Consider the function y = sin⁻¹(2x√(1 - x²)), x ∈ [(-1)/(√2), 1/(√2)]. Find the range of y if x =…

Consider the function $y = \sin^{-1}\left(2x\sqrt{1 - x^2}\right),\ x \in \left[\dfrac{-1}{\sqrt{2}}, \dfrac{1}{\sqrt{2}}\right]$. Find the range of $y$ if $x = 0,\ \dfrac{1}{2}$ and $\dfrac{-1}{2}$.

Answer

$y = \sin^{-1}\left(2x\sqrt{1 - x^2}\right)$. At $x = 0$: $y = \sin^{-1}0 = 0$. At $x = \dfrac{1}{2}$: $y = \sin^{-1}\left(2\cdot\dfrac{1}{2}\cdot\dfrac{\sqrt{3}}{2}\right) = \sin^{-1}\dfrac{\sqrt{3}}{2} = \dfrac{\pi}{3}$. At $x = -\dfrac{1}{2}$: $y = \sin^{-1}\left(-\dfrac{\sqrt{3}}{2}\right) = -\dfrac{\pi}{3}$. So $y = 0,\ \dfrac{\pi}{3},\ -\dfrac{\pi}{3}$ respectively.