The trig functions answer "given an angle, what's the ratio?" Their inverses answer the opposite question — "given a ratio, what's the angle?" The catch: $\sin$, $\cos$, and $\tan$ are periodic and many-to-one, so an honest inverse only exists after we surgically restrict each one to a slice where it's strictly monotone.
13 min readPrereqs: Sine cosine tangent · The unit circle · Graphs of trig functionsUpdated 2026·05·17
What you'll leave with
Why $\sin$, $\cos$, and $\tan$ have no inverse on their full domain — and the standard restriction that fixes each.
The principal-value ranges of $\arcsin$, $\arccos$, $\arctan$ and how to read them off the unit circle.
Why $\sin(\arcsin x) = x$ always, but $\arcsin(\sin\theta) = \theta$ only when $\theta$ is in $[-\pi/2,\,\pi/2]$.
The right-triangle trick for simplifying compositions like $\sin(\arccos x)$ to a clean algebraic expression.
How to use $\arcsin$, $\arccos$, $\arctan$ to find all solutions to equations like $\sin x = 0.3$.
1. Why we have to restrict first
Foundation · inverse functions in one minute
An inverse function $f^{-1}$ undoes $f$: if $f(a) = b$, then $f^{-1}(b) = a$. For this to make sense as a function, $f$ must be one-to-one — no two inputs can share an output, or $f^{-1}(b)$ wouldn't know which $a$ to return. The quick visual test: a horizontal line drawn across the graph of $f$ should hit it at most once. When a function fails that test, you can still build an inverse by chopping the domain down to a slice where it passes — and the value you get back is called the principal value of the inverse.
An inverse function only exists when the original function is one-to-one — every output comes from exactly one input. The trig functions fail this test badly.
Take $\sin$. It hits the value $\tfrac{1}{2}$ at $\theta = \tfrac{\pi}{6}$, but also at $\tfrac{5\pi}{6}$, and then again at $\tfrac{\pi}{6} + 2\pi$, and so on forever. Asking "what angle has $\sin\theta = \tfrac{1}{2}$?" has infinitely many right answers. There is no function that can pick a single one — unless we agree, in advance, on a rule.
The rule is to amputate. We throw away most of the domain of $\sin$ and keep only an interval on which it's strictly monotone. Out of the infinitely many choices, the convention picks the one that's symmetric around the origin and covers the full output range $[-1, 1]$:
On this slice, $\sin$ is strictly increasing. It has a clean inverse, which we call $\arcsin$. Every calculator's sin⁻¹ button returns values from exactly this interval; every textbook problem that asks "find the angle" expects an answer from it by default. The chosen output is the principal value.
Why this slice?
You could have restricted $\sin$ to $[\tfrac{\pi}{2}, \tfrac{3\pi}{2}]$ instead — it's also one-to-one there. The mathematics would work, but every textbook and every calculator would have to agree on the choice. The interval $[-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$ won because it contains $0$, is symmetric, and makes $\arcsin$ an odd function — small aesthetic wins that compound across all of analysis.
Cosine and tangent need their own restrictions, chosen by the same criteria: smallest interval that's monotone and covers the full range, with a preference for "starts at $0$ or is symmetric around $0$".
2. The three inverses and their ranges
$\arcsin$ (inverse sine, also written $\sin^{-1}$)
For $x \in [-1, 1]$, $\arcsin x$ is the unique angle $\theta \in \left[-\tfrac{\pi}{2},\, \tfrac{\pi}{2}\right]$ with $\sin\theta = x$.
$\arccos$ (inverse cosine, $\cos^{-1}$)
For $x \in [-1, 1]$, $\arccos x$ is the unique angle $\theta \in [0,\, \pi]$ with $\cos\theta = x$.
$\arctan$ (inverse tangent, $\tan^{-1}$)
For any real $x$, $\arctan x$ is the unique angle $\theta \in \left(-\tfrac{\pi}{2},\, \tfrac{\pi}{2}\right)$ with $\tan\theta = x$. The interval is open because $\tan$ blows up at $\pm\tfrac{\pi}{2}$.
