FAC1004 L13 — Inverse Trigonometric Functions

Learning Objective

To identify the domains, ranges, and graphs of inverse trigonometric functions and calculate their principal values.

Core Idea

The six basic trigonometric functions are not one-to-one (their values repeat periodically). To define their inverses, we restrict their domains to intervals on which they are one-to-one.

Definitions, Domains, and Ranges

1. Inverse Sine — $\sin^{-1} x$ or $\arcsin x$

Defined as the inverse of the restricted sine function: $$y = \sin x, \quad -\frac{\pi}{2} \le x \le \frac{\pi}{2}$$

  • Domain: $[-1, 1]$
  • Range (Principal Value): $\displaystyle\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

2. Inverse Cosine — $\cos^{-1} x$ or $\arccos x$

Defined as the inverse of the restricted cosine function: $$y = \cos x, \quad 0 \le x \le \pi$$

  • Domain: $[-1, 1]$
  • Range (Principal Value): $[0, \pi]$

3. Inverse Tangent — $\tan^{-1} x$ or $\arctan x$

Defined as the inverse of the restricted tangent function: $$y = \tan x, \quad -\frac{\pi}{2} < x < \frac{\pi}{2}$$

  • Domain: $(-\infty, \infty)$ or $\mathbb{R}$
  • Range (Principal Value): $\displaystyle\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$

4. Inverse Secant — $\sec^{-1} x$ or $\text{arcsec } x$

Defined as the inverse of the restricted secant function: $$y = \sec x, \quad 0 \le x < \frac{\pi}{2}, ; \frac{\pi}{2} < x \le \pi$$

  • Domain: $(-\infty, -1] \cup [1, \infty)$
    (equivalently: $x \ge 1$ or $x \le -1$)
  • Range (Principal Value): $\displaystyle\left[0, \frac{\pi}{2}\right) \cup \left(\frac{\pi}{2}, \pi\right]$

5. Inverse Cosecant — $\csc^{-1} x$ or $\text{arccsc } x$

Defined as the inverse of the restricted cosecant function: $$y = \csc x, \quad -\frac{\pi}{2} \le x < 0, ; 0 < x \le \frac{\pi}{2}$$

  • Domain: $(-\infty, -1] \cup [1, \infty)$
    (equivalently: $x \ge 1$ or $x \le -1$)
  • Range (Principal Value): $\displaystyle\left[-\frac{\pi}{2}, 0\right) \cup \left(0, \frac{\pi}{2}\right]$

6. Inverse Cotangent — $\cot^{-1} x$ or $\text{arccot } x$

Defined as the inverse of the restricted cotangent function: $$y = \cot x, \quad 0 < x < \pi$$

  • Domain: $(-\infty, \infty)$ or $\mathbb{R}$
  • Range (Principal Value): $(0, \pi)$

Note: The lecture definition page explicitly gives Range $(0, \pi)$ for $\cot^{-1} x$ (consistent with the asymptotic behaviour of $\cot x$ at $0$ and $\pi$).

Summary Table

Function Domain Range (Principal Value)
$y = \sin^{-1} x$ $[-1, 1]$ $\displaystyle\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
$y = \cos^{-1} x$ $[-1, 1]$ $[0, \pi]$
$y = \tan^{-1} x$ $\mathbb{R}$ $\displaystyle\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$
$y = \cot^{-1} x$ $\mathbb{R}$ $(0, \pi)$
$y = \sec^{-1} x$ $(-\infty, -1] \cup [1, \infty)$ $\displaystyle\left[0, \frac{\pi}{2}\right) \cup \left(\frac{\pi}{2}, \pi\right]$
$y = \csc^{-1} x$ $(-\infty, -1] \cup [1, \infty)$ $\displaystyle\left[-\frac{\pi}{2}, 0\right) \cup \left(0, \frac{\pi}{2}\right]$

Worked Examples

Example 1 — Principal Values

Find the principal value of each of the following:

(i) $\displaystyle\sin^{-1}\left(\frac{1}{\sqrt{2}}\right)$

Let $\displaystyle\sin^{-1}\left(\frac{1}{\sqrt{2}}\right) = \theta$. Then: $$\sin \theta = \frac{1}{\sqrt{2}} = \sin\left(\frac{\pi}{4}\right)$$ $$\boxed{\theta = \frac{\pi}{4}}$$

(ii) $\displaystyle\cos^{-1}\left(-\frac{1}{2}\right)$

Let $\displaystyle\cos^{-1}\left(-\frac{1}{2}\right) = \theta$. Then: $$\cos \theta = -\frac{1}{2} = \cos\left(\pi - \frac{\pi}{3}\right) = \cos\left(\frac{2\pi}{3}\right)$$ $$\boxed{\theta = \frac{2\pi}{3}}$$

(iii) $\displaystyle\tan^{-1}\left(-\frac{1}{\sqrt{3}}\right)$

Let $\displaystyle\tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) = \theta$. Then: $$\tan \theta = -\frac{1}{\sqrt{3}} = \tan\left(-\frac{\pi}{6}\right)$$ $$\boxed{\theta = -\frac{\pi}{6}}$$

Example 2 — Composition of Functions

Find the value of $\displaystyle\sec\left[\cos^{-1}\frac{\sqrt{3}}{2}\right]$.

Let $\displaystyle\cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \theta$. Then: $$\cos \theta = \frac{\sqrt{3}}{2} = \cos\left(\frac{\pi}{6}\right) \quad\Longrightarrow\quad \theta = \frac{\pi}{6}$$

Therefore: $$\sec\left[\cos^{-1}\frac{\sqrt{3}}{2}\right] = \sec \theta = \sec\left(\frac{\pi}{6}\right) = \frac{2}{\sqrt{3}}$$

Example 3 — Algebraic Simplification

Simplify $\cos\left(\sin^{-1} x\right)$.

Let $\sin^{-1} x = \theta ;\Rightarrow; x = \sin \theta$.

$$\cos\left(\sin^{-1} x\right) = \cos \theta = \sqrt{1 - \sin^{2}\theta} = \sqrt{1 - x^{2}}$$

$$\boxed{\cos\left(\sin^{-1} x\right) = \sqrt{1 - x^{2}}}$$

Simplify $\cot\left(\csc^{-1} x\right)$.

Let $\csc^{-1} x = \theta ;\Rightarrow; x = \csc \theta$.

$$\cot\left(\csc^{-1} x\right) = \cot \theta = \sqrt{\csc^{2}\theta - 1} = \sqrt{x^{2} - 1}$$

$$\boxed{\cot\left(\csc^{-1} x\right) = \sqrt{x^{2} - 1}}$$

Key Takeaways

  1. Inverse trig functions require domain restriction on the original trig function to ensure it is one-to-one.
  2. The principal value is the unique angle in the restricted range that satisfies the equation.
  3. For negative inputs, use the corresponding reference angle and adjust according to the range of the inverse function.
  4. Compositions can be simplified using Pythagorean identities.

Related

Source File

LECTURE_NOTES_2526/L13 Inverse Trigonometric Function Full Version.pdf