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Arccos(x) + arccos(y)

  1. Dec 25, 2005 #1
    arccos(x) + arccos(y) = ?

    How are these added? I can't find it anywhere, and I'm sure there has to be a way...


    Actually, what would be more helpful would be
    arccos(x) + arccos(y) + arccos(z)

    Or even
    cos(x) + cos(y) + cox(z)


    Well... Thanks for your help.
     
    Last edited: Dec 25, 2005
  2. jcsd
  3. Dec 25, 2005 #2

    Tide

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    Please elaborate. Are x and y arbitrary? Do they represent a coordinate pair on the unit circle?
     
  4. Dec 25, 2005 #3
    x and y (and z, if you care to answer the other parts) are arbitrary variables; they have nothing to do with any coordinates.
     
  5. Dec 25, 2005 #4

    Hurkyl

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    Take the cosine (or sine) of both sides of the equation.
     
  6. Dec 25, 2005 #5
    OK if my math is correct then
    arccos(x) + arccos(y) = arccos( xy - [tex]\sqrt{(1-x^2)(1-y^2)}[/tex])

    Yea that should be right.
     
    Last edited: Dec 25, 2005
  7. Dec 25, 2005 #6
    I think it's correct.

    and

    [tex]\arccos (x) + \arccos (y) + \arccos (z) =
    \arccos(xyz - z \sqrt{(1-x^2)(1-y^2)} - x \sqrt{(1-y^2)(1-z^2)} - y \sqrt{(1-z^2)(1-x^2)})[/tex]

    Let [tex]X=\arccos(x), Y=\arccos(y), Z=\arccos(z)[/tex] then

    [tex]\cos(X+Y+Z) = \cos(X+Y) \cos(Z) - \sin(X+Y) \sin(Z) [/tex]
    [tex]
    = (\cos(X) \cos(Y) - \sin(X) \sin(Y)) \cos(Z) - (\sin(X) \cos(Y) + \cos(X) \sin(Y)) \sin(X)
    = ..... [/tex]
    [tex] = xyz - z \sqrt{(1-x^2)(1-y^2)} -y \sqrt{(1-z^2)(1-x^2)} - x \sqrt{(1-y^2)(1-z^2)}[/tex]
     
    Last edited: Dec 26, 2005
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