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

1 Answer | Add Yours

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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.

We’ve answered 318,979 questions. We can answer yours, too.

Ask a question