What is the value of expression cos^2(x-y)+cos^2(x+y)-cos2x*cos2y?
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We need to simplify the expression (cos (x-y))^2 + (cos (x+y))^2 - cos 2x *cos 2y
(cos (x-y))^2 + (cos (x+y))^2 - cos 2x *cos 2y
=> (cos x * cos y + sin x * sin y)^2 + (cos x * cos y - sin x * sin y)^2 - cos 2x *cos 2y
=> (cos x * cos y)^2 + (sin x * sin y)^2 + 2* (cos x * cos y)*(sin x * sin y) + (cos x * cos y)^2 + (sin x * sin y)^2 - 2* (cos x * cos y)*(sin x * sin y) - cos 2x *cos 2y
=> 2*(cos x * cos y)^2 + 2*(sin x * sin y)^2 - cos 2x *cos 2y
=> 2*(cos x * cos y)^2 + 2*(sin x * sin y)^2 - [(cos x)^2 - (sin x)^2][(cos y)^2 - (sin y)^2]
=> 2*(cos x * cos y)^2 + 2*(sin x * sin y)^2 - (cos x)^2*(cos y)^2 +(cos x)^2 * (sin y)^2 + (sin x)^2 *(cos y)^2 - (sin x)^2 *(sin y)^2
=> (cos x * cos y)^2 + (sin x * sin y)^2 +(cos x)^2 * (sin y)^2 + (sin x)^2 *(cos y)^2
=> (cos x)^2 *[(cos y)^2 + (sin y)^2] + (sin x)^2*[(sin y)^2 + (cos y)^2]
=> (cos x)^2 + (sin x)^2
=> 1
This gives (cos (x-y))^2 + (cos (x+y))^2 - cos 2x *cos 2y = 1
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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.
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