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Inverse Trigonometric Functions

Mathematics Algebra Difficulty 2/5

Beginning of the lesson

Inverse Trigonometric Functions

Arcsine, Arccosine, Arctangent & Arccotangent — Definitions, Properties, Compositions, and Applications

1. The Invertibility Problem

Why Can't We Just "Invert" Sine, Cosine, or Tangent?

Recall from our earlier study of inverse functions that a function $ f $ has an inverse $ f^{-1} $ if and only if $ f $ is one-to-one (injective). A function is one-to-one when every horizontal line crosses its graph at most once.

The Problem: The trigonometric functions $ \sin x $, $ \cos x $, $ \tan x $, and $ \cot x $ are all periodic. A periodic function repeats its values, so it is never one-to-one on its full domain. For example, $ \sin 0 = \sin \pi = \sin 2\pi = 0 $. If we tried to define $ \sin^{-1}(0) $, which angle should it return?

The Solution: Domain Restriction

We restrict each trigonometric function to a carefully chosen interval where it is:

  1. Strictly monotonic (either always increasing or always decreasing) — guaranteeing injectivity.
  2. Surjective onto its full range — so every output value is still attainable.

Together, these two conditions make the restricted function a bijection, and a bijection always has an inverse.

Key Principle: The domain of the inverse function equals the range of the original (restricted) function, and the range of the inverse function equals the restricted domain.

The specific intervals are chosen to include the first-quadrant angles $ (0, \pi/2) $ and to be as large as possible while maintaining monotonicity. This is a natural and standard convention used across mathematics.

2. Definitions — Domains and Ranges

Arcsine: $ y = \arcsin x $

Definition: $ y = \arcsin x $ means $ \sin y = x $ and $ y \in \left[-\dfrac{\pi}{2},\, \dfrac{\pi}{2}\right] $.

In words: the unique angle between $ -90° $ and $ 90° $ whose sine equals $ x $.

Domain$ [-1,\, 1] $
Range$ \left[-\dfrac{\pi}{2},\, \dfrac{\pi}{2}\right] $

Why this interval? $ \sin x $ is strictly increasing on $ [-\pi/2,\, \pi/2] $ and takes every value from $ -1 $ to $ 1 $.

Arccosine: $ y = \arccos x $

Definition: $ y = \arccos x $ means $ \cos y = x $ and $ y \in [0,\, \pi] $.

In words: the unique angle between $ 0° $ and $ 180° $ whose cosine equals $ x $.

Domain$ [-1,\, 1] $
Range$ [0,\, \pi] $

Why this interval? $ \cos x $ is strictly decreasing on $ [0,\, \pi] $ and takes every value from $ -1 $ to $ 1 $.

Arctangent: $ y = \arctan x $

Definition: $ y = \arctan x $ means $ \tan y = x $ and $ y \in \left(-\dfrac{\pi}{2},\, \dfrac{\pi}{2}\right) $.

In words: the unique angle strictly between $ -90° $ and $ 90° $ whose tangent equals $ x $.

Domain$ (-\infty,\, +\infty) = \mathbb{R} $
Range$ \left(-\dfrac{\pi}{2},\, \dfrac{\pi}{2}\right) $

Why this interval? $ \tan x $ is strictly increasing on $ (-\pi/2,\, \pi/2) $ and takes every real value. The endpoints are excluded because $ \tan x $ has vertical asymptotes there.

Arccotangent: $ y = \text{arccot}\, x $

Definition: $ y = \text{arccot}\, x $ means $ \cot y = x $ and $ y \in (0,\, \pi) $.

In words: the unique angle strictly between $ 0° $ and $ 180° $ whose cotangent equals $ x $.

Domain$ (-\infty,\, +\infty) = \mathbb{R} $
Range$ (0,\, \pi) $

Why this interval? $ \cot x $ is strictly decreasing on $ (0,\, \pi) $ and takes every real value.

Convention note: Some textbooks use the range $ (-\pi/2, 0) \cup (0, \pi/2] $ for arccotangent. We follow the more common convention $ (0, \pi) $, which mirrors the arccosine range and keeps arccotangent a continuous, strictly decreasing function.

Summary Table

FunctionNotationDomainRangeMonotonicity
Arcsine$ \arcsin x $$ [-1, 1] $$ [-\pi/2,\, \pi/2] $Increasing
Arccosine$ \arccos x $$ [-1, 1] $$ [0,\, \pi] $Decreasing
Arctangent$ \arctan x $$ \mathbb{R} $$ (-\pi/2,\, \pi/2) $Increasing
Arccotangent$ \text{arccot}\, x $$ \mathbb{R} $$ (0,\, \pi) $Decreasing

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