himanshu121
Apr10-04, 07:41 AM
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 \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