Given that the inverse trigonometric function assumes principal values only. Let $$x, y$$ be any two real numbers in $$[-1, 1]$$ such that $$\cos^{-1} x - \sin^{-1} y = \alpha$$, $$\frac{-\pi}{2} \leq \alpha \leq \pi$$. Then, the minimum value of $$x^2 + y^2 + 2xy \sin \alpha$$ is

Let $$\cos^{-1} x = A \implies x = \cos A$$ and $$\sin^{-1} y = B \implies y = \sin B$$.

The given equation is $$A - B = \alpha \implies A = B + \alpha$$.

Substitute $$x$$ and $$y$$ into the expression:

$$E = \cos^2 A + \sin^2 B + 2 \cos A \sin B \sin \alpha$$

Replace $$A$$ with $$B + \alpha$$:

$$E = \cos^2(B + \alpha) + \sin^2 B + 2 \cos(B + \alpha) \sin B \sin \alpha$$

 Use the identity $$\cos(B + \alpha) = \cos B \cos \alpha - \sin B \sin \alpha$$:

$$E = (\cos B \cos \alpha - \sin B \sin \alpha)^2 + \sin^2 B + 2(\cos B \cos \alpha - \sin B \sin \alpha) \sin B \sin \alpha$$

Expanding and simplifying (specifically focusing on the $$\sin B \sin \alpha$$ terms cancelling out), the expression reduces to:

$$E = \cos^2 B \cos^2 \alpha + \sin^2 B \sin^2 \alpha - 2 \sin B \cos B \sin \alpha \cos \alpha + \sin^2 B + 2 \sin B \cos B \sin \alpha \cos \alpha - 2 \sin^2 B \sin^2 \alpha$$

$$E = \cos^2 B \cos^2 \alpha - \sin^2 B \sin^2 \alpha + \sin^2 B = \cos^2 B \cos^2 \alpha + \sin^2 B (1 - \sin^2 \alpha)$$

$$E = \cos^2 B \cos^2 \alpha + \sin^2 B \cos^2 \alpha = \cos^2 \alpha (\cos^2 B + \sin^2 B) = \cos^2 \alpha$$

The minimum value of $$\cos^2 \alpha$$ is 0 (which occurs when $$\alpha = \pi/2$$).