CBSE Class 12 Mathematics Inverse Trigonometric Functions Notes Set 01

These Class 12 CBSE Mathematics notes for Chapter 2, Inverse Trigonometric Functions, cover the definition, domain, range and principal value branches of all six inverse trigonometric functions. Solved examples with marking, Do Yourself practice questions and their answers follow the notes.

Important Notes and Formulas

1. Definition

(a) Inverse of a function: Let f : A → B be a one-one and onto function, then there exists a unique function g : B → A such that f(x) = y ⇔ g(y) = x, ∀ x ∈ A and y ∈ B. Then g is said to be the inverse of f.
Thus \( g = f^{-1} : B \to A = \{(f(x), x) \mid (x, f(x)) \in f\} \)

(b) Inverse trigonometric function: An inverse function which is related to a trigonometric ratio is called an inverse trigonometric function.
If \( \sin\theta = x \Rightarrow \theta = \sin^{-1} x;\ \theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \)
Note: \( (\sin^{-1} x) \ne (\sin x)^{-1} \)

2. Domain, Range and Principal Value Branch

Note-1:
(i) \( y = \sin(\sin^{-1} x) = x,\ x \in [-1, 1],\ y \in [-1, 1] \), y is aperiodic
(ii) \( y = \cos(\cos^{-1} x) = x,\ x \in [-1, 1],\ y \in [-1, 1] \), y is aperiodic
(iii) \( y = \tan(\tan^{-1} x) = x,\ x \in R,\ y \in R \), y is aperiodic
(iv) \( y = \cot(\cot^{-1} x) = x,\ x \in R,\ y \in R \), y is aperiodic
(v) \( y = \text{cosec}(\text{cosec}^{-1} x) = x,\ |x| \ge 1,\ |y| \ge 1 \), y is aperiodic
(vi) \( y = \sec(\sec^{-1} x) = x,\ |x| \ge 1,\ |y| \ge 1 \), y is aperiodic

Note-2:
(i) \( y = \sin^{-1}(\sin x) = x,\ x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \), periodic with period 2π
(ii) \( y = \cos^{-1}(\cos x) = x,\ x \in [0, \pi] \), periodic with period 2π
(iii) \( y = \tan^{-1}(\tan x) = x,\ x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \), periodic with period π
(iv) \( y = \cot^{-1}(\cot x) = x,\ x \in (0, \pi) \), periodic with period π
(v) \( y = \text{cosec}^{-1}(\text{cosec}\, x) = x,\ x \in \left[-\frac{\pi}{2}, 0\right) \cup \left(0, \frac{\pi}{2}\right] \), periodic with period 2π
(vi) \( y = \sec^{-1}(\sec x) = x,\ x \in \left[0, \frac{\pi}{2}\right) \cup \left(\frac{\pi}{2}, \pi\right] \), periodic with period 2π

3. Graphs of Inverse Trigonometric Functions

Each graph shows the principal value branch. Filled dots are end points included in the graph; red dashed lines are horizontal asymptotes that the graph approaches but never meets.

Solved Examples

Illustration 1. Evaluate: \( \sin\left(\tan^{-1} \frac{15}{8}\right) \). [1 Mark]
Answer:
We know that \( \sin(\sin^{-1} x) = x \) for all x ∈ [−1, 1]. So, we will convert each expression in the form \( \sin(\sin^{-1} x) \) by using
\( \cos^{-1} \frac{b}{h} = \sin^{-1} \frac{p}{h},\ \tan^{-1} \frac{p}{b} = \sin^{-1} \frac{p}{h},\ \cot^{-1} \frac{p}{b} = \sin^{-1} \frac{b}{h} \) etc.
where b, p and h denote the base, perpendicular and hypotenuse of a right triangle.

\( = \sin\left(\tan^{-1} \frac{15}{8}\right) \)
\( = \sin\left(\sin^{-1} \frac{15}{17}\right) = \frac{15}{17} \)

Illustration 2. Evaluate: \( \tan^{-1} \sqrt{3} - \sec^{-1}(-2) \). [1 Mark]
Answer:
Let \( \sec^{-1}(-2) = \theta \Rightarrow \sec\theta = -2 \)
\( \Rightarrow \cos\theta = -\frac{1}{2} = \cos \frac{2\pi}{3} \)
\( \Rightarrow \theta = \frac{2\pi}{3} \Rightarrow \sec^{-1}(-2) = \frac{2\pi}{3} \)
\( \therefore \tan^{-1} \sqrt{3} - \sec^{-1}(-2) = \frac{\pi}{3} - \frac{2\pi}{3} = -\frac{\pi}{3} \)

Illustration 3. Evaluate: \( \sin^{-1}\left(\sin \frac{7\pi}{6}\right) \). [1 Mark]
Answer:
\( \sin^{-1}\left(\sin \frac{7\pi}{6}\right) \ne \frac{7\pi}{6} \), because \( \frac{7\pi}{6} \) does not lie between \( -\frac{\pi}{2} \) and \( \frac{\pi}{2} \).
Now, \( \sin^{-1}\left(\sin \frac{7\pi}{6}\right) = \sin^{-1}\left\{\sin\left(\pi + \frac{\pi}{6}\right)\right\} \) \( \left[\because \frac{7\pi}{6} = \pi + \frac{\pi}{6}\right] \)
\( \Rightarrow \sin^{-1}\left(-\sin \frac{\pi}{6}\right) = -\sin^{-1}\left(\sin \frac{\pi}{6}\right) = -\frac{\pi}{6} \) \( [\because \sin(\pi + \theta) = -\sin\theta;\ \sin^{-1}(-\theta) = -\sin^{-1}\theta] \)

Illustration 4. Find the value of \( \sin\left(2\tan^{-1} \frac{2}{3}\right) + \cos(\tan^{-1} \sqrt{3}) \). [4 Marks]
Answer:
Let \( \tan^{-1} \frac{2}{3} = x \) and \( \tan^{-1} \sqrt{3} = y \) so that \( \tan x = \frac{2}{3} \) and \( \tan y = \sqrt{3} \). [1]
Therefore, \( \sin\left(2\tan^{-1} \frac{2}{3}\right) + \cos(\tan^{-1} \sqrt{3}) \)
= sin(2x) + cos y
\( = \frac{2\tan x}{1 + \tan^2 x} + \frac{1}{\sqrt{1 + \tan^2 y}} = \frac{2 \cdot \frac{2}{3}}{1 + \frac{4}{9}} + \frac{1}{\sqrt{1 + (\sqrt{3})^2}} \) [2]
\( = \frac{12}{13} + \frac{1}{2} = \frac{37}{26} \) [1]

Do Yourself

Do yourself–1:
(i) Prove that \( \cos^{-1}\left(\cos\left(\frac{5\pi}{3}\right)\right) = \frac{\pi}{3} \)

Do yourself–2:
(i) Evaluate \( \sin(2\sin^{-1} 0.8) \)

Answers for Do Yourself

Do yourself–2: (i) 0.96