Verified answer

Do you know this identity?

cos(a + b) = cos(a).cos(b) - sin(a).sin(b) → suppose that: a = b = θ

cos(θ + θ) = cos(θ).cos(θ) - sin(θ).sin(θ)

cos(2θ) = cos²(θ) - sin²(θ)

cos(2θ) = [1 - sin²(θ)] - sin²(θ)

cos(2θ) = 1 - sin²(θ) - sin²(θ)

cos(2θ) = 1 - 2.sin²(θ) → given that: sin(θ) = (1/2).[a + (1/a)]

cos(2θ) = 1 - 2.{ (1/2).[a + (1/a)] }²

cos(2θ) = 1 - 2.{ (1/4).[a + (1/a)]² }

cos(2θ) = 1 - (2/4).[a + (1/a)]²

cos(2θ) = 1 - (1/2).[a² + 2 + (1/a²)]

cos(2θ) = 1 - (1/2).a² - 1 - (1/2).(1/a²)

cos(2θ) = - (1/2).a² - (1/2).(1/a²)

cos(2θ) = - (1/2).[a² + (1/a²)]

cos(2θ) + (1/2).[a² + (1/a²)] = 0

cos(2θ) = 1 - 2sin^2(θ)

= 1 - 2[1/2(a + 1/a)^2]

= 1 - 2[ 1/4(a^2 + 2 + 1/a^2)]

= 1 - 1/2(a^2 + 2 + 1/a^2)

= 1 - 1/2(2) - 1/2(a^2 + 1/a^2)

= 1 - 1 - 1/2(a^2 + 1/a^2)

= - 1/2(a^2 + 1/a^2)

cos(2θ) + 1/2(a^2 + 1/a^2) = - 1/2(a^2 + 1/a^2) + 1/2(a^2 + 1/a^2)

= 0

In the absence of proper parenthesis, it is confusing.

Wish your question is:

"If sin(Ө) = (1/2)*(a + 1/a), then prove that

cos(2Ө) + (1/2)*(a² + 1/a²) = 0"

With this assumption, the solution is provided as below:

i) Multiplying both sides by 2, the given one is:

a + 1/a = 2*sin(Ө)

ii) Squaring this both sides,

a² + 1/a² + 2 = 4*sin²Ө

Rearranging, a² + 1/a² + 2 - 4*sin²Ө = 0

==> (a² + 1/a²) + 2(1 - 2*sin²Ө) = 0

==> (a² + 1/a²) + 2*cos(2Ө) = 0

Dividing by 2,

cos(2Ө) + (1/2)*(a² + 1/a²) = 0 [Proved]

Note: If you are satisfied kindly acknowledge.