Let g(x) = sin⁻¹(2x√(1 - x²)) - 2 sin⁻¹x for -1 ≤ x ≤ 1…

Step 1. The regions. $\dfrac{1}{\sqrt{2}} = 0.7071$. So $0.8$ and $0.9$ are in the right region, and $-0.9$ is in the left region. (For a number such as $0.5$, in the middle region, $g = 0$.)

Step 2. At $x = 0.8$. The first term is $\pi - 2\sin^{-1}x$, so $g(x) = \pi - 4\sin^{-1}x = 3.1416 - 4 \times 0.9273 = -0.5676$.

Step 3. At $x = 0.9$. In the same way $g(x) = \pi - 4\sin^{-1}0.9 = 3.1416 - 4 \times 1.1198 = -1.3375$.

Step 4. At $x = -0.9$. In the left region the first term is $-\pi - 2\sin^{-1}x$, so $g(x) = -\pi - 4\sin^{-1}x = -3.1416 + 4 \times 1.1198 = 1.3375$. This is the negative of $g(0.9)$, as expected for an odd function.