arcsin, arccos, and arctan exist because we restrict the domain of each trig function to make it one-to-one. Understanding those restrictions is the key to getting every inverse trig problem right.
Each inverse function undoes the corresponding trig function — but only on a restricted interval. The range of the inverse equals the restricted domain of the original.
arcsin(x)
also written sin⁻¹(x)
At x = 0:
arcsin(0) = 0
Increasing S-curve through origin. Endpoints at (−1, −π/2) and (1, π/2).
arccos(x)
also written cos⁻¹(x)
At x = 0:
arccos(0) = π/2
Decreasing curve from (−1, π) to (1, 0). Passes through (0, π/2).
arctan(x)
also written tan⁻¹(x)
At x = 0:
arctan(0) = 0
Increasing curve with horizontal asymptotes at y = ±π/2. Passes through origin.
A function has an inverse only if it is one-to-one — every output comes from exactly one input. The trig functions are periodic, so they repeat the same output infinitely many times. For example, sin(π/6) = sin(5π/6) = sin(13π/6) = 1/2. Without restriction, “arcsin(1/2)” could mean π/6, or 5π/6, or any of infinitely many angles. We fix this by agreeing on a standard restricted domain for each function.
sin x
restricted to [−π/2, π/2]
One full increasing piece through the origin. Covers all output values [−1, 1].
cos x
restricted to [0, π]
One full decreasing piece. Starts at 1 (x=0) and decreases to −1 (x=π).
tan x
restricted to (−π/2, π/2)
One full increasing piece without the vertical asymptotes at the endpoints.
The convention is universally agreed upon
Every calculator, textbook, and exam uses these same restricted domains. When you press sin⁻¹ on your calculator, it always returns a value in [−π/2, π/2] — not the other possible angles.
All answers are exact radian values. “undef.” means x is outside the domain of that function. “—” means x is a valid input for arctan but not a standard special value shown here.
| x | arcsin(x) | arccos(x) | arctan(x) |
|---|---|---|---|
| −1 | −π/2 | π | −π/4 |
| −√3/2 | −π/3 | 5π/6 | — |
| −√2/2 | −π/4 | 3π/4 | — |
| −1/2 | −π/6 | 2π/3 | — |
| −√3 | undef. | undef. | −π/3 |
| −1/√3 | undef. | undef. | −π/6 |
| 0 | 0 | π/2 | 0 |
| 1/√3 | undef. | undef. | π/6 |
| √3 | undef. | undef. | π/3 |
| 1/2 | π/6 | π/3 | — |
| √2/2 | π/4 | π/4 | — |
| √3/2 | π/3 | π/6 | — |
| 1 | π/2 | 0 | π/4 |
Note: arcsin and arccos only accept inputs in [−1, 1]. arctan accepts all real numbers but the table only shows standard special values.
| Expression | Equals | When? |
|---|---|---|
| sin(arcsin x) | x | for x ∈ [−1, 1] — Always true on the domain. |
| arcsin(sin x) | x | only if x ∈ [−π/2, π/2] — Outside this interval, use the reflection rule. |
| cos(arccos x) | x | for x ∈ [−1, 1] — Always true on the domain. |
| arccos(cos x) | x | only if x ∈ [0, π] — Outside this interval, find the equivalent angle in [0, π]. |
| tan(arctan x) | x | for all real x — Always true since arctan has domain ℝ. |
| arctan(tan x) | x | only if x ∈ (−π/2, π/2) — Adjusts by adding/subtracting π outside this interval. |
When x is outside [−π/2, π/2], arcsin(sin x) does NOT equal x. Use this process:
Example: arcsin(sin(5π/6))
sin(5π/6) = sin(π − 5π/6) reflected... = sin(π/6) = 1/2
arcsin(1/2) = π/6 ∈ [−π/2, π/2] ✓
arcsin(sin(5π/6)) = π/6, not 5π/6
When you see a composition like sin(arccos(x)) or tan(arcsin(x)), the fastest method is to draw a right triangle based on the inner inverse trig function, then read off the outer function.
