Question

Given 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:

• A. -1
• B. 0
• C. \( \frac{-1}{2} \)
• D. \( \frac{1}{2} \)

Answer 1

The Correct Option is B

We start by analyzing the expression \( x^2 + y^2 + 2xy \sin \alpha \). This expression can be recognized as the expansion of \( (x + y \sin \alpha)^2 \), which is always non-negative.

Given that \( \cos^{-1} x - \sin^{-1} y = \alpha \), the values of \( x \) and \( y \) are restricted to the interval \([-1, 1]\), ensuring the values lie within the principal range of the inverse trigonometric functions.

Now, let’s rewrite the expression:

\[ x^2 + y^2 + 2xy \sin \alpha = (x + y \sin \alpha)^2. \]

The minimum value of a square term \( (x + y \sin \alpha)^2 \) is 0, which occurs when \( x + y \sin \alpha = 0 \).

Thus, the minimum value of \( x^2 + y^2 + 2xy \sin \alpha \) is 0.

Answer 2

Step 1: Use given equation
Given: \( \cos^{-1}x - \sin^{-1}y = \alpha, \, -\frac{\pi}{2} \leq \alpha \leq \pi \)
Let \( \sin^{-1}y = \theta \implies y = \sin \theta, \, -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2} \)
So, \( \cos^{-1}x = \alpha + \theta \implies x = \cos(\alpha + \theta) \)

Step 2: Compute required expression
Required minimum value of \( x^2 + y^2 + 2xy \sin \alpha \)
Substitute values:
\( x^2 = \cos^2(\alpha + \theta) \), \( y^2 = \sin^2 \theta \), \( 2xy \sin \alpha = 2 \cos(\alpha+\theta)\sin \theta \sin \alpha \)
So,
\( x^2 + y^2 + 2xy \sin \alpha = \cos^2(\alpha + \theta) + \sin^2 \theta + 2 \sin \theta \cos(\alpha + \theta) \sin \alpha \)

Step 3: Simplify using identity
\( 2 \sin \theta \cos(\alpha+\theta) \sin\alpha = [\sin(\theta + \alpha + \theta) - \sin(\alpha + \theta - \theta)] \)
But better to combine using variable elimination:
Let \( x = r \cos \alpha, y = r \sin \alpha \), by constraint \( x^2 + y^2 = r^2 \leq 1 \).
Now, required value \( x^2 + y^2 + 2xy \sin \alpha = r^2 + 2r^2 \sin\alpha\cos\alpha = r^2(1 + \sin2\alpha) \).
Since minimum is 0 when \( r = 0 \), minimum value is 0.