Question:

If a function f:XYf : X \to Y defined as f(x)=yf(x) = y is one-one and onto, then we can define a unique function f(x)=yf(x) = y such that f(x)=yf(x) = y, where f(x)=yf(x) = y and f(x)=yf(x) = yf(x)=yf(x) = y. Function gg is called the inverse of function ff

The domain of sine function is R\mathbb{R} and function sine : RR\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 AA to [1,1][-1, 1] such that inverse of sine function exists, i.e., sin1x\sin^{-1} x is defined from [1,1][-1, 1] to AA

On the basis of the above information, answer the following questions: 
(i) If AA is the interval other than principal value branch, give an example of one such interval.
(ii) If sin1(x)\sin^{-1}(x) is defined from [1,1][-1, 1] to its principal value branch, find the value of sin1(12)sin1(1)\sin^{-1}\left(-\frac{1}{2}\right) - \sin^{-1}(1).  
(iii) Draw the graph of sin1x\sin^{-1} x from [1,1][-1, 1] to its principal value branch. 
(iv) Find the domain and range of f(x)=2sin1(1x)f(x) = 2 \sin^{-1}(1 - x).

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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.
Updated On: Jan 18, 2025
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Solution and Explanation

Given Information: The sine function y=sinxy = \sin x is defined from R\mathbb{R} to [1,1][-1, 1], but it is neither one-one nor onto over R\mathbb{R}
By restricting the domain to an interval, we can define the inverse sin1x\sin^{-1}x, which is a one-one and onto function. 

(i) If AA 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 [π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}]. Another interval where the sine function is one-one and onto [1,1][-1, 1] is: A=[π2,3π2]. A = [\frac{\pi}{2}, \frac{3\pi}{2}].

(ii) If sin1(x)\sin^{-1}(x) is defined from [1,1][-1, 1] to its principal value branch, find the value of: sin1(12)sin1(1). \sin^{-1}\left(-\frac{1}{2}\right) - \sin^{-1}(1).  

Step 1: Evaluate sin1(12)\sin^{-1}\left(-\frac{1}{2}\right). From the definition of sin1\sin^{-1}, sin1(12)\sin^{-1}\left(-\frac{1}{2}\right) is the angle θ\theta in the interval [π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}] such that: sinθ=12. \sin \theta = -\frac{1}{2}. Thus: sin1(12)=π6. \sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6}.

Step 2: Evaluate sin1(1)\sin^{-1}(1). From the definition of sin1\sin^{-1}, sin1(1)\sin^{-1}(1) is the angle θ\theta in the interval [π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}] such that: sinθ=1. \sin \theta = 1. Thus: sin1(1)=π2. \sin^{-1}(1) = \frac{\pi}{2}.

Step 3: Compute the difference. sin1(12)sin1(1)=π6π2. \sin^{-1}\left(-\frac{1}{2}\right) - \sin^{-1}(1) = -\frac{\pi}{6} - \frac{\pi}{2}. Simplify: sin1(12)sin1(1)=π63π6=4π6=2π3. \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 sin1x\sin^{-1}x from [1,1][-1, 1] to its principal value branch. Step 1: Description of the graph. The graph of sin1x\sin^{-1}x is obtained by reflecting the graph of y=sinxy = \sin x (restricted to [π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}]) across the line y=xy = x

(iii) (b) Find the domain and range of f(x)=2sin1(1x)f(x) = 2 \sin^{-1}(1 - x). Step 1: Domain of f(x)f(x). For f(x)=2sin1(1x)f(x) = 2 \sin^{-1}(1 - x), the argument of sin1\sin^{-1} must lie within [1,1][-1, 1], i.e.: 11x1. -1 \leq 1 - x \leq 1.  
Simplify: 11x11    2x0. -1 - 1 \leq -x \leq 1 - 1 \implies -2 \leq -x \leq 0.  
Multiplying through by 1-1 (and reversing inequalities): 0x2. 0 \leq x \leq 2. Thus, the domain is: x[0,2]. x \in [0, 2].

Step 2: Range of f(x)f(x). The range of sin1(x)\sin^{-1}(x) is [π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}]. Therefore: f(x)=2sin1(1x)    f(x)[2π2,2π2]. 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)[π,π]. f(x) \in [-\pi, \pi].


Final Answers: 
1. Example of an interval other than the principal value branch: [π2,3π2][\frac{\pi}{2}, \frac{3\pi}{2}].
2. sin1(12)sin1(1)=2π3\sin^{-1}\left(-\frac{1}{2}\right) - \sin^{-1}(1) = -\frac{2\pi}{3}.
3. (a) The graph of y=sin1xy = \sin^{-1}x is shown above. 
(b) The domain of f(x)=2sin1(1x)f(x) = 2 \sin^{-1}(1 - x) is [0,2][0, 2], and the range is [π,π][-\pi, \pi].

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