Put them side by side:
Function
Domain (input)
Range (output)
Parent slice
$\arcsin x$
$[-1, 1]$
$[-\tfrac{\pi}{2},\, \tfrac{\pi}{2}]$
$\sin$ on $[-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$ — strictly increasing
$\arccos x$
$[-1, 1]$
$[0,\, \pi]$
$\cos$ on $[0, \pi]$ — strictly decreasing
$\arctan x$
$(-\infty,\, \infty)$
$(-\tfrac{\pi}{2},\, \tfrac{\pi}{2})$
$\tan$ on $(-\tfrac{\pi}{2}, \tfrac{\pi}{2})$ — strictly increasing
Three facts worth burning in:
$\arcsin$ and $\arccos$ refuse inputs with $|x| > 1$ — no real angle has a sine or cosine outside $[-1, 1]$.
$\arctan$ accepts every real number, because $\tan$ runs the full real line as $\theta$ sweeps across its open interval.
$\arccos$ is the odd one out — its range lives entirely above the x-axis. You will never get a negative angle from $\arccos$.
Notation hazard
The notation $\sin^{-1} x$ is universal but unfortunate: it means $\arcsin x$, the inverse function, not $(\sin x)^{-1} = \tfrac{1}{\sin x} = \csc x$. The $-1$ in $\sin^{-1}$ is functional inversion; the $-1$ in $(\sin x)^{-1}$ is reciprocation. Calculator keys labeled "sin⁻¹" mean the former.
3. The graphs of $\arcsin$ and $\arctan$
The restricted range is the single most important thing about each inverse, and a plot makes it visible at a glance. The y-axis tells you the angle you'll get out; notice that it never strays outside the prescribed band.
The output is squeezed into the dashed band; outside $|x| > 1$ the function simply doesn't exist.
$\arctan$ flattens forever — no matter how big $x$ gets, the output sidles asymptotically toward $\pi/2$ but never touches it.
4. Reading values off the unit circle
The unit-circle picture you used to compute $\sin\theta$, $\cos\theta$, $\tan\theta$ runs in reverse for inverse trig. To evaluate $\arcsin(\tfrac{1}{2})$, you ask the circle: which angles have y-coordinate $\tfrac{1}{2}$? Then you pick the one in the principal range.
For each of the three, the principal range corresponds to a different arc of the circle:
$\arcsin$ — the right half of the circle ($-\tfrac{\pi}{2}$ at the bottom, $\tfrac{\pi}{2}$ at the top). Read off the angle whose y-coordinate matches.
$\arccos$ — the top half of the circle ($0$ on the right, $\pi$ on the left). Read off the angle whose x-coordinate matches.
$\arctan$ — same arc as $\arcsin$ (right half, excluding the endpoints). Read off the angle whose slope-from-origin matches.
