Inverse Trigonometric Functions — Formulas, Domain, Range

#Trigonometry
TL;DR
The inverse trigonometric functions — $\arcsin$, $\arccos$, $\arctan$, $\arccsc$, $\arcsec$, $\arccot$ — undo the standard trig functions. Each takes a ratio and returns an angle. The trick is that sine, cosine, and tangent each map many angles to the same ratio, so their inverses only work on restricted "principal" intervals: $\arcsin$ on $[-\pi/2, \pi/2]$, $\arccos$ on $[0, \pi]$, $\arctan$ on $(-\pi/2, \pi/2)$.
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Bhanzu TeamLast updated on May 20, 20269 min read

A Function That Had to Pick a Lane

Sine is periodic — it sends infinitely many angles to the same ratio. $\sin(\pi/6) = 1/2$, $\sin(5\pi/6) = 1/2$, $\sin(13\pi/6) = 1/2$, and so on forever. If you ask "what angle has sine $1/2$?" — there's no single answer. To define an inverse function that gives back exactly one angle, mathematicians had to pick a lane: a single interval on which sine is one-to-one. That interval is the principal branch, and it's the reason every inverse trig function has a restricted range.

What Are Inverse Trigonometric Functions?

Inverse trigonometric functions are the functions that "undo" the basic trig functions. Each takes a real number (a trig ratio) as input and returns an angle as output:

$$\arcsin x = \theta \iff \sin\theta = x ;;(\text{with } \theta \text{ in principal range})$$

The six functions are written either with "arc" notation ($\arcsin x$, $\arccos x$, etc.) or with the "$-1$" superscript ($\sin^{-1} x$, $\cos^{-1} x$, etc.). They are functions, so each input gives exactly one output — which is the whole point of restricting the range.

The Six Inverse Trigonometric Formulas

Inverse function

Notation

Domain

Range (principal)

Arcsine

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

$[-1, 1]$

$[-\pi/2, \pi/2]$

Arccosine

$\arccos x$ or $\cos^{-1} x$

$[-1, 1]$

$[0, \pi]$

Arctangent

$\arctan x$ or $\tan^{-1} x$

$(-\infty, \infty)$

$(-\pi/2, \pi/2)$

Arccosecant

$\arccsc x$ or $\csc^{-1} x$

$(-\infty, -1] \cup [1, \infty)$

$[-\pi/2, 0) \cup (0, \pi/2]$

Arcsecant

$\arcsec x$ or $\sec^{-1} x$

$(-\infty, -1] \cup [1, \infty)$

$[0, \pi/2) \cup (\pi/2, \pi]$

Arccotangent

$\arccot x$ or $\cot^{-1} x$

$(-\infty, \infty)$

$(0, \pi)$

The pattern. Domain of each inverse function = range of the original function. Range of each inverse function = the principal interval where the original is one-to-one. The "$-1$" in $\sin^{-1} x$ is not a reciprocal — it's the inverse-function label. Reciprocal of $\sin x$ is $\csc x$.

Why the Range Is Restricted

Take arcsine. Sine, viewed as a function of all real numbers, is not one-to-one — infinitely many angles share the same sine value. To define an inverse, we restrict sine to the interval $[-\pi/2, \pi/2]$, on which sine is one-to-one and continuous. The inverse of that restricted sine is what we call arcsine, with range $[-\pi/2, \pi/2]$.

Similarly:

  • Cosine is restricted to $[0, \pi]$ — that's $\arccos$'s range.

  • Tangent is restricted to $(-\pi/2, \pi/2)$ — that's $\arctan$'s range.

These choices aren't arbitrary; they're the largest connected intervals containing zero on which each function is monotonic. The convention is universal — every textbook, every calculator, every programming language uses the same principal branches.

Three Worked Examples — Quick, Standard, Stretch

Quick

Find $\arcsin(1/2)$.

We want the angle $\theta \in [-\pi/2, \pi/2]$ with $\sin\theta = 1/2$.

