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Math Help - Proving inverse functions identity

  1. #1
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    Proving inverse functions identity

    Hello everybody and thanks for reading.
    I have been trying to solve the following problem without success, so I am looking for some advices..

    I have the following equation

    arccos(x) = arctan( root(1 - x^2) / x)

    So I have to prove if that 'identity' is valid and say for what values of x is valid.

    Even when I am able to construct the triangle from the data, I don't know how to find the valid values for x.

    So far I have constructed a triangle with opposite root(1-x^2), adjacent x and hypotenuse 1.

    I also restricted the domain so 1-x^2 >= 0, x != 0 and -1 <= x <= 1 (for the root, denominator, and arccos respectively).

    I also tried to prove the identity using the Pythagorean theorem from the data of the arctan ( opposite and adjacent), and I get that the hypotenuse is 1, which is the same the arccos says.

    But I have seen the plot of both functions and I realized they are only equal for 0 < x <= 1 and I don't know how I could get there. Intersecting the restrictions I said I only get [-1,1] - {0}.

    I have even thought I should intersect the ranges of those functions, but I don't think that is valid.

    Thanks in advance and sorry for the long post
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  2. #2
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    Re: Proving inverse functions identity

    Quote Originally Posted by Danielc View Post
    I have the following equation
    arccos(x) = arctan( root(1 - x^2) / x)
    So I have to prove if that 'identity' is valid and say for what values of x is valid.
    Even when I am able to construct the triangle from the data, I don't know how to find the valid values for x.
    So far I have constructed a triangle with opposite root(1-x^2), adjacent x and hypotenuse 1.
    Let \theta=\arccos(x), so \cos(\theta)=\frac{x}{1}.

    Thus \tan(\theta)=\frac{\sqrt{1-x^2}}{x}.

    What don't you follow?
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  3. #3
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    Re: Proving inverse functions identity

    Yes, that is what I did.. But if you try the identity for, say, -1, it is not true... So what I need to do is to find (hopefully algebraically) the values that make that identity true.

    I constructed the triangle doing the same as you did, but I don't know how should i proceed.

    Thank you very much
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    Lightbulb Re: Proving inverse functions identity

    Revised
    Last edited by MaxJasper; September 13th 2012 at 02:05 PM.
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    Re: Proving inverse functions identity

    (deleted due to previous post deletion)
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  6. #6
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    Question Re: Proving inverse functions identity

    How can posts be deleted? There is no delete button at the top of my edit window
    Last edited by MaxJasper; September 13th 2012 at 09:05 AM.
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  7. #7
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    Re: Proving inverse functions identity

    Quote Originally Posted by MaxJasper View Post
    How can posts be deleted?
    EDIT your post.
    At the very top of the edit box there is a delete option.
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  8. #8
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    Re: Proving inverse functions identity

    Hi and thanks for your answer.

    Yes, I can see where it is valid graphically, but I would like to know how to get to that conclusion algebraically

    Thank you very much
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  9. #9
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    Lightbulb Re: Proving inverse functions identity

    If you are dealing with real numbers only then :

    1-x^2\geq 0

    Hence we investigate:

    \cos ^{-1}(-1\leq x\leq 1)=?=\tan ^{-1}\left(\frac{\sqrt{1-(-1\leq x\leq 1)^2}}{-1\leq x\leq 1}\right)

    Attached Thumbnails Attached Thumbnails Proving inverse functions identity-arccos-x-arctan.png  
    Last edited by MaxJasper; September 13th 2012 at 03:28 PM.
    Thanks from Danielc
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  10. #10
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    Re: Proving inverse functions identity

    Let \varphi(x) denote \arccos x. We have the following facts.

    If -1\le x\le 1, then \cos(\arccos x)=\cos\varphi(x)=x (1)
    If -1\le x\le 0, then \pi/2\le \varphi(x)\le\pi (2)
    From (2), if -1\le x\le0, then \sin\varphi(x)\ge0 (3)
    The function arctan is the inverse of tan on [-\pi/2,\pi/2] (4)
    \tan(\alpha)=\tan(\alpha-\pi) (5)
    From (3) and (4), if \pi/2\le\alpha\le\pi, then -\pi/2\le\alpha-\pi\le0<\pi/2, so \arctan(\tan\alpha)=\arctan(\tan(\alpha-\pi))=\alpha-\pi (6)

    So, if -1\le x\le 0, then
    \frac{\sqrt{1-x^2}}{x} = \frac{\sqrt{1-\cos^2\varphi(x)}}{\cos\varphi(x)} by (1)
    = \frac{\sqrt{\sin^2\varphi(x)}}{\cos\varphi(x)} = \frac{|\sin\varphi(x)|}{\cos\varphi(x)} = \frac{\sin\varphi(x)}{\cos\varphi(x)}=\tan\varphi(  x) by (3).

    Therefore,

    \arctan\frac{\sqrt{1-x^2}}{x}=\arctan(\tan\varphi(x))=\varphi(x)-\pi=\arccos x-\pi by (6). For 0\le x\le 1 we simialrly have 0\le\varphi(x)\le\pi/2 and \arctan(\tan\varphi(x))=\varphi(x), so \arctan\frac{\sqrt{1-x^2}}{x}=\arccos x.
    Thanks from Danielc
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    Re: Proving inverse functions identity

    Thank you very much for your answers.

    I wanted something like what emakarov posted, so it helped a lot. I tried to put it in simple terms so concluded:

    - tan (theta) = opposite / adjacent, but as we are talking about lengths, negative values for any of both have no sense. In this case, x was in the denominator, si if it is negative it has no sense.
    - So tan(theta) = sin(theta) / cos(theta) . As the numerator is always non-negative, theta must be an angle between [0, PI] so its sin is positive. Then, for theta to be positive, x >= 0, but as it can't be 0, it must be between (0, 1], considering the restrictions on the domain.

    Thank you very much
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