A small table of the values worth memorising:
$x$
$\arcsin x$
$\arccos x$
$\arctan x$
$-1$
$-\tfrac{\pi}{2}$
$\pi$
$-\tfrac{\pi}{4}$ (at $x = -1$)
$-\tfrac{\sqrt 3}{2}$
$-\tfrac{\pi}{3}$
$\tfrac{5\pi}{6}$
—
$-\tfrac{\sqrt 2}{2}$
$-\tfrac{\pi}{4}$
$\tfrac{3\pi}{4}$
—
$-\tfrac{1}{2}$
$-\tfrac{\pi}{6}$
$\tfrac{2\pi}{3}$
—
$0$
$0$
$\tfrac{\pi}{2}$
$0$
$\tfrac{1}{2}$
$\tfrac{\pi}{6}$
$\tfrac{\pi}{3}$
—
$\tfrac{\sqrt 2}{2}$
$\tfrac{\pi}{4}$
$\tfrac{\pi}{4}$
—
$\tfrac{\sqrt 3}{2}$
$\tfrac{\pi}{3}$
$\tfrac{\pi}{6}$
—
$1$
$\tfrac{\pi}{2}$
$0$
$\tfrac{\pi}{4}$ (at $x = 1$)
$\tfrac{1}{\sqrt 3}$
—
—
$\tfrac{\pi}{6}$
$\sqrt 3$
—
—
$\tfrac{\pi}{3}$
Negative inputs follow two clean rules:
$$ \arcsin(-x) = -\arcsin x, \qquad \arctan(-x) = -\arctan x \quad \text{(both odd functions)} $$
$$ \arccos(-x) = \pi - \arccos x \quad \text{(not odd — the range never goes negative)} $$
5. Direct composition: when undoing is honest
Compose in one order — outer trig, inner inverse — and the inverse really does undo the trig:
$$ \sin(\arcsin x) = x \quad \text{for all } x \in [-1, 1] $$
$$ \cos(\arccos x) = x \quad \text{for all } x \in [-1, 1] $$
$$ \tan(\arctan x) = x \quad \text{for all real } x $$
The logic: $\arcsin x$ is, by definition, the angle whose sine is $x$. Applying $\sin$ then asks for that angle's sine, which is $x$ all over again. As long as you feed $\arcsin$ a value it accepts (i.e., $|x| \le 1$), you get back exactly what you put in.
6. Inverse-of-trig: the projection trap
Reverse the order — inner trig, outer inverse — and the story changes. The composition only returns $\theta$ when $\theta$ is already inside the inverse's range. For angles outside, the inverse silently projects the answer back into its principal range.
$\arcsin\!\left(\tfrac{\sqrt 2}{2}\right) = \tfrac{\pi}{4}$, because $\tfrac{\pi}{4}$ is the one angle in $[-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$ with that sine.
So the answer is $\tfrac{\pi}{4}$, not $\tfrac{3\pi}{4}$.
The original $\tfrac{3\pi}{4}$ got lost. Sine threw away the information "which side of $\tfrac{\pi}{2}$ am I on?" because both $\tfrac{\pi}{4}$ and $\tfrac{3\pi}{4}$ have the same sine. $\arcsin$ can only return one of them, and by convention it returns the one in its range.
Pitfall
The string of symbols $\arcsin(\sin\theta)$ looks like it should always simplify to $\theta$ — but it only does so for angles in the principal range. For any other input, you must reduce to the reference angle and then place it back inside $[-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$. The composition isn't broken; it's just lossy in a specific, predictable way.
7. The triangle trick for mixed compositions
What about $\sin(\arccos x)$? The outer and inner functions don't match, so the easy cancellation doesn't apply. The cleanest way to simplify is to draw a right triangle.
Let $\theta = \arccos x$. By definition $\cos\theta = x$ and $\theta \in [0, \pi]$, which puts $\sin\theta \ge 0$. Build the right triangle that fits:
Set up so $\cos\theta = \tfrac{\text{adj}}{\text{hyp}} = \tfrac{x}{1} = x$. Pythagoras gives the missing side.
From Pythagoras, opposite $= \sqrt{1 - x^2}$. Now read $\sin\theta$ straight off:
The same picture, relabeled, generates the other common compositions:
Expression
Simplifies to
Valid for
$\sin(\arccos x)$
$\sqrt{1 - x^2}$
$|x| \le 1$
$\cos(\arcsin x)$
$\sqrt{1 - x^2}$
$|x| \le 1$
$\tan(\arcsin x)$
$\dfrac{x}{\sqrt{1 - x^2}}$
$|x| < 1$
$\sin(\arctan x)$
$\dfrac{x}{\sqrt{1 + x^2}}$
all real $x$
$\cos(\arctan x)$
$\dfrac{1}{\sqrt{1 + x^2}}$
all real $x$
Why the positive root
Every one of these formulas takes a positive square root. The reason is the principal range: $\arcsin$ and $\arctan$ live where cosine is non-negative; $\arccos$ lives where sine is non-negative. The triangle picture only sees positive side lengths, and the principal-value convention guarantees the trig output you're computing is non-negative on its own.
