CBSE Class 12 Mathematics Inverse Trigonometric Functions Notes Set 02

Important Notes and Formulas

The inverse of a function \( f : A \to B \) exists if f is one-one onto, i.e., a bijection, and is given by \( f(x) = y \Rightarrow f^{-1}(y) = x \).

Consider the sine function with domain R and range [−1, 1]. Clearly this function is not a bijection and so it is not invertible. If we restrict the domain of it in such a way that it becomes one-one, then it would become invertible. If we consider sine as a function with domain \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \) and co-domain [−1, 1], then it is a bijection and therefore, invertible. The inverse of sine function is defined as \( \sin^{-1} x = \theta \Leftrightarrow \sin\theta = x \), where \( \theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \) and \( x \in [-1, 1] \).

(1) Meaning of Inverse Function

(i) \( \sin\theta = x \Rightarrow \sin^{-1} x = \theta \)    (ii) \( \cos\theta = x \Rightarrow \cos^{-1} x = \theta \)    (iii) \( \tan\theta = x \Rightarrow \tan^{-1} x = \theta \)
(iv) \( \cot\theta = x \Rightarrow \cot^{-1} x = \theta \)    (v) \( \sec\theta = x \Rightarrow \sec^{-1} x = \theta \)    (vi) \( \text{cosec}\,\theta = x \Rightarrow \text{cosec}^{-1} x = \theta \)

(2) Domain and Range of Inverse Functions

(i) If \( \sin y = x \), then \( y = \sin^{-1} x \), under certain conditions.
\( -1 \leq \sin y \leq 1 \); but \( \sin y = x \) ∴ \( -1 \leq x \leq 1 \)
Again, \( \sin y = -1 \Rightarrow y = -\frac{\pi}{2} \) and \( \sin y = 1 \Rightarrow y = \frac{\pi}{2} \).
Keeping in mind numerically smallest angles or real numbers, ∴ \( -\frac{\pi}{2} \leq y \leq \frac{\pi}{2} \)
These restrictions on the values of x and y provide us with the domain and range for the function \( y = \sin^{-1} x \).
i.e., Domain: \( x \in [-1, 1] \)
Range: \( y \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \)

(ii) Let \( \cos y = x \), then \( y = \cos^{-1} x \), under certain conditions \( -1 \leq \cos y \leq 1 \)
⇒ \( -1 \leq x \leq 1 \)
\( \cos y = -1 \Rightarrow y = \pi \)
\( \cos y = 1 \Rightarrow y = 0 \)
∴ \( 0 \leq y \leq \pi \) {as cos x is a decreasing function in \( [0, \pi] \); hence \( \cos\pi \leq \cos y \leq \cos 0 \)}
These restrictions on the values of x and y provide us the domain and range for the function \( y = \cos^{-1} x \).
i.e. Domain: \( x \in [-1, 1] \)
Range: \( y \in [0, \pi] \)

(iii) If \( \tan y = x \), then \( y = \tan^{-1} x \), under certain conditions.
Here, \( \tan y \in R \Rightarrow x \in R \), \( -\infty \lt \tan y \lt \infty \Rightarrow -\frac{\pi}{2} \lt y \lt \frac{\pi}{2} \)
Thus, Domain: \( x \in R \);
Range: \( y \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \)

(iv) If \( \cot y = x \), then \( y = \cot^{-1} x \)
under certain conditions, \( \cot y \in R \Rightarrow x \in R \); \( -\infty \lt \cot y \lt \infty \Rightarrow 0 \lt y \lt \pi \)
These conditions on x and y make the function, \( \cot y = x \) one-one and onto so that the inverse function exists, i.e., \( y = \cot^{-1} x \) is meaningful.
⇒ Domain: \( x \in R \)
Range: \( y \in (0, \pi) \)

(v) If \( \sec y = x \), then \( y = \sec^{-1} x \), where \( |x| \geq 1 \) and \( 0 \leq y \leq \pi, y \neq \frac{\pi}{2} \)
Here, Domain: \( x \in R - (-1, 1) \)
Range: \( y \in [0, \pi] - \left\{\frac{\pi}{2}\right\} \)

(vi) If \( \text{cosec}\, y = x \), then \( y = \text{cosec}^{-1} x \)
where \( |x| \geq 1 \) and \( -\frac{\pi}{2} \leq y \leq \frac{\pi}{2}, y \neq 0 \)
Here, Domain: \( x \in R - (-1, 1) \)
Range: \( y \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] - \{0\} \)

(3) \( \sin^{-1}(\sin\theta) = \theta \), provided that \( -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2} \);   \( \cos^{-1}(\cos\theta) = \theta \), provided that \( 0 \leq \theta \leq \pi \)
\( \tan^{-1}(\tan\theta) = \theta \), provided that \( -\frac{\pi}{2} \lt \theta \lt \frac{\pi}{2} \);   \( \cot^{-1}(\cot\theta) = \theta \), provided that \( 0 \lt \theta \lt \pi \)
\( \sec^{-1}(\sec\theta) = \theta \), provided that \( 0 \leq \theta \lt \frac{\pi}{2} \) or \( \frac{\pi}{2} \lt \theta \leq \pi \)
\( \text{cosec}^{-1}(\text{cosec}\,\theta) = \theta \), provided that \( -\frac{\pi}{2} \leq \theta \lt 0 \) or \( 0 \lt \theta \leq \frac{\pi}{2} \)

