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

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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