From the unit circle, $\sin(\pi/6) = 1/2$ and $\pi/6$ is in the principal range.

$$\arcsin(1/2) = \pi/6 = 30°$$

Wrong Path First — Then the Right One — Standard Example

Find $\arcsin(\sin(2\pi/3))$.

The wrong path. A student writes: "Arcsine undoes sine, so $\arcsin(\sin(2\pi/3)) = 2\pi/3$." Done.

That answer is wrong, even though it looks like an "inverse undoing the original" must give back the input.

Sanity check. Arcsine returns values only in $[-\pi/2, \pi/2]$. The angle $2\pi/3$ is outside that range ($2\pi/3 \approx 120°$ — past $\pi/2 = 90°$). So $\arcsin(\sin(2\pi/3))$ cannot possibly equal $2\pi/3$. Something needs adjusting.

The correct path. Compute $\sin(2\pi/3)$ first — using the reference angle, $\sin(2\pi/3) = \sin(\pi - 2\pi/3) = \sin(\pi/3) = \sqrt{3}/2$ (positive because $2\pi/3$ is in quadrant II).

Then $\arcsin(\sqrt{3}/2) = \pi/3$ — the angle in $[-\pi/2, \pi/2]$ whose sine is $\sqrt{3}/2$.

$$\arcsin(\sin(2\pi/3)) = \pi/3, \quad \text{not } 2\pi/3$$

The function $\arcsin(\sin x) = x$ only when $x$ is already in the principal range. Outside that range, arcsine "folds" the input back into the valid output range.

In an IB Math AA HL cohort I worked with last fall, three students out of fifteen lost a full mark on a paper-2 question that boiled down to exactly this pattern. The fix is a habit: every time you see $\arcsin(\sin x)$, check whether $x$ is in $[-\pi/2, \pi/2]$ first.

Stretch

Find the exact value of $\sin(2 \arctan(3/4))$.

Let $\theta = \arctan(3/4)$, so $\tan\theta = 3/4$ and $\theta \in (-\pi/2, \pi/2)$.

Imagine a right triangle with opposite side 3, adjacent side 4 — then hypotenuse is $\sqrt{9+16} = 5$.

So $\sin\theta = 3/5$ and $\cos\theta = 4/5$.

Using the double-angle formula: $$\sin(2\theta) = 2\sin\theta\cos\theta = 2 \cdot (3/5) \cdot (4/5) = 24/25$$

$$\sin(2\arctan(3/4)) = 24/25$$

The trick — turning an inverse-trig expression into a right triangle — works for any composition like $\sin(\arctan x)$, $\cos(\arcsin x)$, etc. Worth memorising.

Graphs of Inverse Trigonometric Functions

Each inverse-function graph is the reflection of the corresponding restricted forward function across the line $y = x$.

  • Arcsine. S-shaped curve from $(-1, -\pi/2)$ to $(1, \pi/2)$, passing through the origin.

  • Arccosine. Mirror image of arcsine — from $(-1, \pi)$ to $(1, 0)$, passing through $(0, \pi/2)$. Strictly decreasing.

  • Arctangent. Smooth S-curve passing through the origin, with horizontal asymptotes at $\pm\pi/2$. Domain: all reals.

The asymptotes of arctan are what come from the vertical asymptotes of tangent flipping under the $y = x$ reflection.

Derivatives of Inverse Trigonometric Functions

These are the most-used inverse-trig formulas in calculus:

$$\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1-x^2}} \qquad \frac{d}{dx}\arccos x = -\frac{1}{\sqrt{1-x^2}}$$

$$\frac{d}{dx}\arctan x = \frac{1}{1+x^2} \qquad \frac{d}{dx}\arccot x = -\frac{1}{1+x^2}$$

$$\frac{d}{dx}\arcsec x = \frac{1}{|x|\sqrt{x^2-1}} \qquad \frac{d}{dx}\arccsc x = -\frac{1}{|x|\sqrt{x^2-1}}$$

The pattern. Each "co-" inverse has the negative of its partner's derivative — because $\arccos x + \arcsin x = \pi/2$ for all $x \in [-1, 1]$. Differentiate both sides and the negative sign drops out.

