Given that x+y=z, what is numerical value of expression cos^2x+cos^2y+cos^2z- 2cosx*cosy*cosz?

We'll re-write the term (cos z)^2 = [cos
(x+y)]^2

cos (x+y) = cos x*cos y - sin x*sin
y

[cos (x+y)]^2 = (cos x*cos y - sin x*sin
y)^2

We'll expand the
square:

[cos (x+y)]^2 = (cos x*cos y)^2 + (sin x*sin y)^2 -
2sin x*sin y*cos x*cos y

We'll re-write the last term of
the givn expression:

2cosx*cosy*cosz =
2cosx*cosy*cos(x+y)

2cosx*cosy*cosz = 2cosx*cosy*(cos x*cos
y - sin x*sin y)

We'll remove the
brakets:

2cosx*cosy*cosz = 2(cos x*cos y)^2 - 2sin x*sin
y*cos x*cos y

We'll re-write the
expression:

E(x) = (cos x)^2 + (cos y)^2 + (cos x*cos y)^2
+ (sin x*sin y)^2 - 2sin x*sin y*cos x*cos y - (cos x*cos y)^2 - (cos x*cos y)^2 + 2sin
x*sin y*cos x*cos y

We'll eliminate like
terms:

E(x) = (cos x)^2 + (cos y)^2 + (sin x*sin y)^2 -
(cos x*cos y)^2

We'll use the Pythagorean
identity:

(sin x)^2 = 1 - (cos
x)^2

(sin y)^2 = 1 - (cos
y)^2

(sin x*sin y)^2 = [1 - (cos x)^2][1 - (cos
y)^2]

We'll remove the
brackets:

(sin x*sin y)^2 = 1 - (cos x)^2 - (cos y)^2 +
(cos x*cos y)^2

We'll re-write the
expression:

E(x) = (cos x)^2 + (cos y)^2 +1 - (cos x)^2 -
(cos y)^2 + (cos x*cos y)^2 - (cos x*cos y)^2

We'll
eliminate like terms:

E(x) =
1

The numerical  value of the given
expression is E(x) = 1.