# Inverse Circular functions

1. Apr 10, 2004

### himanshu121

I dont 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

2. Apr 10, 2004

### Hurkyl

Staff Emeritus
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.