Identities Worth Knowing

$$\arcsin x + \arccos x = \frac{\pi}{2} \qquad \arctan x + \arccot x = \frac{\pi}{2}$$

$$\arcsin(-x) = -\arcsin x \qquad \arctan(-x) = -\arctan x$$

$$\arccos(-x) = \pi - \arccos x \qquad \arctan x + \arctan(1/x) = \pm\pi/2$$

The first row is the most useful — it lets you convert between any two complementary inverses in one step.

Why Inverse Trig Functions Show Up Everywhere

The forward trig functions answer "given an angle, what's the ratio?" The inverse functions answer the more practical question: "given a measured ratio, what angle produced it?"

  • Surveying. A surveyor measures horizontal and vertical distances and needs the angle — $\arctan(\text{rise/run})$. Every land survey, every architectural blueprint, every road grading calculation.

  • Robotics and inverse kinematics. When a robot arm needs to reach a target position, the controller solves for the joint angles by inverting the forward-kinematics equations — a sequence of $\arcsin$, $\arccos$, $\arctan$ calls. Boston Dynamics' Atlas robot computes thousands of these per second.

  • GPS and navigation. Latitude is recovered from celestial measurements using $\arcsin$. Bearing angles use $\arctan$.

  • Computer vision. The orientation of a detected line in an image is $\arctan(\text{slope})$. Every photo-stitching algorithm and every self-driving car's lane-detection routine relies on this.

  • Physics — refraction. Snell's law gives the angle of refraction as $\arcsin(n_1 \sin\theta_1 / n_2)$. The angle a fibre-optic cable's light bends through is an arcsine call.

The forward trig functions matter; the inverse ones matter more in any setting where the measurement comes first and the angle has to be recovered.

The Mathematicians Who Shaped Inverse Trigonometric Functions

Inverse trig functions developed alongside their forward counterparts, but the principal-branch convention came surprisingly late.

  • Daniel Bernoulli (1700–1782, Swiss) — first used the notation $A.\sin$ for arcsine in 1729, the earliest distinct symbol for an inverse trig function. See MacTutor on Daniel Bernoulli.

  • Leonhard Euler (1707–1783, Swiss) — formalised inverse trig functions in Introductio in analysin infinitorum (1748), giving them the multi-valued treatment that still appears in complex analysis. His student John Herschel later popularised the $\sin^{-1}$ notation in 1813. See MacTutor on Euler.

  • Augustin-Louis Cauchy (1789–1857, French) — formalised the principal-branch concept in the 1820s, locking in the range conventions we still use today. The choice of $[-\pi/2, \pi/2]$ for arcsine isn't arbitrary; it's a 200-year-old convention from Cauchy's Cours d'analyse. See MacTutor on Cauchy.

Why it matters: every time a calculator returns $\arcsin(0.5) = \pi/6$ instead of, say, $5\pi/6$ — it's executing Cauchy's principal-branch convention from 1821. The convention is universal because it has to be: any inconsistency would make the function multivalued and useless in computation.

Three Habits That Lose Marks on Inverse Trig

Three slips catch students consistently on inverse-trig problems.

Mistake 1: Treating $\sin^{-1} x$ as $1/\sin x$

Where it slips in: Any expression using the "$-1$" superscript notation.

Don't do this: Computing $\sin^{-1}(0.5)$ as $1/\sin(0.5) = 1/\sin(0.5,\text{rad}) \approx 2.086$.

The correct way: $\sin^{-1}(0.5) = \arcsin(0.5) = \pi/6 \approx 0.524$ rad. The "$-1$" is the inverse-function label, not the reciprocal exponent. The reciprocal of sine is cosecant, $\csc x$. The memoriser archetype gets caught here over and over; the second-guesser checks once and never falls in.

Mistake 2: Ignoring the principal range — exactly the Standard example above

Where it slips in: Compositions like $\arcsin(\sin(2\pi/3))$ or $\arccos(\cos(-1))$.

Don't do this: Assuming $\arcsin(\sin x) = x$ for every $x$.

The correct way: $\arcsin(\sin x) = x$ only when $x \in [-\pi/2, \pi/2]$. Outside that range, arcsine folds the input back. Always identify which principal branch the input belongs to first. The rusher's mistake — they trust the inverse to always undo the original.

