How Do You Add Arccos(x), Arccos(y), and Arccos(z)?

[Trepidation]

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.

 

[Tide]

Please elaborate. Are x and y arbitrary? Do they represent a coordinate pair on the unit circle?

 

[Trepidation]

x and y (and z, if you care to answer the other parts) are arbitrary variables; they have nothing to do with any coordinates.

 

[Hurkyl]

Take the cosine (or sine) of both sides of the equation.

 

[d_leet]

OK if my math is correct then
arccos(x) + arccos(y) = arccos( xy - \sqrt{(1-x^2)(1-y^2)})

Yea that should be right.

 

[maverick6664]

I think it's correct.

and

\arccos (x) + \arccos (y) + \arccos (z) =<br /> \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)})

Let X=\arccos(x), Y=\arccos(y), Z=\arccos(z) then

\cos(X+Y+Z) = \cos(X+Y) \cos(Z) - \sin(X+Y) \sin(Z)
<br /> = (\cos(X) \cos(Y) - \sin(X) \sin(Y)) \cos(Z) - (\sin(X) \cos(Y) + \cos(X) \sin(Y)) \sin(X) <br /> = ...
= 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)}