1. Rewrite the condition using principal values

Let $A=\cos^{-1}x,\qquad B=\sin^{-1}y.$ Then, by principal value ranges, $A\in[0,\pi],\qquad B\in\left[-\frac\pi2,\frac\pi2\right],$ and the given relation becomes $A-B=\alpha.$ Also, $x=\cos A,\qquad y=\sin B.$

Since $A=\alpha+B$, we can express $x$ in terms of $B$: $x=\cos(\alpha+B)=\cos\alpha\cos B-\sin\alpha\sin B.$ Because $y=\sin B$, let $\cos B=t.$ Since $B\in\left[-\frac\pi2,\frac\pi2\right]$, we have $t=\cos B\ge 0,$ and $y=\sin B,\qquad y^2+ t^2=1 \Rightarrow t^2=1-y^2.$ Thus $x=t\cos\alpha-y\sin\alpha.$

2. Evaluate the required expression

We need the minimum of $E=x^2+y^2+2xy\sin\alpha.$ Substitute $x=t\cos\alpha-y\sin\alpha.$ Then $E=(t\cos\alpha-y\sin\alpha)^2+y^2+2y(t\cos\alpha-y\sin\alpha)\sin\alpha.$ Expand: $E=t^2\cos^2\alpha-2ty\cos\alpha\sin\alpha+y^2\sin^2\alpha+y^2+2ty\cos\alpha\sin\alpha-2y^2\sin^2\alpha.$ The mixed terms cancel, so $E=t^2\cos^2\alpha+y^2-y^2\sin^2\alpha.$ That is, $E=t^2\cos^2\alpha+y^2(1-\sin^2\alpha)=t^2\cos^2\alpha+y^2\cos^2\alpha.$ Hence $E=\cos^2\alpha\,(t^2+y^2).$ But $t^2+y^2=1.$ Therefore, $E=\cos^2\alpha.$

3. Find the minimum possible value

Since $E=\cos^2\alpha,$ and $\alpha\in\left[-\frac\pi2,\pi\right]$, the minimum of $\cos^2\alpha$ is clearly $0,$ attained when $\cos\alpha=0\quad\Rightarrow\quad \alpha=\frac\pi2\text{ or }-\frac\pi2,$ which lie in the given interval.

So the minimum value is $\boxed{0}.$

4. Check options

• A: $0$ ✅
• B: $-1$ ❌
• C: $\frac12$ ❌
• D: $-\frac12$ ❌

Hence the correct option is A.