(4) \( \sin(\sin^{-1} x) = x \), provided that \( -1 \leq x \leq 1 \);   \( \cos(\cos^{-1} x) = x \), provided that \( -1 \leq x \leq 1 \)
\( \tan(\tan^{-1} x) = x \), provided that \( -\infty \lt x \lt \infty \);   \( \cot(\cot^{-1} x) = x \), provided that \( -\infty \lt x \lt \infty \)
\( \sec(\sec^{-1} x) = x \), provided that \( -\infty \lt x \leq -1 \) or \( 1 \leq x \lt \infty \)
\( \text{cosec}(\text{cosec}^{-1} x) = x \), provided that \( -\infty \lt x \leq -1 \) or \( 1 \leq x \lt \infty \)

Important Tips
Here, \( \sin^{-1} x, \text{cosec}^{-1} x, \tan^{-1} x \) belong to I and IV quadrant.
Here, \( \cos^{-1} x, \sec^{-1} x, \cot^{-1} x \) belong to I and II quadrant.
I quadrant is common to all the inverse functions.
III quadrant is not used in inverse function.
IV quadrant is used in the clockwise direction, i.e., \( -\frac{\pi}{2} \leq y \leq 0 \).

(5) Principal Values for Inverse Circular Functions

Thus \( \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6} \), not \( \frac{5\pi}{6} \); \( \cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3} \), not \( \frac{4\pi}{3} \); \( \tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3} \), not \( \frac{2\pi}{3} \); \( \cot^{-1}(-1) = \frac{3\pi}{4} \), not \( -\frac{\pi}{4} \) etc.
Note: \( \sin^{-1} x, \cos^{-1} x, \tan^{-1} x \) are also written as arc sin x, arc cos x and arc tan x respectively.
It should be noted that if not otherwise stated only principal values of inverse circular functions are to be considered.

Solved Examples

Example 1. The principal value of \( \sin^{-1}\left(\frac{-\sqrt{3}}{2}\right) \) is [Roorkee 1992]
(a) \( -\frac{2\pi}{3} \)
(b) \( -\frac{\pi}{3} \)
(c) \( \frac{4\pi}{3} \)
(d) \( \frac{5\pi}{3} \)
Answer:
Ans. (b) \( -\frac{\pi}{3} \)
\( \sin^{-1}\left[\sin\left(-\frac{\pi}{3}\right)\right] = -\frac{\pi}{3} \)   \( \left(\because -\frac{\pi}{2} \leq \sin^{-1} x \leq \frac{\pi}{2}\right) \)

Example 2. \( \sec^{-1}[\sec(-30^\circ)] = \) [MP PET 1992]
(a) −60°
(b) −30°
(c) 30°
(d) 150°
Answer:
Ans. (c) 30°
\( \sec^{-1}[\sec(-30^\circ)] = \sec^{-1}(\sec 30^\circ) = 30^\circ \).

Example 3. The principal value of \( \sin^{-1}\left(\sin\frac{5\pi}{3}\right) \) is [MP PET 1996]
(a) \( \frac{5\pi}{3} \)
(b) \( -\frac{5\pi}{3} \)
(c) \( -\frac{\pi}{3} \)
(d) \( \frac{4\pi}{3} \)
Answer:
Ans. (c) \( -\frac{\pi}{3} \)
\( \sin^{-1}\left(\sin\frac{5\pi}{3}\right) = \sin^{-1}\left(-\frac{\sqrt{3}}{2}\right) = -\frac{\pi}{3} \).

Example 4. The principal value of \( \sin^{-1}\left[\sin\left(\frac{2\pi}{3}\right)\right] \) is [IIT 1986]
(a) \( -\frac{2\pi}{3} \)
(b) \( \frac{2\pi}{3} \)
(c) \( \frac{4\pi}{3} \)
(d) None of these
Answer:
Ans. (d) None of these
The principal value of \( \sin^{-1}\left[\sin\left(\pi - \frac{2\pi}{3}\right)\right] = \sin^{-1}\left[\sin\left(\frac{\pi}{3}\right)\right] = \frac{\pi}{3} \).

Example 5. If \( \theta = \sin^{-1}[\sin(-600^\circ)] \), then one of the possible values of θ is
(a) \( \frac{\pi}{3} \)
(b) \( \frac{\pi}{2} \)
(c) \( \frac{2\pi}{3} \)
(d) \( -\frac{2\pi}{3} \)
Answer:
Ans. (a) \( \frac{\pi}{3} \)
\( \theta = \sin^{-1}[\sin(-600^\circ)] = \sin^{-1}[-\sin(360^\circ + 240^\circ)] \)
⇒ \( \theta = \sin^{-1}[-\sin 240^\circ] = \sin^{-1}[-\sin(180^\circ + 60^\circ)] \) ⇒ \( \theta = \sin^{-1}(\sin 60^\circ) = \sin^{-1}\left[\sin\left(\frac{\pi}{3}\right)\right] = \frac{\pi}{3} \).

Example 6. Value of \( \cos^{-1}\left(\cos\frac{5\pi}{3}\right) + \sin^{-1}\left(\sin\frac{5\pi}{3}\right) \) is [Roorkee 2000]
(a) 0
(b) \( \frac{\pi}{2} \)
(c) \( \frac{2\pi}{3} \)
(d) \( \frac{10\pi}{3} \)
Answer:
Ans. (a) 0
\( \cos^{-1}\left(\cos\frac{5\pi}{3}\right) + \sin^{-1}\left(\sin\frac{5\pi}{3}\right) = \cos^{-1}\left[\cos\left(2\pi - \frac{\pi}{3}\right)\right] + \sin^{-1}\left[\sin\left(2\pi - \frac{\pi}{3}\right)\right] \)
\( = \frac{\pi}{3} - \frac{\pi}{3} = 0 \).