Arccos(x) + arccos(y)

1. Dec 25, 2005

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)

Last edited: Dec 25, 2005
2. Dec 25, 2005

Tide

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

3. Dec 25, 2005

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.

4. Dec 25, 2005

Hurkyl

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

5. Dec 25, 2005

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.

Last edited: Dec 25, 2005
6. Dec 25, 2005

maverick6664

I think it's correct.

and

$$\arccos (x) + \arccos (y) + \arccos (z) = \arccos(xyz - z \sqrt{(1-x^2)(1-y^2)} - x \sqrt{(1-y^2)(1-z^2)} - y \sqrt{(1-z^2)(1-x^2)})$$

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

$$\cos(X+Y+Z) = \cos(X+Y) \cos(Z) - \sin(X+Y) \sin(Z)$$
$$= (\cos(X) \cos(Y) - \sin(X) \sin(Y)) \cos(Z) - (\sin(X) \cos(Y) + \cos(X) \sin(Y)) \sin(X) = .....$$
$$= xyz - z \sqrt{(1-x^2)(1-y^2)} -y \sqrt{(1-z^2)(1-x^2)} - x \sqrt{(1-y^2)(1-z^2)}$$

Last edited: Dec 26, 2005