Inverse trig functions run trigonometry backwards: arcsin x (also written sin⁻¹ x) is the angle whose sine is x, and the same goes for arccos x and arctan x. Because sine, cosine and tangent repeat forever, each inverse returns only one angle from a fixed range: [−π/2, π/2] for arcsin, [0, π] for arccos and (−π/2, π/2) for arctan. That range is why sin⁻¹(sin x) equals x only when x already sits inside it.
Below you'll see why those ranges were chosen, how to find exact values without a calculator, how to undo a composition like arccos(cos(7π/4)), and how a quick triangle turns tan(arcsin(−5/13)) into a fraction.
What Inverse Trig Functions Do
A trig function takes an angle and gives a number: sin(π/6) = 1/2. Its inverse takes the number and gives back an angle: arcsin(1/2) = π/6. You already use this when you find a missing angle in a right triangle with SOH CAH TOA: the sin⁻¹ key on your calculator is arcsin.
The two notations mean the same thing. sin⁻¹ x matches the general f⁻¹ notation for an inverse function, and arcsin x avoids a trap: the −1 is not an exponent. The reciprocal of sine is (sin x)⁻¹ = 1/sin x = csc x, which is a different function entirely.
| Expression | Meaning | Value at x = 1/2 |
|---|---|---|
sin⁻¹ x or arcsin x | the angle whose sine is x | π/6 ≈ 0.5236 |
(sin x)⁻¹ or csc x | 1 / sin x, with x as the angle | 1 / sin(0.5) ≈ 2.0858 |
Why the Domains Are Restricted
A function has an inverse only if it is one-to-one: every output comes from exactly one input. Sine fails badly. The horizontal line y = 1/2 crosses the sine wave at π/6, 5π/6, 13π/6 and infinitely many more angles, so "the angle whose sine is 1/2" has no single answer.
The fix, explained in the inverse functions guide for x², is to keep one piece of the graph that passes the horizontal line test and throw the rest away. For each trig function, the chosen piece has to do three jobs:
- Hit every output once
Sine must reach every value in
[−1, 1]exactly one time on the piece. - Stay in one connected interval
No gaps, so the inverse is a single smooth curve.
- Include the first-quadrant angles
Angles from 0 to
π/2are the ones right triangles use, so they must be inside.
For sine, [−π/2, π/2] does all three: the wave climbs steadily from −1 to 1 there. Cosine is decreasing on [0, π], running from 1 down to −1, so that is its piece. On [−π/2, π/2] cosine would be no good, because cos(−π/3) and cos(π/3) are both 1/2. Tangent climbs from −∞ to ∞ on the open interval (−π/2, π/2); the endpoints are left out because tangent is undefined there.
y = arcsin x means sin y = x and −π/2 ≤ y ≤ π/2. y = arccos x means cos y = x and 0 ≤ y ≤ π. y = arctan x means tan y = x and −π/2 < y < π/2.The domain of each inverse is the range of the original piece, and the other way round. These are the standard conventions used in OpenStax Precalculus, section 6.3.
| Function | Domain (inputs) | Range (output angles) | Quadrants of the output |
|---|---|---|---|
arcsin x | [−1, 1] | [−π/2, π/2] (−90° to 90°) | I for x > 0, IV for x < 0 |
arccos x | [−1, 1] | [0, π] (0° to 180°) | I for x > 0, II for x < 0 |
arctan x | all real numbers | (−π/2, π/2) (open) | I for x > 0, IV for x < 0 |
That last column is the shortcut for negative inputs. Arcsin and arctan answer a negative input with a negative angle (quadrant IV). Arccos never gives a negative angle; a negative input lands in quadrant II, between π/2 and π. It also explains why arcsin(2) is undefined: no angle has a sine of 2, so 2 is outside the domain, and your calculator shows an error.
How to Find Exact Values Without a Calculator
Exact values come from the special angles of the unit circle. Ask "which angle in the allowed range has this sine (or cosine, or tangent)?" and read it off. For negative inputs, these three rules save you a trip around the circle:
arcsin(−x) = −arcsin x, arctan(−x) = −arctan x, but arccos(−x) = π − arccos x.Why does arccos break the pattern? Its range has no negative angles, so flipping the sign of the angle can't work. Instead the answer reflects across the y-axis, from angle θ to π − θ, which keeps the same height on the circle but flips the x-coordinate (the cosine).
