How Are Limits for Inverse Trigonometric Functions Derived?

[himanshu121]

I don't know how will i get these Limits for inverse Trigonometric functions for eg

$$2 \sin^{-1}x = - \pi - sin^{-1} [2x \sqrt{1-x^2}] for x \leq -\frac{1}{\sqrt{2}}$$
=$$sin^{-1} [2x \sqrt{1-x^2}] -\frac{1}{\sqrt{2}} \leq x \leq \frac{1}{\sqrt{2}}$$
=$$\pi - sin^{-1} [2x \sqrt{1-x^2}] x \geq \frac{1}{\sqrt{2}}$$


I want to know how we arrive at these values Or intervals

 

[Hurkyl]

The easiset way, I think, to do these kinds of problems is to (reversibly) convert it from an inverse trig function identity to a trig function identity.

 

[tacman]

for x

To understand how these limits for inverse trigonometric functions are derived, we first need to understand the concept of inverse circular functions. Inverse circular functions are the inverse operations of circular functions (such as sine, cosine, and tangent). They are used to find the angle measure given the ratio of sides in a right triangle.

To find the limits for inverse trigonometric functions, we need to consider the domain and range of the original function. For example, in the case of 2 \sin^{-1}x, the domain of the function is -1 \leq x \leq 1. This means that the possible values of x for which the function is defined are between -1 and 1.

Now, let's look at the different intervals for x in the given problem. For x \leq -\frac{1}{\sqrt{2}}, the value of 2x \sqrt{1-x^2} is negative. This means that when we take the inverse sine of this value, we will get a negative angle. However, the domain of inverse sine is between -\frac{\pi}{2} and \frac{\pi}{2}. So, in order to get a valid angle measure, we need to add a negative sign to the inverse sine value, giving us - \pi - sin^{-1} [2x \sqrt{1-x^2}].

Similarly, for x \geq \frac{1}{\sqrt{2}}, the value of 2x \sqrt{1-x^2} is positive, and we can directly take the inverse sine without adding a negative sign. This gives us \pi - sin^{-1} [2x \sqrt{1-x^2}].

For the middle interval, -\frac{1}{\sqrt{2}} \leq x \leq \frac{1}{\sqrt{2}}, the value of 2x \sqrt{1-x^2} is between -1 and 1, which is the domain of inverse sine. Therefore, we can directly take the inverse sine without any additional modifications.

In summary, the limits for inverse trigonometric functions are derived by considering the domain and range of the original function and making adjustments to the angle measure to ensure that it falls within the valid domain of inverse circular functions.