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Trigonometry Problem

  1. Sep 3, 2004 #1
    Hello all

    In my textbook I encountered the following problem:

    Find the six trigonometric values of cos^ -1 (3/7). They must be exact. I gather what they mean is that I find arccos (3/7). I tried applying basic identities, but didn't work. Any help would be appreciated.

    Thanks
     
  2. jcsd
  3. Sep 3, 2004 #2
    If cos^-1 (3/7) = x, do they allow you to give x a value of more than 360 degrees?
     
  4. Sep 3, 2004 #3

    HallsofIvy

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    3/7?? I'm going to have to think about that!
     
  5. Sep 3, 2004 #4
    in the answer book it says cos (theta) = 3/7
     
  6. Sep 3, 2004 #5

    BobG

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    [tex]cos^{-1}(\frac{3}{7})[/tex] is saying "the angle whose cosine is 3/7", so the cosine is already given. You don't need to actually figure out the angle, since the sine, cosine, tangent, etc. all have set relationships between each other.

    It would help to have a range for the angle, though, since the sign of the sine, tangent, and cosecant are all going to depend on whether

    [tex]cos^{-1}(\frac{3}{7})[/tex] lies between [tex]0[/tex] and [tex]\frac{\pi}{2}[/tex] or between [tex]0[/tex] and [tex]-\frac{\pi}{2}[/tex]
     
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