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Thread: Inverse circular functions

  1. #1
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    Inverse circular functions

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

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

    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!
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  2. #2
    Super Member Quacky's Avatar
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    Quote Originally Posted by Joker37 View Post
    State (i) the implied domain and (ii) the range of each of the following.

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

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

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

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

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

    $\displaystyle Sin^{-1}(1)=90$
    $\displaystyle Sin^{-1}(0)=0$

    So $\displaystyle 0\leq~y\leq90$

    Not entirely certain on this but that was my attempt!
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