The graph of an inverse trig function is the graph of a restricted trig function reflected across the line y = x. That simple idea explains every feature: why arcsin is a short curve, why arctan flattens out, and why arcsec comes in two pieces. This tool plots any of the six inverse functions, shades its principal range, draws its asymptotes and endpoints, and marks the point for an x value you choose, with the exact angle when there is one.
How to use the inverse trig graphs tool
- Choose the function: arcsin, arccos, arctan, arccsc, arcsec or arccot.
- Optionally enter an x value to highlight, as a number or an expression such as
sqrt(2)/2. Leave it blank to see the plain graph. - Choose whether to label the y-axis in radians or degrees.
- Read the point on the tape and the graph, then the domain, range, asymptotes, symmetry and behavior of the function. If x lies outside the domain, the tool says so and explains why.
Domain, range and asymptotes
| Function | Domain | Range | Horizontal asymptotes |
|---|---|---|---|
| arcsin x | [−1, 1] | [−π/2, π/2] | none |
| arccos x | [−1, 1] | [0, π] | none |
| arctan x | all reals | (−π/2, π/2) | y = −π/2, y = π/2 |
| arccot x | all reals | (0, π) | y = 0, y = π |
| arcsec x | x ≤ −1 or x ≥ 1 | [0, π/2) ∪ (π/2, π] | y = π/2 |
| arccsc x | x ≤ −1 or x ≥ 1 | [−π/2, 0) ∪ (0, π/2] | y = 0 |
Each inverse undoes its function only on the restricted piece:
Worked example: arcsec(−2)
Check the domain. −2 ≤ −1, so −2 is a valid input.
Rewrite. arcsec(−2) = arccos(1/(−2)) = arccos(−1/2).
Find the angle. The angle between 0 and π whose cosine is −1/2 is 2π/3, or 120°.
Locate the point. (−2, 2π/3) sits on the left branch of the graph, above the asymptote y = π/2. As x moves further left the branch sinks toward π/2; as x approaches −1 it rises to the closed endpoint (−1, π).
Reflecting across y = x
Swapping x and y turns any graph into the graph of its inverse, which visually is a reflection across the diagonal line y = x. Take y = sin x on the interval −π/2 ≤ x ≤ π/2: it rises from (−π/2, −1) to (π/2, 1). Reflect it, and you get arcsin rising from (−1, −π/2) to (1, π/2). The same move explains the asymptotes. tan x has vertical asymptotes at x = ±π/2; after reflection they become the horizontal asymptotes y = ±π/2 of arctan.
The reflection also shows why a restriction is needed. The full sine wave fails the horizontal line test; its reflection would fail the vertical line test and would not be a function. Restricting to a piece where the function is one-to-one fixes that.
Symmetry and shape
- Odd functions: arcsin, arctan and arccsc satisfy f(−x) = −f(x). Their graphs are symmetric about the origin.
- Supplement symmetry: arccos, arccot and arcsec satisfy f(−x) = π − f(x). Their graphs are symmetric about the point (0, π/2).
- Direction: arcsin, arctan and arcsec increase; arccos, arccot and arccsc decrease.
- Steepness: arcsin and arccos become vertical at x = ±1, since their derivatives are ±1/√(1 − x²), which blow up at the endpoints. arctan is steepest at the origin, where its slope is 1.
A note on conventions
The ranges for arcsin, arccos and arctan are universal. For arccot this page uses (0, π), the most common choice in US textbooks, which makes the graph one continuous decreasing curve. Some software defines arccot x as arctan(1/x), whose graph jumps at x = 0. To evaluate any of these functions with full steps and every solution of the equation, use the inverse trig functions calculator.
Frequently asked questions
What are the domain and range of arcsin?
The domain is −1 ≤ x ≤ 1 and the range is −π/2 ≤ y ≤ π/2 (−90° to 90°). The graph is a short S-shaped curve from (−1, −π/2) to (1, π/2) that passes through the origin.
Does arctan have asymptotes?
Yes, two horizontal ones: y = −π/2 and y = π/2. As x grows large, arctan x creeps toward π/2 but never reaches it. Those lines are the vertical asymptotes of tan x, turned sideways.
Why is the graph of arcsin so short?
Only inputs from −1 to 1 are allowed, because sine never produces anything outside that interval. The curve is the mirror image, across the line y = x, of sin x restricted to −π/2 ≤ x ≤ π/2.
Why does the arcsec graph have a gap?
Secant takes no values strictly between −1 and 1, so arcsec has no inputs there. Its output also never equals π/2, because sec(π/2) is undefined, so the line y = π/2 is a horizontal asymptote.