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Evaluate cos(arccos((32pi)/3)) (I think I know how, just need someone to confirm)

  1. Feb 26, 2012 #1
    Evaluate cos(arccos((32pi)/3))

    From my understanding, I need to subtract (32pi)/3 by pi until the answer falls within the domain of [-1, 1]. Is this the correct way to solve this problem?
     
  2. jcsd
  3. Feb 26, 2012 #2

    D H

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    Are you sure you don't mean [itex]\arccos(\cos(\frac{32\pi}3))[/itex] instead of [itex]\cos(\arccos(\frac{32\pi}3))[/itex] ?
     
  4. Feb 27, 2012 #3
    Yes, I am sure. It IS possible, however, that the problem has no solution, if that's what you're getting at.
     
  5. Feb 27, 2012 #4
    What is the domain of arccos? Also, this shouldn't be in the "Calculus and Beyond" section of the forum...
     
  6. Feb 27, 2012 #5

    SammyS

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    No. That's not the way to solve this.

    See Ansatz7's post above.
     
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