Question

If sin^-1x + sin^-1y + sin^-1z = $-\frac{3\mathrm{\pi }}{2}$, then xyz = __________________.

Answer 1

We know

$-\frac{\mathrm{\pi }}{2}\le {\sin }^{-1}a\le \frac{\mathrm{\pi }}{2}$, for all a ∈ [−1, 1]

So, the minimum value of sin⁻¹a is $-\frac{\mathrm{\pi }}{2}$.

Now,

${\sin }^{-1}x+{\sin }^{-1}y+{\sin }^{-1}z=-\frac{3\mathrm{\pi }}{2}$ (Given)

This is possible if

${\sin }^{-1}x=-\frac{\mathrm{\pi }}{2},{\sin }^{-1}y=-\frac{\mathrm{\pi }}{2}$ and ${\sin }^{-1}z=-\frac{\mathrm{\pi }}{2}$

⇒ x = −1, y = −1 and z = −1

∴ xyz = (−1) × (−1) × (−1) = −1

If sin⁻¹x + sin⁻¹y + sin⁻¹z = $-\frac{3\mathrm{\pi }}{2}$, then xyz = ___−1___.