8. The reciprocal inverses
Cotangent, secant, and cosecant each get their own inverses. They show up less often in introductory work, but they exist and you should at least recognise them. Conventions vary slightly between sources, but the most common choices are:
In practice you almost always reach for the identity
$$ \operatorname{arcsec} x = \arccos\!\left(\tfrac{1}{x}\right) $$
(and similarly $\operatorname{arccsc} x = \arcsin\!\left(\tfrac{1}{x}\right)$, $\operatorname{arccot} x = \arctan\!\left(\tfrac{1}{x}\right)$ for $x > 0$) and reuse what you already know about $\arcsin$, $\arccos$, $\arctan$.
Convention varies
Some textbooks define $\operatorname{arcsec}$ on $[0, \pi/2) \cup [\pi, 3\pi/2)$ instead, because that interval keeps certain calculus identities simpler. If you're working from formulas in a calculus book, check which convention it uses before trusting the formula for an $\operatorname{arcsec}$ derivative.
9. Solving $\sin x = k$ and friends
Inverse trig is the start of solving a trig equation, not the end. $\arcsin$ returns one solution — the principal one — but periodic functions have infinitely many. The job is to use the inverse to anchor the first solution, then exploit the symmetry of the parent function to find the rest.
Three recipes, one for each function. The pattern is always: principal value, then a second one inside the period, then add $2\pi k$ for general $k \in \mathbb{Z}$.
For $\sin x = k$ (with $|k| \le 1$), use the supplementary-angle identity $\sin(\pi - \alpha) = \sin\alpha$:
Principal solution: $x_1 = \arcsin k$.
Supplementary solution: $x_2 = \pi - x_1$.
General solutions: $x = x_1 + 2\pi n$ or $x = x_2 + 2\pi n$ for $n \in \mathbb{Z}$.
Example. $\sin x = 0.3$. Principal: $x_1 = \arcsin(0.3) \approx 0.3047$. Supplementary: $x_2 = \pi - 0.3047 \approx 2.8369$. Both lie in $[0, 2\pi)$; everything else is $+\, 2\pi n$.
For $\cos x = k$ (with $|k| \le 1$), use the even-function identity $\cos(-\alpha) = \cos\alpha$:
General solutions: $x = \pm x_1 + 2\pi n$ for $n \in \mathbb{Z}$.
Example. $\cos x = \tfrac{1}{2}$. Principal: $x_1 = \arccos(\tfrac{1}{2}) = \tfrac{\pi}{3}$. Reflected: $-\tfrac{\pi}{3}$ (or $\tfrac{5\pi}{3}$ in $[0, 2\pi)$).
For $\tan x = k$ (any real $k$), use the period-$\pi$ identity $\tan(\alpha + \pi) = \tan\alpha$:
Principal solution: $x_1 = \arctan k$.
General solutions: $x = x_1 + \pi n$ for $n \in \mathbb{Z}$.
Example. $\tan x = -1$. Principal: $x_1 = \arctan(-1) = -\tfrac{\pi}{4}$. In $[0, 2\pi)$: take $n = 1$ to get $\tfrac{3\pi}{4}$, and $n = 2$ for $\tfrac{7\pi}{4}$.
Sanity check
Always plot the parent function and the horizontal line $y = k$ mentally. The intersections are the solutions. If your algebra is finding more or fewer than the picture, something is wrong — probably a missing or extra reflection.
10. Common pitfalls
$\sin^{-1}$ is not $\csc$
$\sin^{-1} x = \arcsin x$ (the inverse function). $(\sin x)^{-1} = \csc x = \tfrac{1}{\sin x}$ (the reciprocal). The notation is a historical accident; if you see the $-1$ as a superscript on the function name, it means inverse, not reciprocal.
