Inverse Trigonometric Functions

You’ll be able to

State the restricted domains that make sine, cosine and tangent invertible
Give the exact range of each inverse trigonometric function
Evaluate compositions such as sin(arccos x) and explain when arcsin(sin x) ≠ x

Restriction is what makes an inverse possible

sin x = 0.5 has infinitely many solutions, so sine as a whole has no inverse — it fails the horizontal line test everywhere. The fix is to restrict the domain to a stretch on which the function is one-to-one and still covers the full range. For sine the agreed choice is [−π/2, π/2]; for cosine it is [0, π]; for tangent it is (−π/2, π/2). Those choices then become the ranges of the inverse functions, and they are conventions worth memorizing exactly.

Domains and ranges of the inverses

arcsin: domain [−1, 1], range [−π/2, π/2] · arccos: domain [−1, 1], range [0, π] · arctan: domain all reals, range (−π/2, π/2)

arcsin and arctan return values in quadrants I and IV (so negative inputs give negative angles). arccos returns quadrants I and II (so it never returns a negative angle).

The output is always a single angle

arcsin(0.5) is π/6 and nothing else. It is not "π/6 or 5π/6" — the whole point of restricting the domain was to make the answer unique. When a problem asks for all solutions to sin x = 0.5, the inverse function gives one of them and you generate the rest yourself using symmetry and periodicity. Confusing "evaluate the inverse" with "solve the equation" is the central difficulty of this topic.

Worked example

Evaluate arcsin(−1/2), arccos(−1/2) and arctan(−1) exactly.

1.arcsin(−1/2): need an angle in [−π/2, π/2] whose sine is −1/2. That is −π/6.
2.arccos(−1/2): need an angle in [0, π] whose cosine is −1/2. Cosine is negative in quadrant II, so 2π/3.
3.arctan(−1): need an angle in (−π/2, π/2) whose tangent is −1. That is −π/4.
4.Note the contrast: arcsin and arctan returned negative angles, arccos returned a positive one in quadrant II.

Answer: arcsin(−1/2) = −π/6, arccos(−1/2) = 2π/3, arctan(−1) = −π/4. arccos can never return a negative value — for a negative input it goes into quadrant II instead, which is the asymmetry students most often get wrong.

Watch out

sin⁻¹(x) means arcsin(x), the inverse function. It does not mean 1/sin(x), which is csc(x). The −1 in this notation is not an exponent, and (sin x)⁻¹ means something entirely different from sin⁻¹(x).

Checkpoint

What is arccos(−√3/2)?

Compositions and when they collapse

sin(arcsin x) = x for every x in [−1, 1] — the inner function returns an angle in the restricted domain, and sine undoes it cleanly. But arcsin(sin x) = x only when x is already in [−π/2, π/2]. For x = 3π/4, sin x = √2/2, and arcsin(√2/2) = π/4 ≠ 3π/4. The composition returns the co-terminal or reflected angle that lives inside the restricted range. Which order you compose in decides whether the identity holds unconditionally.

Worked example

Evaluate cos(arcsin(3/5)) without a calculator.

1.Let θ = arcsin(3/5), so sin θ = 3/5 and θ lies in [−π/2, π/2].
2.Draw a right triangle with opposite 3 and hypotenuse 5; the adjacent side is √(25 − 9) = 4.
3.So cos θ = 4/5.
4.Sign check: since sin θ = 3/5 > 0, θ is in quadrant I, where cosine is positive. So +4/5 is correct.

Answer: cos(arcsin(3/5)) = 4/5. The triangle method works for any such composition; the sign is settled by which quadrant the inverse function's range places the angle in — a step that matters whenever the input is negative.

On the exam

Exact-value questions on inverse trig almost always use the special values 0, ±1/2, ±√2/2, ±√3/2, ±1 for sine and cosine, and 0, ±√3/3, ±1, ±√3 for tangent. Recognizing them removes the need for a calculator entirely.

Checkpoint

What is arcsin(sin(5π/6))?

Answer the 2 checkpoints as you read.