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Show that f is constant

  1. Feb 10, 2009 #1
    1. The problem statement, all variables and given/known data
    Show that f(x) = arcsin(x) + arccos(x) = constant, for [tex] -1 \leq x \leq 1[/tex]


    2. Relevant equations
    Some theorem involving continuity or inverse functions


    3. The attempt at a solution
    The theorem is easy for the interval with strict bounds (differentiability on open interval and we get f'(x) = 0, so f constant on that interval), but how do I show this for the end points? The hint is to "show and use continuity" but I'm pretty sure I need to use a theorem about inverse functions. I'm just not sure how to deal with the end points -1 and 1. Any help is appreciated.
     
  2. jcsd
  3. Feb 10, 2009 #2

    Office_Shredder

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    Gold Member

    arcsin(x) and arccos(x) are continuous on the closed interval [-1,1] so arcsin(x) + arccos(x) is too. Then you can take a sequence of points converging to, say, 1 (you do -1 as a separate case). You use the fact that arcsin(x) + arccos(x) is continuous plus the fact that you know the value of arcsin(x) + arccos(x) for each point in the sequence
     
  4. Feb 11, 2009 #3

    Mark44

    Staff: Mentor

    Here is a simpler approach that uses only basic trig and the ideas of inverse functions.
    Let w = arcsin(y)

    Draw a right triangle with angle w at left, and right angle at right, with altitude y and hypotenuse 1. The remaining angle, at the top, is (pi/2 - y).
    Then sin(w) = y and cos(pi/2 -w) = y.
    So w = arcsin(y) and pi/2 - w = arccos(y), hence arcsin(y) + arccos(y) = pi/2, a constant.

    The range that is common to both arcsin and arcsin is [0, pi/2], so the representation of w as an angle in the first quadrant is, I believe, all we need.
     
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