Question

If a function \( f : X \to Y \) defined as \( f(x) = y \) is one-one and onto, then we can define a unique function \( g : Y \to X \) such that \( g(y) = x \), where \( x \in X \) and \( y = f(x) \), \( y \in Y \). Function \( g \) is called the inverse of function \( f \). \vspace{0.3cm} The domain of the sine function is \( \mathbb{R} \) and function sine: \( \mathbb{R} \to \mathbb{R} \) is neither one-one nor onto. The following graph shows the sine function. \includegraphics[width=\linewidth]{latex4.png} Let sine function be defined from set \( A \) to \( [-1, 1] \) such that the inverse of the sine function exists, i.e., \( \sin^{-1} x \) is defined from \( [-1, 1] \) to \( A \). \vspace{0.3cm} On the basis of the above information, answer the following questions: \begin{enumerate} If \( A \) is the interval other than the principal value branch, give an example of one such interval. If \( \sin^{-1}(x) \) is defined from \( [-1, 1] \) to its principal value branch, find the value of \( \sin^{-1}\left(-\frac{1}{2}\right) - \sin^{-1}(1) \). \begin{enumerate} Draw the graph of \( \sin^{-1} x \) from \( [-1, 1] \) to its principal value branch. OR Find the domain and range of \( f(x) = 2 \sin^{-1}(1 - x) \). \end{enumerate} \end{enumerate}

Hint:

To find the domain of functions involving inverse trigonometric functions, ensure the argument satisfies the range of the trigonometric function. For the range, scale the interval accordingly.

Answer 1

Step 1: Understanding the sine function and its inverse. The sine function \( y = \sin x \) is defined from \( \mathbb{R} \) to \( [-1, 1] \), but it is neither one-one nor onto over \( \mathbb{R} \). By restricting the domain, the inverse \( \sin^{-1}x \) can be defined. Step 2: Answering the given questions. (i) Interval other than the principal value branch: The principal value branch is the interval \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \). Another interval where sine is one-one and onto is: \[ \left[\frac{\pi}{2}, \frac{3\pi}{2}\right] \] (ii) Evaluating \( \sin^{-1}\left(-\frac{1}{2}\right) - \sin^{-1(1) \):} \[ \sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6}, \quad \sin^{-1}(1) = \frac{\pi}{2} \] \[ \sin^{-1}\left(-\frac{1}{2}\right) - \sin^{-1}(1) = -\frac{\pi}{6} - \frac{\pi}{2} = -\frac{2\pi}{3} \] (iii) (a) Graph of \( \sin^{-1} x \): The graph of \( \sin^{-1} x \) is obtained by reflecting \( y = \sin x \) across the line \( y = x \). (iii) (b) Domain and range of \( f(x) = 2 \sin^{-1(1 - x) \):} For \( \sin^{-1}(1 - x) \) to be defined: \[ -1 \leq 1 - x \leq 1 \quad \Rightarrow \quad 0 \leq x \leq 2 \] Thus, the domain is \( [0, 2] \). Since \( \sin^{-1}(x) \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \), \[ f(x) \in \left[-\pi, \pi \right] \] Final Answers: \begin{enumerate} Example interval: \( \left[\frac{\pi}{2}, \frac{3\pi}{2}\right] \). \( \sin^{-1}\left(-\frac{1}{2}\right) - \sin^{-1}(1) = -\frac{2\pi}{3} \). (a) Graph is shown above. (b) Domain: \( [0, 2] \), Range: \( [-\pi, \pi] \). \end{enumerate}