1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Inverse Circular functions
  1. Inverse functions (Replies: 3)

  2. Inverse functions (Replies: 3)

  3. Inverse function (Replies: 12)

  4. Inverse of a function (Replies: 6)

  5. Inverse function (Replies: 5)

Loading...