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Inverse trig functions

  1. Jan 28, 2010 #1
    Evaluate the following expressions. Your answer must be in radians.

    a) arctan(-(sqrt3)/3)

    b) arctan((sqrt3)/3)

    c) arctan(-sqrt3)

    What I got are:

    a) 5pi/6

    b) pi/6

    c) 2pi/3

    Do I got these answers right?
  2. jcsd
  3. Jan 28, 2010 #2


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    Hi shiri! :smile:

    (have a pi: π and a square-root: √ :wink:)

    Yes and no.

    The tangents of those angles are the numbers given.

    But "arctan" mean the principal value, which is usually taken to be between -π/2 and π/2 …

    see http://en.wikipedia.org/wiki/Arctan" [Broken].

    (The reason why that range is chosen (and not the [0,2π) for arccos and arcsin) is because tan goes smoothly from -∞ to ∞ in that range.)

    You've chosen the range from 0 to 2π (of course, if your professor told you to do so, ignore wikipedia and me :wink:).
    Last edited by a moderator: May 4, 2017
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