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Inverse Trig Functions: Why arcsin, arccos and arctan Have Restricted Ranges
Inverse trig functions are the first place many students meet a definition that looks arbitrary. arcsin returns angles between −90° and 90°, arccos between 0° and 180°, arctan strictly between −90° and 90°. Three different ranges, no obvious pattern, and an exam question that punishes you for forgetting which is which.
The ranges are not arbitrary. Each one is the unique choice that makes an inverse possible at all while still reaching every value the original function produces. Understanding why they had to be chosen that way is much faster than memorising three intervals, and it is what makes composition questions such as arcsin(sin 5π/6) straightforward instead of a trap.
The problem an inverse has to solve
A function has an inverse only if it is one-to-one: each output must come from exactly one input. Sine fails this badly. sin 30° = 1/2, and so does sin 150°, and so does sin 390°, and so on forever. Asking "which angle has sine 1/2" has infinitely many correct answers, so there is no function that can return the answer.
The standard fix is to restrict the domain of the original function to a stretch on which it is one-to-one, and then invert that restricted version. The restricted stretch has to satisfy two competing requirements: the function must be strictly increasing or strictly decreasing on it, so no output repeats, and it must still produce every output the unrestricted function could, so nothing is lost.
For sine, the natural choice is [−π/2, π/2]. Sine climbs steadily from −1 to 1 across that interval, hitting every value in [−1, 1] exactly once, and it contains the origin, which keeps sin θ ≈ θ behaviour intact near zero. That restricted sine is what arcsin inverts.
The three principal ranges. Note the round brackets on arctan - its endpoints are excluded.
Why cosine gets a different interval
It would be tidier if arccos used [−π/2, π/2] too, and it cannot. Cosine is an even function, so cos(−θ) = cos θ, which means every value it takes on [0, π/2] it takes again on [−π/2, 0]. That interval is not one-to-one for cosine at all - it fails the requirement immediately.
Worse, on [−π/2, π/2] cosine only ever outputs values in [0, 1]. It never becomes negative there, so half of the possible inputs to arccos would have no answer.
The interval [0, π] fixes both problems at once. Cosine decreases steadily from 1 to −1 across it, so it is one-to-one, and it covers the full output range [−1, 1]. That is why arccos returns angles in the first and second quadrants while arcsin returns first and fourth.
Tangent needs a third choice for a different reason again. It is one-to-one on (−π/2, π/2) and takes every real value there, so its domain is all of ℝ - but the endpoints must be excluded, because tan is undefined at ±π/2 where cosine vanishes. That is why arctan's range uses round brackets, and why the graph of arctan has horizontal asymptotes at ±π/2 that it approaches but never reaches.
- arcsin outputs land in Quadrants I and IV - that is, the sign of the output matches the sign of the input.
- arccos outputs land in Quadrants I and II - a negative input gives an obtuse angle.
- arctan outputs land in Quadrants I and IV, and unlike the other two it accepts any real number.
- arccos is the only one of the three whose output is never negative.
- The notation sin⁻¹x means arcsin x, not 1/(sin x). The reciprocal is csc x. This clash of conventions is why many texts prefer the arc- names.
Evaluating an inverse function
The method is always the same two questions. First, which angle has the required sine, cosine or tangent? Second, which of the infinitely many such angles lies in the principal range? Only the second question has a unique answer, and that is the value you report.
Example 1 - arcsin of a negative value
Evaluate arcsin(−√2/2) exactly.
- Find the reference angle from the magnitude. sin 45° = √2/2, so the reference angle is π/4.
- The input is negative, and arcsin's range is [−π/2, π/2], which covers Quadrants I and IV. A negative sine there means a negative angle.
- Check the answer lies in range: −π/4 is between −π/2 and π/2. It does.
The angle 5π/4 also has sine −√2/2, but it is outside the principal range, so it is not the value of arcsin.
Answer. arcsin(−√2/2) = −π/4, or −45°.
Example 2 - arccos of a negative value
Evaluate arccos(−1/2) exactly.
- The reference angle comes from cos 60° = 1/2, so it is π/3.
- arccos's range is [0, π], covering Quadrants I and II. A negative cosine puts the angle in Quadrant II, so take the supplement of the reference angle.
- Check: cos(2π/3) = −1/2, and 2π/3 lies in [0, π]. Correct.
This is the difference that catches people out. A negative input gives arcsin a negative answer but gives arccos an obtuse positive one.
Answer. arccos(−1/2) = 2π/3, or 120°.
Compositions: the order matters
The two compositions of a function with its inverse behave very differently here, and knowing which is which is worth a lot of marks.
Taking the inverse first and the function second always returns the input: sin(arcsin x) = x for every x in [−1, 1], and likewise for the other two. This works because arcsin only ever produces angles inside the restricted domain, so sine is genuinely invertible there.
Going the other way, arcsin(sin θ) = θ only when θ already lies in [−π/2, π/2]. For any other θ, the composition returns the angle in the principal range that has the same sine. It is a folding operation, not an identity.