$\arccos$ never returns a negative angle
The range is $[0, \pi]$ — entirely non-negative. If your work produces $\arccos x = -\tfrac{\pi}{3}$, you've made a sign error somewhere upstream. The correct value would be $\pi - \tfrac{\pi}{3} = \tfrac{2\pi}{3}$.
Out-of-range composition
$\arcsin(\sin\theta) = \theta$ is only true when $\theta$ is already in $[-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$. For other angles, the composition projects into the principal range — and you have to compute the projection by reducing to the reference angle. Same trap exists for $\arccos\circ\cos$ and $\arctan\circ\tan$.
Forgetting the other solutions
When solving $\sin x = k$, $\arcsin k$ gives only one answer. There's almost always at least one more inside $[0, 2\pi)$ (the supplementary one for sine, the reflection for cosine), plus the periodic copies. Don't write down the principal value and stop.
Radians vs degrees
Inverse trig outputs are angles, and the units depend entirely on what your calculator or library was configured to use. A function returning $0.5236$ might mean $30°$ (radians) or $0.5236°$ (degrees) — wildly different. Make the mode explicit before trusting any numerical answer.
$\arctan(y/x)$ in code is almost always wrong
For finding an angle from a point $(x, y)$ in the plane, $\arctan(y/x)$ only returns values in $(-\tfrac{\pi}{2}, \tfrac{\pi}{2})$ — it can't tell quadrant II from IV, or III from I. Use $\operatorname{atan2}(y, x)$ instead; it inspects the signs of both coordinates and returns the correct angle in $(-\pi, \pi]$.
11. Worked examples
Try each one yourself first. The point is to check whether your steps match the recipe, not just to match the final number.
Example 1 · Evaluate $\arcsin\!\left(-\tfrac{\sqrt 2}{2}\right)$
Step 1. Find a familiar angle whose sine is $\tfrac{\sqrt 2}{2}$. That's $\tfrac{\pi}{4}$.
Step 2. $\arcsin$ is odd, so $\arcsin\!\left(-\tfrac{\sqrt 2}{2}\right) = -\tfrac{\pi}{4}$.
Step 2. Tangent has period $\pi$. Add $\pi$: $-\tfrac{\pi}{4} + \pi = \tfrac{3\pi}{4}$. ✓ in range.
Step 3. Add another $\pi$: $\tfrac{3\pi}{4} + \pi = \tfrac{7\pi}{4}$. ✓ in range.
Step 4. Adding once more exceeds $2\pi$ — stop.
Answer. $x = \tfrac{3\pi}{4}$ or $\tfrac{7\pi}{4}$.
Example 7 · Ladder angle: ladder 10 ft, reaches 8 ft up the wall
Setup. Right triangle: hypotenuse $= 10$ (the ladder), vertical side $= 8$ (the wall reach). Let $\theta$ be the angle between the ladder and the ground.
Sanity. The angle is well within $[0, \tfrac{\pi}{2}]$, as expected for a ladder leaning against a wall — $\arcsin$'s principal range matches the physical reality automatically.
Sources & further reading
The treatment above synthesises standard precalculus presentations of inverse trig. If anything reads ambiguously, the primary sources are the ground truth — and the "going deeper" links pick up where this page ends.
Peer-reviewed, openly licensed textbook section. The closest thing to a canonical, rigorous source for inverse trig at the precalculus level — domain restrictions, principal ranges, and composition identities all stated carefully.
Bite-sized video lessons with practice problems and instant feedback. Best for drilling the mechanics of evaluating $\arcsin$, $\arccos$, $\arctan$ on special-angle values until they're automatic.
Concise, worked-example-heavy notes. Particularly clear on the composition cases — both the "easy" direct identities and the trickier $\arcsin(\sin\theta)$ projection.
Formal mathematical reference. Short, dense, and precise — go here for the exact statement of an identity, including the branch-cut conditions usually glossed over in introductory texts.
Broad overview covering the topic from one-variable inverses through complex-analytic extensions and the alternative conventions for $\operatorname{arcsec}$ and $\operatorname{arccsc}$. Useful for placing the topic in the wider landscape.
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