Chapter 2: Inverse Trigonometric Functions
This chapter extends the concept of inverse functions to trigonometric functions by defining the principal value branches.
Domain and Range of Inverse Trigonometric Functions
Trigonometric functions are not one-one and onto over their natural domains and ranges. Therefore, their inverses do not exist. But if we appropriately restrict their domains, they become bijections and thus invertible. The restricted domains and ranges are called principal value branches.
| Function | Domain | Range (Principal Value Branch) |
|---|---|---|
| $y = \sin^{-1} x$ | $[-1, 1]$ | $[-\pi/2, \pi/2]$ |
| $y = \cos^{-1} x$ | $[-1, 1]$ | $[0, \pi]$ |
| $y = \tan^{-1} x$ | $\mathbb{R}$ | $(-\pi/2, \pi/2)$ |
| $y = \cot^{-1} x$ | $\mathbb{R}$ | $(0, \pi)$ |
| $y = \sec^{-1} x$ | $\mathbb{R} - (-1, 1)$ | $[0, \pi] - \{\pi/2\}$ |
| $y = \text{cosec}^{-1} x$ | $\mathbb{R} - (-1, 1)$ | $[-\pi/2, \pi/2] - \{0\}$ |
Graphs of Inverse Trigonometric Functions
The graph of an inverse trigonometric function can be obtained from the graph of the original restricted trigonometric function by interchanging $x$ and $y$ axes, which geometrically translates to reflecting the graph across the line $y = x$.
Fig 1. Graph of Principal Value Branch of $\sin^{-1}(x)$
Competency-Based Questions
1. [CBSE 2020] Write the principal value of $\cos^{-1}\left(\cos\left(\frac{7\pi}{6}\right)\right)$.
Solution:
We know that the principal value branch of $\cos^{-1} x$ is $[0, \pi]$.
$\frac{7\pi}{6}$ does not belong to $[0, \pi]$.
Therefore, $\cos^{-1}\left(\cos\left(\frac{7\pi}{6}\right)\right) = \cos^{-1}\left(\cos\left(2\pi - \frac{5\pi}{6}\right)\right)$
$= \cos^{-1}\left(\cos\left(\frac{5\pi}{6}\right)\right)$
Since $\frac{5\pi}{6} \in [0, \pi]$, we have $\cos^{-1}\left(\cos\left(\frac{5\pi}{6}\right)\right) = \frac{5\pi}{6}$.
2. [Sample Paper 2022] Evaluate: $\tan^{-1}(1) + \cos^{-1}(-1/2) + \sin^{-1}(-1/2)$.
Solution:
Let $\tan^{-1}(1) = x \implies \tan x = 1 \implies x = \frac{\pi}{4} \in (-\pi/2, \pi/2)$
Let $\cos^{-1}(-1/2) = y \implies \cos y = -1/2 = -\cos(\pi/3) = \cos(\pi - \pi/3) = \cos(2\pi/3) \implies y = \frac{2\pi}{3} \in [0, \pi]$
Let $\sin^{-1}(-1/2) = z \implies \sin z = -1/2 = -\sin(\pi/6) = \sin(-\pi/6) \implies z = -\frac{\pi}{6} \in [-\pi/2, \pi/2]$
Therefore, the sum is $\frac{\pi}{4} + \frac{2\pi}{3} - \frac{\pi}{6}$
$= \frac{3\pi + 8\pi - 2\pi}{12} = \frac{9\pi}{12} = \frac{3\pi}{4}$.
3. [CBSE 2018] Find the domain of the function $f(x) = \sin^{-1}(x^2 - 4)$.
Solution:
The domain of $\sin^{-1} x$ is $[-1, 1]$.
Therefore, for $f(x)$ to be defined, $-1 \le x^2 - 4 \le 1$.
Adding $4$ to the inequality:
$3 \le x^2 \le 5$.
This implies $x \in [-\sqrt{5}, -\sqrt{3}] \cup [\sqrt{3}, \sqrt{5}]$.
So, the domain of $f(x)$ is $[-\sqrt{5}, -\sqrt{3}] \cup [\sqrt{3}, \sqrt{5}]$.