Set the inner inverse function equal to θ.
Example: let θ = arccos(3/5), so cos θ = 3/5.
Draw a right triangle from the trig ratio.
cos θ = adjacent/hypotenuse = 3/5. Label: adjacent = 3, hypotenuse = 5.
Find the missing side with the Pythagorean theorem.
opposite = √(5² − 3²) = √(25 − 9) = √16 = 4.
Read off the outer trig function.
sin(arccos(3/5)) = sin θ = opposite/hypotenuse = 4/5.
Key insight: the sign of the missing side
Since arccos returns angles in [0, π], we know θ ∈ [0, π], so sin θ ≥ 0 — use the positive root. Since arcsin returns angles in [−π/2, π/2], cos θ ≥ 0 — use the positive root. Since arctan returns angles in (−π/2, π/2), cos θ > 0 — use the positive root. The right triangle method always gives positive side lengths, and the range of the inverse function tells you the sign of the result.
y = arcsin(x)
Increasing. Starts at lower-left (−1, −π/2), passes through origin at 45°, ends at (1, π/2). Resembles a stretched S between the endpoint values.
y = arccos(x)
Decreasing. Starts at upper-left (−1, π), passes through (0, π/2) at the midpoint, ends at (1, 0). Reflection of arcsin about the horizontal line y = π/2.
y = arctan(x)
Increasing. Horizontal asymptotes at y = π/2 (top) and y = −π/2 (bottom). Passes through origin. S-shaped but never reaches the asymptotes.
The output of an inverse trig function must land in its restricted range. If your answer falls outside [−π/2, π/2] for arcsin or [0, π] for arccos, you have the wrong answer. Always check.
arcsin(sin(5π/6)) = π/6 (reflect to [−π/2, π/2]). arccos(cos(5π/6)) = 5π/6 (already in [0, π], so no adjustment needed). Always ask which restricted domain applies before simplifying.
When you see sin(arccos(x)) or tan(arcsin(x)), draw a right triangle. Label the sides from the inner function, use the Pythagorean theorem to find the missing side, then read off the outer function. This is faster than using identities.
A function must be one-to-one (pass the horizontal line test) to have an inverse. The original sin, cos, and tan functions are periodic and fail this test — each output value repeats infinitely many times. To define arcsin, arccos, and arctan, we restrict the domain of each trig function to one interval where it is one-to-one: sin is restricted to [−π/2, π/2], cos to [0, π], and tan to (−π/2, π/2). The inverse then maps that restricted range back to that interval.
The range of arcsin(x) is [−π/2, π/2] (outputs between −90° and 90°). The range of arccos(x) is [0, π] (outputs between 0° and 180°). The range of arctan(x) is (−π/2, π/2) (outputs strictly between −90° and 90°, never reaching the endpoints because tan has vertical asymptotes there). These ranges come directly from the restricted domains of the original trig functions.
arcsin(sin x) = x only when x is in [−π/2, π/2]. Outside that interval, the answer must be adjusted. For example, arcsin(sin(5π/4)): since 5π/4 is in Quadrant III, sin(5π/4) = −√2/2, and arcsin(−√2/2) = −π/4. The rule is: arcsin(sin x) equals the reference angle in [−π/2, π/2] with the same sine value. Similarly, arccos(cos x) = x only on [0, π], and arctan(tan x) = x only on (−π/2, π/2).
Unit circle, right triangle trig, and graph transformations
Pythagorean, double angle, half angle, and sum/difference formulas
SOH-CAH-TOA, special triangles, Law of Sines and Cosines
All angle values — the foundation for evaluating inverse trig
Interactive problems covering arcsin, arccos, arctan — with step-by-step solutions, composition identities, and right triangle method exercises. Free to try.
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