The inverse trig functions arcsin x, arccos x and arctan x undo sine, cosine and tangent, but only after each original function is restricted to a slice where it is one-to-one. That gives arcsin a range of −π/2 to π/2, arccos a range of 0 to π, and arctan a range strictly between −π/2 and π/2. Their graphs are the restricted curves reflected in the line y = x.
Domains and ranges of the inverse trig functions
- arcsin x: domain −1 ≤ x ≤ 1, range −π/2 ≤ y ≤ π/2
- arccos x: domain −1 ≤ x ≤ 1, range 0 ≤ y ≤ π
- arctan x: domain all real x, range −π/2 < y < π/2
sin(arcsin x) = x for every x in the domain, but arcsin(sin x) = x only when x is already between −π/2 and π/2.
Why do the trig functions need a restricted domain?
A function only has an inverse if every output comes from exactly one input. A horizontal line cuts y = sin x infinitely many times, so “the angle whose sine is 0.5” has no single answer. The fix is to cut each curve down to a slice where it only ever increases or only ever decreases.
The slices are −π/2 ≤ x ≤ π/2 for sine, 0 ≤ x ≤ π for cosine, and −π/2 < x < π/2 for tangent. Swapping inputs and outputs then gives the range of each inverse. arctan uses strict inequalities because its range ends at asymptotes, which are approached but never reached.
How do you sketch arcsin, arccos and arctan?
Each inverse graph is the restricted base curve reflected in y = x. Reflection swaps coordinates, so every feature swaps with it:
- Restrict the base curve to its one-way slice.
- Mark the key points on that slice.
- Swap every coordinate, so (a, b) becomes (b, a).
- Swap the asymptotes: vertical becomes horizontal.
- Swap the domain and range.
Three things worth remembering: arcsin and arccos stop at x = ±1 with endpoints and no asymptotes, arctan runs forever with horizontal asymptotes at y = ±π/2, and arccos is the only one of the three that decreases.
Transformations of inverse trig graphs
Treat them like any other graph, keeping one distinction clear. Anything outside the function acts on the output, so it changes the range. Anything inside the bracket acts on the input, so it changes the domain. y = 2 arcsin x doubles the range; y = arcsin(2x) halves the domain.
How do you find the domain of something like arcsin(2x − 1)?
Whatever is inside arcsin or arccos must lie between −1 and 1. Write that as a three-part inequality, −1 ≤ (inside) ≤ 1, and solve for x, operating on all three parts at once. If you divide by a negative, flip both inequality signs. arctan has no such restriction: its domain is all real numbers.
The composite trap: arcsin(sin x)
Going one way is safe: sin(arcsin x) = x. Going the other way is not. arcsin can only return an angle inside its own range, so arcsin(sin x) gives back x only when x is already between −π/2 and π/2. Otherwise it returns the angle in its range with the same sine. Always ask whether the angle is inside the inverse's range before writing the answer.
Worked examples
Example 1. Find the exact value of arcsin(½), in radians.
- Ask which angle has a sine of ½: both π/6 and 5π/6 do.
- arcsin must return a value between −π/2 and π/2.
- Only π/6 lies in that range.
Answer: arcsin(½) = π/6.
Marks: B1 for the exact value.
Example 2. Sketch y = arccos x, stating its endpoints and where it crosses the y-axis.
- Restrict y = cos x to 0 ≤ x ≤ π, where it falls from 1 to −1.
- Key points on that slice: (0, 1), (π/2, 0), (π, −1).
- Swap the coordinates: (1, 0), (0, π/2), (−1, π).
- The cosine slice has no asymptotes, so arccos has endpoints instead.
- Domain −1 ≤ x ≤ 1, range 0 ≤ y ≤ π. The curve decreases from (−1, π) to (1, 0).
Answer: Endpoints (−1, π) and (1, 0); crosses the y-axis at (0, π/2).
Marks: B1 for the endpoints, B1 for the y-intercept.
Example 3. Find the domain and range of y = arcsin(2x − 1).
- The inside must satisfy −1 ≤ 2x − 1 ≤ 1.
- Add 1 to all three parts: 0 ≤ 2x ≤ 2.
- Divide all three parts by 2: 0 ≤ x ≤ 1. That is the domain.
- Nothing acts outside arcsin, so the range is unchanged: −π/2 ≤ y ≤ π/2.
Answer: Domain 0 ≤ x ≤ 1; range −π/2 ≤ y ≤ π/2.
Marks: M1 for the inequality, A1 for the domain, B1 for the range.
Example 4. Find the exact value of arcsin(sin 3π/4).
- It is tempting to answer 3π/4, but that is outside the range of arcsin.
- Evaluate the inside first: sin(3π/4) = √2/2.
- The angle between −π/2 and π/2 with sine √2/2 is π/4.
Answer: arcsin(sin 3π/4) = π/4.
Marks: M1 for evaluating the inside, A1 for applying the range.
Common mistakes
- Giving an angle outside the range. arcsin and arctan return values between −π/2 and π/2, and arccos between 0 and π. Any other angle cannot be the answer.
- Confusing arcsin x with 1 / sin x. arcsin x is the inverse function, sometimes written sin⁻¹x. The reciprocal 1 / sin x is cosec x.
- Mixing up domain and range changes. A number multiplying outside the function stretches the range; a change inside the bracket affects the domain.
- Drawing asymptotes on arcsin or arccos. These graphs have endpoints at x = ±1. Only arctan has asymptotes.
- Forgetting to flip the inequality. When finding a domain by dividing all three parts by a negative number, both inequality signs reverse.
How inverse trig functions are examined
In Edexcel International A-Level, inverse trig functions are Pure 3 content (paper WMA13), in the trigonometric functions chapter. In Edexcel A-Level (9MA0) they are part of Year 2 Pure. Questions tend to ask you to:
- sketch y = arcsin x, arccos x or arctan x, or a transformation of one, with key coordinates
- state the domain and range of a transformed graph
- find exact values, including composites such as arctan(tan 5π/6)
- solve simple equations such as arccos(3x) = 2π/3
Practise with the Pure 3 trigonometric functions exam questions and their mark schemes, or the trigonometric functions worksheet. The trigonometry revision engine has unlimited extra questions.
This topic sits alongside sec, cosec and cot and builds on composite and inverse functions.
Frequently asked questions
Is arcsin the same as sin⁻¹?
Yes. arcsin x and sin⁻¹x mean the same inverse function. Neither means 1 / sin x, which is cosec x.
Why is the range of arccos 0 to π and not −π/2 to π/2?
Cosine is not one-to-one between −π/2 and π/2, because it rises then falls. Between 0 and π it only decreases, so that is the slice used, and it becomes the range of arccos.
Does arctan have a restricted domain?
No. tan x takes every real value on −π/2 < x < π/2, so arctan x is defined for all real x. Its range is restricted instead.
Why is arcsin(sin x) not always equal to x?
arcsin only returns angles between −π/2 and π/2. If x is outside that interval, arcsin(sin x) gives the angle in the interval with the same sine instead.