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Inverse Trig Function Problem

  1. Jan 27, 2008 #1
    1. The problem statement, all variables and given/known data
    The problem here involves inverse trig functions. The question here is why doesn't arccos(cos x) yield the equation y = x?


    2. Relevant equations
    y = arccos(cos x)


    3. The attempt at a solution
    I assume the reason is due to restrictions on domain and/or range of the trig functions. I graphed it out from -4pi to 4pi, and noticed that from 0 to pi, 2pi to 3pi, etc., it does display a function similar y = x (0 to pi it IS y = x). However, on intervals such as -pi to 0, it is the same as the function y = -x. Now, I looked at the unit circle and can see that, if it is the arccos (cos x), the cos of radian measurements (essentially the sine when reading on the unit circle) is strictly positive. However, between -pi and 0, the values would be negative. They increase and decrease graphically, respectively. Anyone have any input or guiding suggestions here? Thanks.
     
  2. jcsd
  3. Jan 27, 2008 #2

    jambaugh

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    Science Advisor
    Gold Member

    Start by looking at the graph of y=cos(x) and asking if it is an invertible function.
    As such it is not! (Horizontal line test!)
    What then does arccos mean?
     
  4. Jan 27, 2008 #3
    That means arccos x can only exist under certain restrictions, since an equation does not have an inverse unless its graph passes the horizontal line test. The domain of arccos x is [-1, 1] and the range is [0, pi]. Ok, so if the range can only be 0 to pi, then the graph of arccos (cos x) cannot have a range outside of that. And since the cos x only ranges between -1 and 1, then arccos (cos x) can be defined for any value of x, since the cos x is between -1 and 1, which is the domain of arccos x?
     
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