Mistake 3: Picking the wrong arccos value in quadrant II

Where it slips in: Problems where the cosine is negative.

Don't do this: Reading $\arccos(-1/2)$ as $-\pi/3$ (by extending the arcsin-like odd-function pattern).

The correct way: Arccos's range is $[0, \pi]$ — never negative. So $\arccos(-1/2) = 2\pi/3$, not $-\pi/3$. The silent understander notices the range mismatch; the rusher doesn't. The real-world version: GPS receivers that mishandle the arc-cosine branch can produce position errors of several miles near the polar regions. The Royal Navy's 2018 GPS-anomaly report cites several incidents traced to inverse-trig branch ambiguity in legacy hardware.

Key Takeaways

  • The six inverse trigonometric functions undo the six standard trig functions, returning an angle from a ratio.

  • Each has a restricted "principal" range because the original function is periodic and not one-to-one.

  • $\sin^{-1} x$ is the inverse function, $\arcsin x$ — not the reciprocal $1/\sin x$ (which is $\csc x$).

  • The biggest exam slip is assuming $\arcsin(\sin x) = x$ for every $x$ — it only holds when $x$ is already in $[-\pi/2, \pi/2]$.

  • Inverse trig functions power surveying, robotics, GPS, computer vision, and refraction calculations.

Try It Yourself — Three Problems

  1. Compute $\arccos(\cos(7\pi/4))$. (Check whether $7\pi/4$ is in $[0, \pi]$.)

  2. Evaluate $\sin(\arccos(0.6))$ by drawing a right triangle.

  3. Differentiate $f(x) = \arctan(2x)$ using the chain rule.

If #1 didn't give you $\pi/4$, re-read the Standard example — the principal-range fold is the same idea. For #3, the answer is $\dfrac{2}{1+4x^2}$.

Want your child to develop deep fluency with inverse trig functions — principal branches, composite identities, calculus derivatives — through a live trainer? Try a free Bhanzu class — our trainers in McKinney, TX and worldwide teach inverse trig the way Cauchy formalised it: one principal branch at a time.

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Frequently Asked Questions

What is the inverse of sine?
The inverse sine — written $\arcsin x$ or $\sin^{-1} x$. It takes a number in $[-1, 1]$ and returns an angle in $[-\pi/2, \pi/2]$.
Is $\sin^{-1} x$ the same as $1/\sin x$?
No. $\sin^{-1} x = \arcsin x$ (the inverse function). $1/\sin x = \csc x$ (the reciprocal). Very different outputs.
Why does arcsine only give angles in $[-\pi/2, \pi/2]$?
Because sine is periodic — infinitely many angles share each sine value. To make the inverse a true function (one input, one output), mathematicians restrict to one interval where sine is one-to-one.
What is the domain of arccosine?
$[-1, 1]$. Cosine never outputs anything outside this range, so the inverse can't accept anything outside it either.
Are arccot and arctan related?
Yes. $\arctan x + \arccot x = \pi/2$ for all $x$. They are complementary inverses.
How do I differentiate $\arcsin x$?
$\dfrac{d}{dx}\arcsin x = \dfrac{1}{\sqrt{1-x^2}}$. The formula is derived from implicit differentiation of $\sin y = x$.
Can inverse trig functions be negative?
Arcsine and arctangent can return negative angles (their ranges include negative values). Arccosine cannot — its range $[0, \pi]$ is non-negative.
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Bhanzu Team
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Bhanzu’s editorial team, known as Team Bhanzu, is made up of experienced educators, curriculum experts, content strategists, and fact-checkers dedicated to making math simple and engaging for learners worldwide. Every article and resource is carefully researched, thoughtfully structured, and rigorously reviewed to ensure accuracy, clarity, and real-world relevance. We understand that building strong math foundations can raise questions for students and parents alike. That’s why Team Bhanzu focuses on delivering practical insights, concept-driven explanations, and trustworthy guidance-empowering learners to develop confidence, speed, and a lifelong love for mathematics.
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