Question

If (sin^–1x)² + (sin^–1y)² + (sin^–1z)² = 3/4 π². Find x² + y² + z².

Answer 1

Range of sin^-1x is \([-\frac{\pi}2,\frac{\pi}2].\)

Given that

(sin^-1x)² + (sin^-1y)² + (sin^-1z)² = \(\frac 34\)π²

Each of sin^-1x, sin^-1y and sin^-1z takes the value of \(\frac{\pi}2.\)

x = 1, y = 1, and z = 1.

Hence,

= x² + y² + z²

= 1 + 1 + 1

= 3.