If cos⁻¹x - cos⁻¹(y/2) = α, where -1 ≤ x ≤ 1, -2 ≤ y ≤ 2, x ≤ y/2, then for all 4x² - 4xycosα + y² is equal to

cos(cos⁻¹x - cos⁻¹(y/2)) = cosα
cosα ⇒ x(y/2) + √(1 - x²)√(1 - y²/4) = cosα
⇒ (cosα - xy/2)² = (1 - x²)(1 - y²/4)
x² + y²/4 - xycosα = 1 - cos²α = sin²α
4x² + y² - 4xycosα = 4sin²α