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

Answer: (B)

0

Answer 2

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.