Given that the inverse trigonometric function assumes principal values only. Let x, y be any two real numbers in [-1,1] such that cos ⁻¹ x-sin ⁻¹ y=α, -π/2 ≤ α ≤ π. Then, the minimum value of x²+y²+2 x y sin α is

Solution

Maths sir · 4 steps

Write x = cos θ and y = sin φ, and the whole expression folds up into cos² α.

1. Name the angles. Let θ = cos⁻¹ x ∈ [0, π] and φ = sin⁻¹ y ∈ [-π/2, π/2]. Then x = cos θ, y = sin φ and α = θ - φ.
2. Expand the cross term. Use sin α = sin θ cos φ - cos θ sin φ. Then 2xy sin α = 2 sin θ cos θ sin φ cos φ - 2 cos² θ sin² φ.
3. Regroup. Split -2 cos² θ sin² φ between cos² θ and sin² φ. This gives cos² θ(1-sin² φ) + sin² φ(1-cos² θ), which is cos² θ cos² φ + sin² θ sin² φ, plus the leftover cross term. That is a perfect square.

   x²+y²+2xy sin α = (cos θ cos φ + sin θ sin φ)² = cos²(θ-φ) = cos² α

4. So. For -π/2 ≤ α ≤ π, the smallest value of cos² α is 0, reached at α = π/2. For example, x = 0 and y = 0 give α = π/2. The minimum is 0, option (a).

The idea to keep. When cos⁻¹ x and sin⁻¹ y appear together, put x = cos θ and y = sin φ. The algebra usually turns into a compound-angle identity in θ - φ.

Answer from the NTA final answer key. Solution by PrepWiser.