Unit 3 · Topic 3.9

3.9 Inverse Trigonometric Functions

The inverse trig functions arcsin, arccos and arctan take a value and return an angle. Because sine, cosine and tangent repeat, they only have inverses on restricted domains, so each inverse returns angles from one specific interval.

Key terms

arcsin
arccos
arctan
restricted domain
principal value

Why the domains are restricted

A function needs to be one-to-one to have an inverse (topic 2.8). Sine, cosine and tangent are periodic, so each output comes from infinitely many angles. sin θ = 1/2 at π/6, 5π/6, 13π/6, and so on.

To fix this, each function is restricted to one interval where it hits every output exactly once. The inverse then returns the single angle from that interval.

The three inverse functions

For inverse trig functions the input is a value of sine, cosine or tangent (a ratio), and the output is an angle.

Function	Input (domain)	Output (range)	Comes from
arcsin x	−1 ≤ x ≤ 1	−π/2 ≤ θ ≤ π/2	sin θ on [−π/2, π/2]
arccos x	−1 ≤ x ≤ 1	0 ≤ θ ≤ π	cos θ on [0, π]
arctan x	All real numbers	−π/2 < θ < π/2	tan θ on (−π/2, π/2)

Notation and graphs

arcsin x is also written sin⁻¹ x, and the same goes for the others. The −1 means “inverse,” not a reciprocal: sin⁻¹ x is not 1/sin x.

Each graph is the reflection of the restricted trig graph over the line y = x. arcsin increases from (−1, −π/2) to (1, π/2). arccos decreases from (−1, π) to (1, 0). arctan increases across all real inputs and has horizontal asymptotes y = −π/2 and y = π/2.

Compositions

sin(arcsin x) = x for every x in [−1, 1], because arcsin returns an angle whose sine is x.

But arcsin(sin θ) = θ only when θ is already in [−π/2, π/2]. Otherwise arcsin returns a different angle with the same sine. Likewise, arccos(cos θ) = θ only for θ in [0, π].

A quick way to remember the ranges: arcsin and arctan give angles in quadrants I and IV (the right half), while arccos gives angles in quadrants I and II (the top half).

Angles from ratios

In a right triangle with acute angle θ, sin θ = opposite/hypotenuse, cos θ = adjacent/hypotenuse and tan θ = opposite/adjacent. When you know two sides, an inverse function gives the angle.

For example, a ramp that rises 1 foot for every 12 feet of horizontal distance makes an angle of arctan(1/12) ≈ 0.083 radians with the ground. Because acute angles lie inside the ranges of all three inverse functions, they cause no trouble here.

Worked examples

Try each one yourself first, then open the solution.

Example 1
Exact inverse values
Evaluate arcsin(−1/2), arccos(−1/2), arctan(√3) and arccos(√2/2).
Show the solutionHide the solution
1. Step 1: arcsin(−1/2): which angle in [−π/2, π/2] has sine −1/2? −π/6.
2. Step 2: arccos(−1/2): which angle in [0, π] has cosine −1/2? 2π/3 (quadrant II, reference angle π/3).
3. Step 3: arctan(√3): which angle in (−π/2, π/2) has tangent √3? π/3.
4. Step 4: arccos(√2/2): π/4.
Answer: −π/6, 2π/3, π/3 and π/4.
Example 2
Trap: undoing a sine outside the range
Evaluate arcsin(sin(5π/6)) and arccos(cos(−π/3)).
Show the solutionHide the solution
1. Step 1: sin(5π/6) = 1/2. Then arcsin(1/2) = π/6, because arcsin must return an angle in [−π/2, π/2].
2. Step 2: So arcsin(sin(5π/6)) = π/6, not 5π/6. The functions don't cancel, because 5π/6 is outside arcsin's range.
3. Step 3: cos(−π/3) = 1/2, and arccos(1/2) = π/3, which is in [0, π].
Answer: arcsin(sin(5π/6)) = π/6 and arccos(cos(−π/3)) = π/3.
Example 3Calculator allowed
An angle from a ratio, with a calculator
A 6-meter ladder leans against a wall and reaches 5 meters up the wall. Find the angle between the ladder and the ground, in radians.
Show the solutionHide the solution
1. Step 1: The ladder is the hypotenuse and the wall height is opposite the angle θ at the ground, so sin θ = 5/6.
2. Step 2: θ is between 0 and π/2, which is inside arcsin's range, so θ = arcsin(5/6).
3. Step 3: In radian mode, arcsin(5/6) ≈ 0.985.
Answer: θ = arcsin(5/6) ≈ 0.985 radians (about 56.4°).

Common mistakes

Giving an angle outside the range, like arccos(−1/2) = −π/3 or arcsin(−1/2) = 7π/6.
Assuming arcsin(sin θ) always equals θ.
Reading sin⁻¹ x as 1/sin x.
Taking arcsin or arccos of a number bigger than 1 or less than −1. Those are undefined.

On the exam

Expect exact-value questions on the no-calculator section and questions about the domains and ranges of the inverse functions.
Inverse trig functions are the main tool for solving trig equations in topic 3.10, where you'll need to find the other solutions arcsin and arccos don't give you.

Connected topics

Videos

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