Outside the principal range, the composition returns a different angle with the same sine.
Example 3 - a composition that is not the identity
Evaluate arcsin(sin(5π/6)).
- Work from the inside out. 5π/6 is 150°, in Quadrant II, with reference angle π/6.
- Now apply arcsin to that value, and answer inside the principal range.
- The answer is π/6, not 5π/6, because 5π/6 is outside [−π/2, π/2].
Writing arcsin(sin θ) = θ without checking the range is one of the most common errors in this topic.
Answer. arcsin(sin(5π/6)) = π/6.
Example 4 - a mixed composition
Evaluate tan(arccos(−1/3)) exactly.
- Name the inner angle: let θ = arccos(−1/3), so cos θ = −1/3 with θ in [0, π]. Since the cosine is negative, θ is in Quadrant II.
- Find sin θ from the Pythagorean identity. In Quadrant II sine is positive, so take the positive root.
- Form the tangent as the quotient.
- The sign is a check, not an accident: tangent is negative in Quadrant II, which is where arccos placed the angle.
Answer. tan(arccos(−1/3)) = −2√2 ≈ −2.828.
Example 5 - an algebraic identity from a composition
Simplify cos(arctan x) to an expression in x, valid for all real x.
- Let θ = arctan x, so tan θ = x and θ lies in (−π/2, π/2). Crucially, cosine is positive across that whole interval.
- Build a right triangle with opposite side x and adjacent side 1, so that the tangent is x/1. The hypotenuse follows from Pythagoras.
- Read the cosine off the triangle.
- The result needs no absolute value or sign case, because arctan's range guarantees a positive cosine.
The same triangle method gives sin(arctan x) = x/√(x²+1) and cos(arcsin x) = √(1 − x²).
Answer. cos(arctan x) = 1/√(x² + 1), for every real x.
Where the restriction bites in practice
The restricted range is not just a technicality on composition questions. It is the reason your calculator gives incomplete answers when you solve equations, and the reason the law of sines has an ambiguous case.
Solving sin θ = 0.6 on [0°, 360°) with a calculator returns 36.87°, and that is genuinely all arcsin is allowed to return. The second solution, 180° − 36.87° = 143.13°, has the same sine but sits outside the principal range, so you must supply it yourself. Every trig equation you solve involves this step.
In triangle work the consequence is sharper. Using the law of sines to find an angle calls arcsin, which can never return an obtuse value - so if the angle you are solving for is actually obtuse, the calculator hands back its acute supplement without any warning. Using the law of cosines instead calls arccos, whose range spans the full 0 to π, which is why it is the safer tool whenever an obtuse angle is possible.
- arcsin and arctan can never return an obtuse angle. arccos can.
- After any arcsin, ask whether a supplementary solution also fits the problem.
- arcsin and arccos reject inputs outside [−1, 1]. If your calculation produces one, the geometry you assumed is impossible - for example a triangle whose sides violate the triangle inequality.
- arctan alone cannot recover a bearing or a direction in the plane, because it cannot distinguish the third quadrant from the first. Programming languages provide atan2(y, x) precisely to resolve that.
Frequently asked questions
Why does arcsin use [−π/2, π/2] rather than [0, π]?
Because sine is not one-to-one on [0, π] - it rises to 1 at π/2 and comes back down, so sin(π/6) and sin(5π/6) are equal. On [−π/2, π/2] sine increases strictly from −1 to 1, hitting every possible output exactly once, which is precisely what an invertible restriction requires.
Does sin⁻¹x mean 1 divided by sin x?
No. In this notation the −1 marks the inverse function, not a reciprocal, so sin⁻¹x is arcsin x. The reciprocal of sine is cosecant. Confusingly, sin²x does mean (sin x)², which is why many writers prefer arcsin to sin⁻¹ altogether.
What is the domain of each inverse function?
arcsin and arccos both accept only [−1, 1], because sine and cosine never produce anything outside that. arctan accepts every real number, since tangent takes every real value within a single period.
How do I evaluate arccos(cos θ) for an arbitrary θ?
Reduce θ to the equivalent angle in [0, π] with the same cosine. For θ in [0, π] the answer is θ itself; for θ in [π, 2π] it is 2π − θ; for negative θ use cos(−θ) = cos θ first. For example arccos(cos(−π/3)) = π/3, and arccos(cos(4π/3)) = 2π/3.
Are there inverses for secant, cosecant and cotangent?
Yes, and they are defined the same way by restricting the original function, though the conventional ranges vary between textbooks - which is why they appear far less often. In practice you can always rewrite them: arcsec x = arccos(1/x), and arccsc x = arcsin(1/x), for |x| ≥ 1.
Keep going
Practise it
Reading a method is not the same as running it under time pressure. The workspace drills this material in the quiz, and the triangle solver shows a worked solution for any triangle you type in.