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Proof of Arccos

  1. Oct 16, 2014 #1
    1. The problem statement, all variables and given/known data

    Show cos(cos-1(x) lies in range [-1,1]
    2. Relevant equations

    I know the derivative of cos(x) is arcos(x).


    3. The attempt at a solution

    D/dx(cos(x)) = arcos(x)

    However i am stuck how to prove cos(arcos(x)) has a domain of [-1,1]..
     
  2. jcsd
  3. Oct 16, 2014 #2
    I think it is taken in context of a trig function in which cosine is described by a circle with a radius equal to one. Then since the cos and it's inverse cancel each other out leave the answer is as follows

    cos(cos-1(x)) = x

    This would simply mean that the line would run along the x axis, in terms of trig terminology this would be [-1,1].
     

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  4. Oct 16, 2014 #3
    You might want to check again.
     
  5. Oct 16, 2014 #4
    Sorry I didn't read the rest, arcos(x) is the inverse of cos(x) not the derivative. Derivative is as follows

    d/dx cos(x) = -sin(x)
    d/dx sin(x) = cos(x)

    This is because sin(x) and cos(x) are essentially the same function except the phase between the two is pi/2 (90 degrees or 1/4 revolution of a full circle), hence one can be derived by the other.
     
    Last edited: Oct 16, 2014
  6. Oct 16, 2014 #5

    SteamKing

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    You should forget this bit of mathematical misinformation immediately.

    The derivative of cos(x) = -sin(x)

    Here is a list of other derivatives:

    http://en.wikipedia.org/wiki/Derivative

    You should check what you 'know' about derivatives of other functions against this list.

    The 'arccos (x)' is the inverse of cosine (x), not the derivative, which has a specific meaning in mathematics.

    arccos(cosine(x)) = x
     
  7. Oct 16, 2014 #6

    RUber

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    For what values of x is ##\cos^{-1} x ## defined ( domain ).
    X can only be in that interval, other than that--
    ## f(f^{-1}(x))=x## all the time by definition.
     
  8. Oct 16, 2014 #7

    RUber

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    Hint:arccos (x)=y means cos(y) = x.
     
  9. Oct 16, 2014 #8

    vela

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    This should really be the other way around: cos(arccos(x)) = x. It's possible for arccos(cos(x)) to differ from x, when x isn't in the range of the arccosine function.
     
  10. Oct 17, 2014 #9

    Ray Vickson

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    Have you looked at a graph of ##y = \cos(\theta)## over a broad range of ##\theta##? I suggest you do that first, before trying to use fancy identities or derivatives and the like. Then, just remember what ##\arccos(x)## stands for.
     
  11. Oct 19, 2014 #10
    In the problem you're analyzing the image of cos(arccos(x)).
    But then you say you're trying to figure out the domain of cos(arccos(x)). Which is it?
     
  12. Oct 20, 2014 #11

    RUber

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    Before you can know the range (image) of a function, you must know which inputs would be valid.
    For example, you could look at cos(arccos(x)) and say, hey, that's a function and its inverse, so the output is just x.
    Then, you would say, hey, that doesn't make sense, since I know cosine has a range between -1 and 1...
    So, you need to know the possible input values for x (domain of arccos(x)) in order to properly resolve this apparent contradiction.
     
  13. Oct 20, 2014 #12

    Mark44

    Staff: Mentor

    Let's hold off on any more replies until the OP comes back.
     
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