It is easy to guess how this works out. That is the angles x,y,z must be all be at mutual angles of 120 degrees. For example x = 0, y = 120, z = 240. Then cos(y-z) + cos(z-x) + cos(x-y) does equal -3/2 and
cosx + cosy + cos(z) =0 = sinx + siny + sinz. If the same arbitrary angle is added to x and to y and to z, the formulas work out the same.
That guess / observation is not a proof. The only proof that I can think of involves some calculus.
The starting point is to note that the expression cos(y-z) + cos(z-x) + cos(x-y) is unchanged in value if the coordinate system is rotated by a fixed angle. This is the same as adding the same arbitrary angle to x and y and z.
This means that we can take x = 0 without loss of generality.
The next step is to consider the function of the remaining two angles (y and z)
f(y,z) = cos(y-z) + cos(z) + cos(-y) .
We can show that the global minimum of this function (with respect to y and z) happens when y = 120 and z = 240 (or alternatively when y= 240 and and z = 120 and for no other angle pairs in in [0 ,360] ); and further that f(120,240) = -3/2. This is equivalent to saying that only y =120 and z = 240 (or equivalently vice versa) will satisfy f(y,z) = -3/2. Since with these values and x =0 we have cosx + cosy + cos(z) =0 = sinx + siny + sinz ,
this completes the proof.
To show the global minimum we require that the partial derivative of f(y,z) with respect to y is zero AND the partial derivative of f(y,z) with respect to z is zero. Using the rules of differential calculus this leads to the equations
-sin(y-z) + sin(-y) = 0 and
sin(y-z) - sin(z) = 0
Adding the two equations leads to sin(-y) = sin(z) or sin(y) = -sin(z)
This means that z = 360 -y. Substituting this into the first equation gives sin(2y) = -sin(y) . The solution to this is y = 120 or y = 240 as claimed above.
