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.