Results 1 to 2 of 2

Math Help - Inverse circular functions

  1. #1
    Member
    Joined
    Aug 2008
    Posts
    218

    Inverse circular functions

    State (i) the implied domain and (ii) the range of each of the following.

    a) y=sin^{-1}(x^2)
    b)  y=tan^{-1}(2x)
    c) y=tan^{-1}(4-x)
    d)  y=cos^{-1}(x^2-1)

    I don't know how to do all of the above. What are the steps for undertaking the processes?

    Perhaps someone can explain and then attempt one or two and then I could try to do the rest and they someone else could check whether it's right or wrong?

    Anyway, whatever the case, how do you do the above?

    Any help/guidance to do the above will be appreciated!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Quacky's Avatar
    Joined
    Nov 2009
    From
    Windsor, South-East England
    Posts
    901
    Quote Originally Posted by Joker37 View Post
    State (i) the implied domain and (ii) the range of each of the following.

    a) y=sin^{-1}(x^2)
    b)  y=tan^{-1}(2x)
    c) y=tan^{-1}(4-x)
    d)  y=cos^{-1}(x^2-1)

    I don't know how to do all of the above. What are the steps for undertaking the processes?

    Perhaps someone can explain and then attempt one or two and then I could try to do the rest and they someone else could check whether it's right or wrong?

    Anyway, whatever the case, how do you do the above?

    Any help/guidance to do the above will be appreciated!
    The implied domain: the real x-values for which the function is valid.
    The range: The possible y values that the function can take.

    a) y=sin^{-1}(x^2)
    Firstly, we know that in any situation, -1\leq~Sin(\theta)\leq~1
    The arcsin function takes the value obtained for Sin(\theta) and generates the \theta value.
    If Sin(\theta)=x then \theta=Sin^{-1}(x)

    So, as -1\leq~Sin(\theta)\leq~1

    Sin^{-1}(\theta) Is only defined for -1<\theta<1
    -1\leq~x^2\leq 1
    -1\leq~x^2\leq 1
    x^2<1
    -1<x<1

    To find the range then. You know that the minimum value obtained for x^2=0 and the maximum for x^2=1

    Sin^{-1}(1)=90
    Sin^{-1}(0)=0

    So 0\leq~y\leq90

    Not entirely certain on this but that was my attempt!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Inverse circular relations problems
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: April 23rd 2010, 11:34 PM
  2. Replies: 3
    Last Post: February 23rd 2010, 05:54 PM
  3. Inverse Circular Function, help!!!
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: August 23rd 2008, 06:45 PM
  4. circular functions help.
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: July 6th 2008, 07:45 PM
  5. Circular Functions
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: June 5th 2008, 04:33 AM

Search Tags


/mathhelpforum @mathhelpforum