1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.


    [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]
    = (\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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook