trigonometry

Why sine doesn't depend on the size of the triangle

The height of a tower, the width of a river, the distance to a ship offshore — all of it can be computed without going anywhere near the thing being measured. The whole of trigonometry rests on one observation: the ratio of two sides in a right triangle depends only on the angle, not on how big the triangle is. That is what turns an angle into a number — and lets it be tabulated once and for all.

13 min