Find arcsin(√3/2), arcsin(−1/2), arccos(−√3/2) and arctan(−1).
sin(π/3) = √3/2 and π/3 is in [−π/2, π/2], so arcsin(√3/2) = π/3.
arcsin(1/2) = π/6, so by symmetry arcsin(−1/2) = −π/6. Not 7π/6 or 11π/6: both have sine −1/2, but neither is in the range.
arccos(√3/2) = π/6, so arccos(−√3/2) = π − π/6 = 5π/6, a quadrant II angle as the table predicts.
arctan(1) = π/4, so arctan(−1) = −π/4.
π/3, −π/6, 5π/6, −π/4Writing arccos(−1/2) = −π/3 by copying the arcsin symmetry rule. A negative angle can never come out of arccos. The right answer is π − π/3 = 2π/3. The slip happens because arcsin and arctan both work by "just add a minus sign", so students expect arccos to as well. Check the range column before you write the answer.
Why sin⁻¹(sin x) Is Not Always x
Inverse functions cancel, so it is tempting to say arcsin(sin x) = x every time. There are two compositions, though, and they behave differently.
| Composition | Equals the input when | Otherwise |
|---|---|---|
sin(arcsin x) | −1 ≤ x ≤ 1 | undefined (arcsin cannot take x) |
arcsin(sin x) | −π/2 ≤ x ≤ π/2 | a different angle with the same sine |
arccos(cos x) | 0 ≤ x ≤ π | a different angle with the same cosine |
arctan(tan x) | −π/2 < x < π/2 | a different angle with the same tangent |
The outer arcsin can only ever output an angle in [−π/2, π/2]. If you feed it sin(5π/6), it can't return 5π/6; it returns the angle in its range with the same sine. So the job is always: find a partner angle with the same trig value that lies in the range. Each function has its own partner rule.
| Outer function | Angles with the same value as x | Pick the one in |
|---|---|---|
| arcsin | x + 2πk or π − x + 2πk | [−π/2, π/2] |
| arccos | x + 2πk or −x + 2πk | [0, π] |
| arctan | x + πk (tangent repeats every π) | (−π/2, π/2) |
Evaluate arcsin(sin(5π/6)), arccos(cos(7π/4)), arctan(tan(5π/6)) and arcsin(sin 2).
5π/6 is outside [−π/2, π/2]. Its partner is π − 5π/6 = π/6, which is inside. So arcsin(sin(5π/6)) = π/6. Check: sin(5π/6) = sin(π/6) = 1/2.
7π/4 is outside [0, π]. Try −x + 2π = −7π/4 + 2π = π/4, which is inside. So arccos(cos(7π/4)) = π/4. Check: both cosines equal √2/2.
For tangent, subtract π: 5π/6 − π = −π/6, inside (−π/2, π/2). So arctan(tan(5π/6)) = −π/6. Check: both tangents equal −√3/3.
The angle 2 (radians) is about 114.6°, past π/2 ≈ 1.5708. Its partner is π − 2 ≈ 1.1416, inside the range. So arcsin(sin 2) = π − 2.
π/6, π/4, −π/6, π − 2If you graph y = arcsin(sin x) you see this rule at work: it follows the line y = x between −π/2 and π/2, then turns down with slope −1, and the pattern zigzags forever as a sawtooth.
Compositions with a Right Triangle: sin(arccos x) and Friends
When the inverse is on the inside, as in sin(arccos(3/5)), you don't need the angle at all. Name it, draw a triangle, and read the other ratio.
- Name the inner angle
Let
θ = arccos(3/5), socos θ = 3/5andθis in[0, π]. - Draw the triangle
Adjacent 3, hypotenuse 5. The Pythagorean theorem gives the opposite side:
√(5² − 3²) = 4. - Fix the sign with the quadrant
Use the range table to decide which quadrant
θis in, then give the answer the sign the outer function has there.
Find the exact value of tan(arcsin(−5/13)).
Let θ = arcsin(−5/13). Then sin θ = −5/13, and because the input is negative, θ is in quadrant IV (between −π/2 and 0).
Triangle sides: opposite 5, hypotenuse 13, adjacent √(13² − 5²) = √144 = 12.
In quadrant IV, sine is negative and cosine is positive, so cos θ = 12/13 and tan θ = sin θ / cos θ = (−5/13) / (12/13) = −5/12.
tan(arcsin(−5/13)) = −5/12The same idea works with a variable, which is how calculus textbooks simplify these expressions.
Write cos(arctan x) as an algebraic expression in x.
Let θ = arctan x, so tan θ = x = x/1. Opposite x, adjacent 1, hypotenuse √(x² + 1).
So cos θ = 1/√(x² + 1).
Sign check: θ is in (−π/2, π/2), where cosine is always positive, and 1/√(x² + 1) is always positive. They agree for every x.
