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Inverse Circular functions

  1. Apr 10, 2004 #1
    I dont know how will i get these Limits for inverse Trigonometric functions for eg

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

    I want to know how we arrive at these values Or intervals
  2. jcsd
  3. Apr 10, 2004 #2


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    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.
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