Inverse Trigonometric Functions¶

Definition¶

Where this appears in the course

Lecture 7 states that the derivatives of the inverse trigonometric functions are derived in that week's tutorial (Tutorial 4) rather than in the lecture itself. The functions and their derivatives are still part of the course.

The inverse sine function $\arcsin \colon [-1,1] \to \left[-\pi/2, \pi/2\right]$ is defined as the inverse of the restriction of the sine function to the interval $\left[-\pi/2, \pi/2\right]\text{.}$ That is, for each $x \in [-1,1]\text{,}$ $\arcsin x$ is the unique number $y \in \left[-\pi/2, \pi/2\right]$ such that $\sin y = x\text{.}$

The inverse cosine function $\arccos \colon [-1,1] \to [0,\pi]$ is defined as the inverse of the restriction of the cosine function to the interval $[0,\pi]\text{.}$ That is, for each $x \in [-1,1]\text{,}$ $\arccos x$ is the unique number $y \in [0,\pi]$ such that $\cos y = x\text{.}$

The inverse tangent function $\arctan \colon \mathbb{R} \to \left(-\pi/2, \pi/2\right)$ is defined as the inverse of the restriction of the tangent function to the interval $\left(-\pi/2, \pi/2\right)\text{.}$ That is, for each $x \in \mathbb{R}\text{,}$ $\arctan x$ is the unique number $y \in \left(-\pi/2,\pi/2\right)$ such that $\tan y = x\text{.}$

Interpretation¶

Inverse trigonometric functions recover an angle from the value of its sine, cosine or tangent. The trigonometric functions are periodic, so each value is taken infinitely often; they are therefore first restricted to an interval on which they are injective, and only then inverted. Thus $\arcsin x$ is the one angle in $[-\pi/2,\pi/2]$ whose sine is $x$ .

Two panels: arcsin x and arccos x on the interval from -1 to 1, and arctan x approaching the horizontal lines at minus pi/2 and pi/2.

Each graph is the reflection in the line $y=x$ of the restricted sine, cosine or tangent. On $[-1,1]$ , $\arcsin$ increases from $-\pi/2$ to $\pi/2$ and $\arccos$ decreases from $\pi$ to $0$ ; together they satisfy $\arcsin x+\arccos x=\pi/2$ $π /2$ . The function $\arctan$ is defined on all of $\mathbb{R}$ and has horizontal asymptotes at $y=\pm\pi/2$ $\pm π /2$ , the mirror images of the vertical asymptotes of $\tan$ at $x=\pm\pi/2$ $\pm π /2$ .

Switch the restriction off: on all of $\mathbb{R}$ the level $c$ is hit infinitely often. Restricted to the standard interval, each function is a bijection and $\arcsin$ , $\arccos$ and $\arctan$ are its mirror images.

The restriction fixes which angle is returned. So $\sin(\arcsin x)=x$ for every $x\in[-1,1]$ $[- 1, 1]$ , but $\arcsin(\sin x)=x$ only for $x\in[-\pi/2,\pi/2]$ $[- π /2, π /2]$ ; for example, $\arcsin(\sin\pi)=\arcsin 0=0$ $0$ . The notation $\sin^{-1}x$ means $\arcsin x$ , not $1/\sin x$ .

The derivatives follow from the inverse function theorem. For $y=\arcsin x$ with $-1<x<1$ $1$ , the angle $y$ lies in $(-\pi/2,\pi/2)$ , where $\cos y>0$ $0$ , so $(\arcsin)'(x)=1/\cos y=1/\sqrt{1-x^2}$ $1/ 1 - x^{2}$ . In the same way $(\arctan)'(x)=\cos^2 y=1/(1+x^2)$ $x^{2})$ . At $x=\pm1$ the sine has derivative $\cos(\pm\pi/2)=0$ $0$ , and $\arcsin$ has a vertical tangent there. Math II returns to the arctangent in the 2-argument arctangent, which recovers angles in all four quadrants.

Key Results¶

Domain restrictions are essential for defining inverses.
Inverse trigonometric functions are continuous on their domains.

Examples¶

$\arcsin 0=0\text{.}$
$\arctan 1=\pi/4\text{.}$

Prerequisites¶

Sources¶

Math I: Calculus · Tutorial 4

Aliases: inverse trig functions, arcus functions, cyclometric functions Keywords: arcsine, arccosine, arctangent Symbols: element of, pi, real numbers, square root

Last updated: 20 October 2026 03:02