Inverse trigonometric functions

Concept 4.9.1. Inverse trigonometric functions.

We remark here that there is no universal convention for the choice of principal domain for $\sec x$ and $\csc x$ to define their inverse functions. For instance, in Section 2.12 of CLP1, a different choice is made (see Definition 2.12.4). The principal domain of $\sec x$ is chosen to be $x \in [0,\pi/2) \cup (\pi/2,\pi]\text{,}$ while the principal domain of $\csc x$ is chosen to be $x \in [-\pi/2, 0) \cup (0, \pi/2]\text{.}$ The domains of both inverse functions $\sec^{-1} y$ and $\csc^{-1} y$ remain $y \in (-\infty, -1] \cup [1,\infty)\text{.}$

Those choices may look simpler than ours, but there is a price to pay. With this choice of principal domains, the derivatives of the inverse secant and cosecant functions look a bit more complicated. Using implicit differentiation, we would then obtain:

$\begin{equation*} \frac{d}{dx} \sec^{-1} x = \frac{1}{|x| \sqrt{x^2-1}}, \end{equation*}$

and

$\begin{equation*} \frac{d}{dx} \csc^{-1} x = - \frac{1}{|x| \sqrt{x^2-1}}. \end{equation*}$

(See Theorem 2.12.8 in CLP1.) Note the appearance of absolute values here, which are not present with our choice of conventions.

In the end, it does not matter which choice of convention is used to define the inverse secant and cosecant functions. But one must be consistent. In this course we will use the choice of conventions presented in the main text, which results in derivatives without absolute values.

Inverse sin function:
$\begin{align*} y = \sin x \qquad \Leftrightarrow \qquad x = \sin^{-1}y \qquad \text{for}~\ \amp x \in [-\pi/2,\pi/2],\\ \amp y \in [-1,1]. \end{align*}$
Inverse cos function:
$\begin{align*} y = \cos x \qquad \Leftrightarrow \qquad x = \cos^{-1}y \qquad \text{for}~ \amp x \in [0,\pi],\\ \amp y\in[-1,1]. \end{align*}$
Inverse tan function:
$\begin{align*} y = \tan x \qquad \Leftrightarrow \qquad x = \tan^{-1}y \qquad \text{for}~\amp x \in (-\pi/2,\pi/2),\\ \amp y \in \mathbb{R}. \end{align*}$
Inverse cotan function:
$\begin{align*} y = \cot x \qquad \Leftrightarrow \qquad x = \cot^{-1}y \qquad \text{for}~\amp x \in (0,\pi),\\ \amp y \in \mathbb{R}. \end{align*}$
Inverse sec function:
$\begin{align*} y = \sec x \qquad \Leftrightarrow \qquad x = \sec^{-1}y \qquad \text{for} ~ \amp x \in [0,\pi/2) \cup [\pi, 3\pi/2),\\ \amp y \in (-\infty, -1] \cup [1, \infty). \end{align*}$
Inverse cosec function:
$\begin{align*} y = \csc x \qquad \Leftrightarrow \qquad x = \csc^{-1}y \qquad \text{for}~\amp x \in (0,\pi/2] \cup (\pi, 3\pi/2]\\ \amp y \in (-\infty, -1] \cup [1, \infty). \end{align*}$

Note that inverse trig functions are also denoted by $\arcsin(x)\text{,}$ $\arccos(x)\text{,}$ etc., which is often preferred. And, very importantly, remark that

$\begin{equation*} \sin^{-1}(x) \neq \frac{1}{\sin(x)}\ ! \end{equation*}$

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Concept 4.9.2. Derivatives of inverse trigonometric functions.

$\begin{align*} \frac{d}{dx} \sin^{-1}(x) =\amp \frac{1}{\sqrt{1-x^2}}, \amp\frac{d}{dx} \cos^{-1}(x) =\amp - \frac{1}{\sqrt{1-x^2}},\\ \frac{d}{dx} \tan^{-1}(x) =\amp \frac{1}{1+x^2}, \amp\frac{d}{dx} \cot^{-1}(x) =\amp - \frac{1}{1+x^2},\\ \frac{d}{dx} \sec^{-1}(x) =\amp \frac{1}{x \sqrt{x^2-1}}, \amp\frac{d}{dx} \csc^{-1}(x) =\amp - \frac{1}{x \sqrt{x^2-1}}. \end{align*}$

MATH 144: Calculus for the Physical Sciences I

Section 4.9 Inverse trigonometric functions

Objectives

Subsection 4.9.1 Instructional video

Subsection 4.9.2 Key concepts

Concept 4.9.1. Inverse trigonometric functions.

Concept 4.9.2. Derivatives of inverse trigonometric functions.

Further readings 4.9.3 Further readings