What is the value of expression cos^2(x-y)+cos^2(x+y)-cos2x*cos2y?

We'll use the formula of the cosine of sum or difference of
angles and the identity of double angles.

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

We'll raise to square both
sides:

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

We'll expand the
binomial:

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

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

We'll raise to square both
sides:

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

We'll expand the
binomial:

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

We'll add (1) and (2) and we'll
get:

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

We'll write the double angle
identity:

cos 2x = (cos x)^2 - (sin x)^2
(3)

cos2y = (cos y)^2 - (sin y)^2
(4)

We'll multiply (3) and (4) and we'll
get:

cos2x*cos2y = (cos x*cos y)^2 - (cos x)^2*(sin y)^2 - (sin
x)^2*(cos y)^2 + (sin x*sin y)^2 (5)

The expresison will
become:

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

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

We'll factorize and we'll get:

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

But, the Pythagorean identity states
that:

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

E =
(cos y)^2 + (sin y)^2

E =
1

The value of the expression is E =
1.