Question:
If
\[ (\sin^{-1}x)^2 + (\sin^{-1}y)^2 + (\sin^{-1}z)^2 = \frac{3\pi^2}{4} \]
Find \(x^2 + y^2 + z^2\).
Concept:
The principal range:
\[ -\frac{\pi}{2} \leq \sin^{-1}x \leq \frac{\pi}{2} \]
Maximum square:
\[ (\sin^{-1}x)^2 \leq \left(\frac{\pi}{2}\right)^2 = \frac{\pi^2}{4} \]
—Solution:
Given:
\[ (\sin^{-1}x)^2 + (\sin^{-1}y)^2 + (\sin^{-1}z)^2 = \frac{3\pi^2}{4} \]
Maximum possible sum:
\[ 3 \times \frac{\pi^2}{4} = \frac{3\pi^2}{4} \]
So equality holds only when:
\[ \sin^{-1}x = \sin^{-1}y = \sin^{-1}z = \pm \frac{\pi}{2} \]
Thus:
\[ x = y = z = \pm 1 \]
Step 2: Compute required value
\[ x^2 + y^2 + z^2 = 1 + 1 + 1 = 3 \]
—Final Answer:
\[ \boxed{3} \]