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 \(f(x) = y\) such that \(f(x) = y\), where \(f(x) = y\) and \(f(x) = y\), \(f(x) = y\). Function \(g\) is called the inverse of function \(f\). 

The domain of 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. Let sine function be defined from set \(A\) to \([-1, 1]\) such that inverse of sine function exists, i.e., \(\sin^{-1} x\) is defined from \([-1, 1]\) to \(A\). 

On the basis of the above information, answer the following questions: 
(i) If \(A\) is the interval other than principal value branch, give an example of one such interval.
(ii) If \(\sin^{-1}(x)\) is defined from \([-1, 1]\) to its principal value branch, find the value of \(\sin^{-1}\left(-\frac{1}{2}\right) - \sin^{-1}(1)\).  
(iii) Draw the graph of \(\sin^{-1} x\) from \([-1, 1]\) to its principal value branch. 
(iv) Find the domain and range of \(f(x) = 2 \sin^{-1}(1 - x)\).

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

Given Information: The sine function \(y = \sin x\) is defined from \(\mathbb{R}\) to \([-1, 1]\). It is neither one-to-one nor onto over \(\mathbb{R}\). By restricting the domain to a specific interval, we can define the inverse sine function, \(\sin^{-1}x\), which is a one-to-one and onto function.

(i) Provide an example of an interval, other than the principal value branch, where the sine function is one-to-one and onto \([-1, 1]\). The principal value branch for the sine function is the interval \([-\frac{\pi}{2}, \frac{\pi}{2}]\). Another such interval is: \[ A = [\frac{\pi}{2}, \frac{3\pi}{2}]. \]

(ii) If \(\sin^{-1}(x)\) is defined from \([-1, 1]\) to its principal value branch, calculate the value of: \[ \sin^{-1}\left(-\frac{1}{2}\right) - \sin^{-1}(1). \]

Step 1: Evaluate \(\sin^{-1}\left(-\frac{1}{2}\right)\). According to the definition of \(\sin^{-1}\), \(\sin^{-1}\left(-\frac{1}{2}\right)\) is the angle \(\theta\) within the interval \([-\frac{\pi}{2}, \frac{\pi}{2}]\) such that \(\sin \theta = -\frac{1}{2}\). Therefore: \[ \sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6}. \]

Step 2: Evaluate \(\sin^{-1}(1)\). By the definition of \(\sin^{-1}\), \(\sin^{-1}(1)\) is the angle \(\theta\) in the interval \([-\frac{\pi}{2}, \frac{\pi}{2}]\) for which \(\sin \theta = 1\). Thus: \[ \sin^{-1}(1) = \frac{\pi}{2}. \]

Step 3: Compute the difference. The difference is: \[ \sin^{-1}\left(-\frac{1}{2}\right) - \sin^{-1}(1) = -\frac{\pi}{6} - \frac{\pi}{2}. \] Simplifying the expression: \[ \sin^{-1}\left(-\frac{1}{2}\right) - \sin^{-1}(1) = -\frac{\pi}{6} - \frac{3\pi}{6} = -\frac{4\pi}{6} = -\frac{2\pi}{3}. \]

(iii) (a) Graph the function \(\sin^{-1}x\) from \([-1, 1]\) to its principal value branch. Step 1: Graph description. The graph of \(\sin^{-1}x\) is obtained by reflecting the graph of \(y = \sin x\) (restricted to the interval \([-\frac{\pi}{2}, \frac{\pi}{2}]\)) across the line \(y = x\).

(iii) (b) Determine the domain and range of the function \(f(x) = 2 \sin^{-1}(1 - x)\). Step 1: Domain of \(f(x)\). For \(f(x) = 2 \sin^{-1}(1 - x)\), the argument of the \(\sin^{-1}\) function, \(1 - x\), must be within the interval \([-1, 1]\). This implies: \[ -1 \leq 1 - x \leq 1. \]
Solving for \(x\): \[ -1 - 1 \leq -x \leq 1 - 1 \implies -2 \leq -x \leq 0. \]
Multiplying by \(-1\) and reversing the inequalities: \[ 0 \leq x \leq 2. \] Therefore, the domain is: \[ x \in [0, 2]. \]

Step 2: Range of \(f(x)\). The range of the inverse sine function, \(\sin^{-1}(x)\), is \([-\frac{\pi}{2}, \frac{\pi}{2}]\). Consequently, for \(f(x) = 2 \sin^{-1}(1 - x)\): \[ f(x) \in \left[2 \cdot -\frac{\pi}{2}, 2 \cdot \frac{\pi}{2}\right]. \] Simplifying this gives: \[ f(x) \in [-\pi, \pi]. \]


Final Answers:
1. An interval other than the principal value branch where the sine function is one-to-one and onto \([-1, 1]\) is \([\frac{\pi}{2}, \frac{3\pi}{2}]\).
2. The value of \(\sin^{-1}\left(-\frac{1}{2}\right) - \sin^{-1}(1)\) is \(-\frac{2\pi}{3}\).
3. (a) The graph of \(y = \sin^{-1}x\) is depicted above.
(b) The domain of \(f(x) = 2 \sin^{-1}(1 - x)\) is \([0, 2]\), and its range is \([-\pi, \pi]\).