Mathematics.

inverse functions

Inverse Trigonometric Functions

Trigonometry25 minDifficulty3 out of 10

You should know: trigonometric functions

Overview

The inverse trigonometric functions — arcsin, arccos, and arctan (also written sin⁻¹, cos⁻¹, tan⁻¹) — undo the trigonometric functions, returning the angle whose sine, cosine, or tangent is a given value. Because sine, cosine, and tangent are periodic and not one-to-one over their full domains, each inverse must be restricted to a specific range to be well-defined: arcsin has range [−90°, 90°], arccos has range [0°, 180°], and arctan has range (−90°, 90°). These functions are essential for solving trigonometric equations, finding angles in applied problems (navigation, physics, engineering), and appear in calculus as antiderivatives of certain rational and algebraic functions.

Intuition

Because sin(30°) = sin(150°) = 0.5, the equation sinθ = 0.5 has infinitely many solutions, so 'the' inverse sine can't just undo sine everywhere — it must pick one canonical answer. Restricting sine to [−90°, 90°], where it is strictly increasing and hits every value in [−1, 1] exactly once, makes it invertible on that slice; arcsin is defined to return the angle from that slice. The same idea — pick the interval where the function is monotonic and covers its whole range — gives arccos (using [0°, 180°], where cosine is decreasing) and arctan.

Formal Definition

Definition

For x in the appropriate domain:

y=arcsin⁡x  ⟺  sin⁡y=x,y∈[−π2,π2], x∈[−1,1]y = \arcsin x \iff \sin y = x, \quad y \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right],\ x\in[-1,1]
Arcsine
y=arccos⁡x  ⟺  cos⁡y=x,y∈[0,π], x∈[−1,1]y = \arccos x \iff \cos y = x, \quad y \in [0, \pi],\ x\in[-1,1]
Arccosine
y=arctan⁡x  ⟺  tan⁡y=x,y∈(−π2,π2), x∈Ry = \arctan x \iff \tan y = x, \quad y \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right),\ x\in\mathbb{R}
Arctangent

Worked Examples

  1. Find the angle in [−90°,90°] whose sine is 1/2.

    arcsin⁡(12)=30°\arcsin\left(\frac{1}{2}\right) = 30°

  2. Find the angle in [0°,180°] whose cosine is −1/2.

    arccos⁡(−12)=120°\arccos\left(-\frac{1}{2}\right) = 120°

Answer: arcsin(1/2) = 30°, arccos(−1/2) = 120°

Practice Problems

Difficulty 2/10

Evaluate arctan(1) in degrees.

Difficulty 4/10

If θ = arccos(5/13) (θ acute), find sin θ.

Difficulty 5/10

A ship sails 30 km east and 40 km north. Using arctan, find the bearing angle θ (measured from east toward north) of its position from the start.

Quiz

The range of arccos(x) is:
arcsin(1/2) equals:
Inverse trig functions must be restricted to a specific range because:

Summary

  • arcsin, arccos, and arctan undo sine, cosine, and tangent on restricted domains where each is one-to-one.
  • Ranges: arcsin → [−90°,90°], arccos → [0°,180°], arctan → (−90°,90°).
  • A right-triangle picture (SOH-CAH-TOA) quickly converts an inverse trig expression like arcsin(3/5) into other trig ratios without finding the angle itself.

References

  1. WebsiteWikipedia — Inverse trigonometric functions