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 \([-1, 1]\), but it is neither one-one nor onto over \(\mathbb{R}\). 
By restricting the domain to an interval, we can define the inverse \(\sin^{-1}x\), which is a one-one and onto function. 

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

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

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

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

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

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

(iii) (b) Find the domain and range of \(f(x) = 2 \sin^{-1}(1 - x)\). Step 1: Domain of \(f(x)\). For \(f(x) = 2 \sin^{-1}(1 - x)\), the argument of \(\sin^{-1}\) must lie within \([-1, 1]\), i.e.: \[ -1 \leq 1 - x \leq 1. \] 
Simplify: \[ -1 - 1 \leq -x \leq 1 - 1 \implies -2 \leq -x \leq 0. \] 
Multiplying through by \(-1\) (and reversing inequalities): \[ 0 \leq x \leq 2. \] Thus, the domain is: \[ x \in [0, 2]. \]

Step 2: Range of \(f(x)\). The range of \(\sin^{-1}(x)\) is \([-\frac{\pi}{2}, \frac{\pi}{2}]\). Therefore: \[ f(x) = 2 \sin^{-1}(1 - x) \implies f(x) \in \left[2 \cdot -\frac{\pi}{2}, 2 \cdot \frac{\pi}{2}\right]. \] Simplify: \[ f(x) \in [-\pi, \pi]. \]


Final Answers: 
1. Example of an interval other than the principal value branch: \([\frac{\pi}{2}, \frac{3\pi}{2}]\).
2. \(\sin^{-1}\left(-\frac{1}{2}\right) - \sin^{-1}(1) = -\frac{2\pi}{3}\).
3. (a) The graph of \(y = \sin^{-1}x\) is shown above. 
(b) The domain of \(f(x) = 2 \sin^{-1}(1 - x)\) is \([0, 2]\), and the range is \([-\pi, \pi]\).