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Homework Help: Simple inverse function.

  1. Nov 7, 2012 #1
    1. The problem statement, all variables and given/known data

    There were other questions before this one but i solved them all.

    Find the inverse function of [tex]f(x)=arctan(\sqrt{1+x^{2}}-x)[/tex] for every x in the interval ]0,pi/2[ .That's the interval that I found when counting f(R) because f is a bijection from R to f(R). Hence f(R)=J=]0,pi/2[. ( just in case someone was wondering where i got it from?

    3. The attempt at a solution

    For every x in J we get :

    x=arctan(\sqrt{1+y^{2}}-y)\Rightarrow tan(x)=tan(arctan(\sqrt{1+y^{2}}-y) & \Rightarrow & tan(x)=\sqrt{1+y^{2}}-y\\
    & \Rightarrow & tan(x)+y=\sqrt{1+y^{2}}\\
    & \Rightarrow & tan^{2}(x)+2ytan(x)+y^{2}=1+y^{2}

    then [tex]tan^{2}(x)+2ytan(x)+y^{2}=1+y^{2}\Rightarrow tan(x)[tan(x)+2y]=1\Rightarrow tan(x)+2y=\frac{1}{tan(x)}\Rightarrow tan(x)+2y=tan(\frac{\pi}{2}-x)[/tex]

    then [tex]tan(x)+2y=tan(\frac{\pi}{2}-x)\Longrightarrow2y=tan(\frac{\pi}{2}-2x)\Rightarrow y=\frac{1}{2}[tan(\frac{\pi}{2}-2x)][/tex]

    Is that correct, because the given answer here is: [tex]f^{-1}(x)=tan(\frac{\pi}{2}-2x)[/tex]
    Last edited by a moderator: Nov 7, 2012
  2. jcsd
  3. Nov 7, 2012 #2
    I'm not really seeing how you did this step.
  4. Nov 7, 2012 #3
    I did 2y= tan(pi/2-x)-tan(x) and i did 2y=tan(pi/2-x-x) and then i got 2y=tan(pi/2-2x)
  5. Nov 7, 2012 #4
    And I don't see why

    [tex]\tan(\pi/2 - x)-\tan(x)=\tan(\pi/2 - x-x)[/tex]
  6. Nov 7, 2012 #5
    Why not??
  7. Nov 7, 2012 #6
    If you think the equality holds, then it's up to you to prove it. I just don't see where it comes form. Do you have a justification for the equality?
    If you do not find a justification, then you might think that it is not true. Try some easy values for x and see if the equality holds. If you found one value for which it does not hold, then you have disproven the equality.
  8. Nov 7, 2012 #7
    I'm sorry I was lost when thinking about that. It should be [tex]y=\frac{1}{2}[tan(\frac{\pi}{2}-x)-tan(x)][/tex] but then when you want to simplify what can you do??
  9. Nov 7, 2012 #8
    You need to prove somehow that

    [tex]2\tan(\pi/2 -2x)=\tan(\pi/2 -x) - \tan(x)[/tex]

    So you see formula's that allow you to do that??
  10. Nov 7, 2012 #9


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    The y2's cancel, leaving


    Solve that for y.
  11. Nov 8, 2012 #10
  12. Nov 8, 2012 #11


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    Right, and that's also the reciprocal of the identity for tan(2x) .
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