Given that the inverse trigonometric function assumes principal values only. Let $x$, $y$ be any two real numbers in $[-1,1]$ such that $\cos^{-1}x - \sin^{-1}y = \alpha$, $\dfrac{-\pi}{2}\leq\alpha\leq\pi$. Then, the minimum value of $x^2+y^2+2xy\sin\alpha$ is
Step-by-Step Solution
Key Concept: Use $\sin^{-1}y=\frac{\pi}{2}-\cos^{-1}y$, so $\cos^{-1}x+\cos^{-1}y=\frac{\pi}{2}+\alpha$. Apply the cosine addition formula for $\cos^{-1}$ to get $xy-\sqrt{1-x^2}\sqrt{1-y^2}=-\sin\alpha$. Square and simplify.
Step 1: Introduce angle substitutions and use the given relation.
Let $\cos^{-1}x = \theta_1$ and $\sin^{-1}y = \theta_2$.
Based on the principal value ranges of inverse trigonometric functions:
$x = \cos\theta_1$, where $\theta_1 \in [0, \pi]$.
$y = \sin\theta_2$, where $\theta_2 \in [-\pi/2, \pi/2]$.
The given condition is $\cos^{-1}x - \sin^{-1}y = \alpha$, which can be written as:
$$\theta_1 - \theta_2 = \alpha$$
From this, we can express $\theta_1$ as:
$$\theta_1 = \alpha + \theta_2$$
Step 2: Substitute the trigonometric expressions for $x$ and $y$ into the given algebraic expression.
We need to find the minimum value of the expression $x^2+y^2+2xy\sin\alpha$.
Substitute $x = \cos\theta_1$ and $y = \sin\theta_2$:
$$x^2+y^2+2xy\sin\alpha = (\cos\theta_1)^2 + (\sin\theta_2)^2 + 2(\cos\theta_1)(\sin\theta_2)\sin\alpha$$
Now, substitute $\theta_1 = \alpha + \theta_2$ into the expression:
$$x^2+y^2+2xy\sin\alpha = \cos^2(\alpha+\theta_2) + \sin^2\theta_2 + 2\cos(\alpha+\theta_2)\sin\theta_2\sin\alpha$$
Step 3: Expand and simplify the trigonometric expression.
We use the angle addition formula for cosine, $\cos(A+B) = \cos A\cos B - \sin A\sin B$, to expand $\cos(\alpha+\theta_2)$:
$$\cos(\alpha+\theta_2) = \cos\alpha\cos\theta_2 - \sin\alpha\sin\theta_2$$
Substitute this expansion back into the expression from Step 2:
\begin{align*} x^2+y^2+2xy\sin\alpha &= (\cos\alpha\cos\theta_2 - \sin\alpha\sin\theta_2)^2 + \sin^2\theta_2 + 2(\cos\alpha\cos\theta_2 - \sin\alpha\sin\theta_2)\sin\theta_2\sin\alpha \\ &= (\cos^2\alpha\cos^2\theta_2 + \sin^2\alpha\sin^2\theta_2 - 2\sin\alpha\cos\alpha\sin\theta_2\cos\theta_2) + \sin^2\theta_2 \\ & \quad + (2\sin\alpha\cos\alpha\cos\theta_2\sin\theta_2 - 2\sin^2\alpha\sin^2\theta_2)\end{align*}
The terms $-2\sin\alpha\cos\alpha\sin\theta_2\cos\theta_2$ and $+2\sin\alpha\cos\alpha\sin\theta_2\cos\theta_2$ cancel each other out.
$$x^2+y^2+2xy\sin\alpha = \cos^2\alpha\cos^2\theta_2 + \sin^2\alpha\sin^2\theta_2 + \sin^2\theta_2 - 2\sin^2\alpha\sin^2\theta_2$$
Combine the terms involving $\sin^2\theta_2$:
$$x^2+y^2+2xy\sin\alpha = \cos^2\alpha\cos^2\theta_2 + \sin^2\theta_2(1 + \sin^2\alpha - 2\sin^2\alpha)$$
$$x^2+y^2+2xy\sin\alpha = \cos^2\alpha\cos^2\theta_2 + \sin^2\theta_2(1 - \sin^2\alpha)$$
Using the trigonometric identity $1 - \sin^2\alpha = \cos^2\alpha$:
$$x^2+y^2+2xy\sin\alpha = \cos^2\alpha\cos^2\theta_2 + \sin^2\theta_2\cos^2\alpha$$
Factor out $\cos^2\alpha$:
$$x^2+y^2+2xy\sin\alpha = \cos^2\alpha(\cos^2\theta_2 + \sin^2\theta_2)$$
Since $\cos^2\theta_2 + \sin^2\theta_2 = 1$:
$$x^2+y^2+2xy\sin\alpha = \cos^2\alpha$$
Step 4: Determine the minimum value of the expression.
We have found that $x^2+y^2+2xy\sin\alpha = \cos^2\alpha$.
The problem states that $-\dfrac{\pi}{2} \leq \alpha \leq \pi$.
The function $\cos^2\alpha$ is always non-negative, meaning $\cos^2\alpha \ge 0$.
The minimum value of $\cos^2\alpha$ occurs when $\cos\alpha = 0$.
Within the given interval $[-\pi/2, \pi]$, $\cos\alpha = 0$ for $\alpha = -\pi/2$ and $\alpha = \pi/2$.
At these values of $\alpha$, $\cos^2\alpha = 0^2 = 0$.
Therefore, the minimum value of $\cos^2\alpha$ in this interval is $0$.
Step 5: Conclude the final answer.
The minimum value of the expression $x^2+y^2+2xy\sin\alpha$ is $0$.
The final answer is $\boxed{0}$.
Correct Answer: 1