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 $[-\pi, \pi]$.