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Question: How do you find the domain and range of \(\arcsin (1 - {x^2})\)?...

How do you find the domain and range of arcsin⁡(1−x2)\arcsin (1 - {x^2})?

Explanation

Solution

First we know that the inverse functions. A function f(x)f(x) has an inverse if and only if it is one-to-one and onto.
Let us consider the function of arcsin⁡(1−x2)\arcsin (1 - {x^2}).
It is denoted by yy. And then constrained be in (−π2,π2)\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right).
We know that illustrates restricted domains, ranges of trigonometric functions and the domains, ranges, of the corresponding inverse functions.
Then find the xx domains.

Complete step-by-step solution:
The given function is arcsin⁡(1−x2)\arcsin (1 - {x^2}).
Let consider, y=arcsin⁡(1−x2)y = \arcsin (1 - {x^2}) is constrained be in (−π2,π2)\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right) and,
We shall illustrate below the restricted domains, ranges of trigonometric functions and the domains, ranges of the corresponding inverse functions.
sin⁡x:[−π2,π2]→[−1,1];\sin x:\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right] \to \left[ { - 1,1} \right]; and
sin⁡−1x:[−1,1]→[−π2,π2]{\sin ^{ - 1}}x:\left[ { - 1,1} \right] \to \left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]
As (1−x2)(1 - {x^2}) is a sine value, −1⩽1−x2⩽1 - 1 \leqslant 1 - {x^2} \leqslant 1
Now, we interchange the values in LHS (Left Hand Side)
⇒−1⩽1−x2⇒x2⩽1+1\Rightarrow - 1 \leqslant 1 - {x^2} \Rightarrow {x^2} \leqslant 1 + 1
Add on RHS,
x2⩽2{x^2} \leqslant 2
Now take square root on both sides,
∣x∣⩽2\left| x \right| \leqslant \sqrt 2

Then, the domain is
−2⩽x⩽2- \sqrt 2 \leqslant x \leqslant \sqrt 2 .
The range is
(−π2,π2)\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)

Note: The inverse functions sin⁡−1x,cos⁡−1x,tan⁡−1x,cos⁡ec−1x,sec⁡−1x,cot⁡−1x{\sin ^{ - 1}}x,{\cos ^{ - 1}}x,{\tan ^{ - 1}}x,\cos e{c^{ - 1}}x,{\sec ^{ - 1}}x,{\cot ^{ - 1}}x are called inverse circular functions. For the function y=sin⁡xy = \sin x, there are infinitely many angles xx that satisfy sin⁡x=t,−1⩽t⩽1.\sin x = t, - 1 \leqslant t \leqslant 1. Of these infinite sets of values, there is one which lies in the interval(−π2,π2)\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right). This angle is called the Principal angle and is denoted bysin⁡−1t{\sin ^{ - 1}}t. The principal value of an inverse function is that value of the general value which is numerically least. It may be positive or negative. When there are two values, one is positive and the other is negative such that they are numerically equal, then the principal value is the positive one.
sin⁡−1x{\sin ^{ - 1}}x does not mean 1sin⁡x\dfrac{1}{{\sin x}}
Another notation for sin⁡−1x{\sin ^{ - 1}}xisarcsin⁡x\arcsin x.
While discussing the inverse of the sine function, we confined to y=sin⁡x,−π2⩽x⩽π2y = \sin x,\dfrac{{ - \pi }}{2} \leqslant x \leqslant \dfrac{\pi }{2} and x=sin⁡−1y,−1⩽y⩽1x = {\sin ^{ - 1}}y, - 1 \leqslant y \leqslant 1.