if cos x+cosy+cosz=0 and sinx+siny+sinz=0 then find the value of cos2x +cos2y+cos2z

cos(x) + cos(y) = -cos(z) [Since given cos(x) + cos(y) + cos(z) = 0]
Squaring both sides, cos²x + cos²y + 2cos(x)*cos(y) = cos²z ----------- (1)

Similarly from the given second data, we have
sin²x + sin²y + 2sin(x)*sin(y) = sin²z ------------ (2)

Adding (1) & (2):
1 + 1 + 2{cos(x)*cos(y) + sin(x)*sin(y)} = 1 [Since by identity cos²x + sin²x = 1]

==> 2*cos(x - y) = -1

==> cos(x - y) = -1/2 --------- (3)

I am in confusion what you need? Whether {cos(x - y)}/2 (or) cos{(x - y)/2}

If it is: {cos(x - y)}/2, then it is = -1/4

If it is: cos{(x - y)/2}, then it is computed as below:

Let (x - y)/2 = u; so cos{(x - y)/2} = cos(u) = (1/2)√{1 + cos(2u)},
[Since by Multiple angle identity 2*cos²u = 1 + cos(2u)]

So, 2cos²{(x - y)/2} = [1+ cos{(x - y)}] = [1 + (-1/2)] = 1/2

==> cos²{(x - y)/2} = 1/4

So cos{(x - y)/2} = ±(1/2)