Spot check with x = 1: cos(arctan 1) = cos(π/4) = √2/2, and 1/√2 = √2/2. ✓
cos(arctan x) = 1/√(x² + 1)Using Inverse Trig to Solve Problems
In applied problems the inverse gives you an angle from a measured ratio. Put your calculator in degree mode when the question wants degrees; in radian mode sin⁻¹(0.5) shows 0.5236, which is π/6, not 30.
A 6 m ladder leans against a wall with its foot 1.5 m from the wall. What angle does the ladder make with the ground?
The ladder is the hypotenuse, and the 1.5 m gap is the side adjacent to the angle at the ground. Adjacent over hypotenuse is cosine.
cos θ = 1.5/6 = 0.25, so θ = arccos(0.25) ≈ 75.52°.
Reasonableness: the foot is close to the wall, so the ladder is steep, and an angle well above 45° fits.
θ ≈ 75.5°Solving a trig equation is where the restricted range bites. The calculator hands you one solution; the others come from the partner rules in the table above. For more equation types, see how to solve trigonometric equations.
Solve 5 sin x − 2 = 0 for 0 ≤ x < 2π, to 4 decimal places.
Isolate the sine: sin x = 2/5 = 0.4.
Calculator (radian mode): x = arcsin(0.4) ≈ 0.4115. That is the quadrant I solution.
Sine is also positive in quadrant II, at the partner angle π − 0.4115 ≈ 2.7301.
Check: sin(2.7301) ≈ 0.4000, so 5(0.4) − 2 = 0. ✓
x ≈ 0.4115 or x ≈ 2.7301Graphs and the Other Three Inverses
Each inverse graph is its restricted piece reflected over the line y = x, because reflecting swaps inputs and outputs.
y = arcsin xrises from(−1, −π/2)through the origin to(1, π/2).y = arccos xfalls from(−1, π)through(0, π/2)to(1, 0).y = arctan xpasses through the origin and flattens toward the horizontal asymptotesy = π/2andy = −π/2, never reaching them.
Most calculators have no keys for arcsec, arccsc or arccot, so check which range your course uses for them. With the common choice of [0, π] without π/2 for arcsec, it reduces to arccos: arcsec x = arccos(1/x) for |x| ≥ 1, as listed on Wolfram MathWorld's inverse secant page. So arcsec 2 = arccos(1/2) = π/3.
If you're heading into calculus, the inverses have derivatives that turn up in integrals: d/dx arcsin x = 1/√(1 − x²), d/dx arccos x = −1/√(1 − x²) and d/dx arctan x = 1/(1 + x²). The derivative rules guide lists them with the rest.
How to Check Your Answer
Every inverse trig answer can be tested in two quick moves:
- Apply the original function
If you claim
arccos(−√3/2) = 5π/6, checkcos(5π/6) = −√3/2. If the value doesn't come back, the angle is wrong. - Check the range
Is the angle inside the allowed range?
7π/6has the right sine forarcsin(−1/2)but fails this test. Both checks must pass.
For triangle compositions, also confirm the sign against the quadrant, and test one easy value of the variable, as in Example 4. To compare your full working with a worked solution, Solver AI can scan a printed or handwritten problem and show the steps, so you can see where your method and its method part ways after you've tried it yourself.
An inverse trig function gives one angle, chosen from a fixed range. Learn the three ranges and the quadrant column, and the hard questions (negative inputs, compositions, extra equation solutions) all become "find the angle with the same value that lives in the range".
Practice Problems
Work each one, then check your answers below.
- Exact value
Find
arccos(−√2/2). - Composition
Evaluate
arcsin(sin(4π/3)). - Triangle method
Find
cos(arcsin(−8/17)). - Equation
Solve
tan x = −3for0 ≤ x < 2π, to 4 decimal places. - Challenge
Find the exact value of
sin(2 arctan(3/4)).
Answers
1. arccos(√2/2) = π/4, so arccos(−√2/2) = π − π/4 = 3π/4.
2. sin(4π/3) = −√3/2, and the angle in [−π/2, π/2] with that sine is −π/3. Partner check: π − 4π/3 = −π/3. Answer: −π/3.
3. Opposite 8, hypotenuse 17, adjacent √(289 − 64) = 15. The angle is in quadrant IV, where cosine is positive. Answer: 15/17.
4. arctan(−3) ≈ −1.2490, which is outside the interval. Tangent repeats every π, so add π and 2π: x ≈ 1.8925 or x ≈ 5.0341.
5. Let θ = arctan(3/4): opposite 3, adjacent 4, hypotenuse 5, all in quadrant I. The double-angle identity from the trig identities cheat sheet gives sin 2θ = 2 sin θ cos θ = 2(3/5)(4/5) = 24/25.