Question

Find the range of ${\sin }^{-1}\left(x\right)-{\cos }^{-1}\left(x\right)$

Answer 1

Find the range of the given function

Given: ${\sin }^{-1}\left(x\right)-{\cos }^{-1}\left(x\right)$

Range of ${\sin }^{-1}\left(x\right)=\left[\frac{-\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]$

Range of ${\cos }^{-1}\left(x\right)=\left[0,\mathrm{\pi }\right]$

Therefore, $f\left(x\right)={\sin }^{-1}\left(x\right)-{\cos }^{-1}\left(x\right)={\sin }^{-1}x+{\cos }^{-1}\left(x\right)-2{\cos }^{-1}\left(x\right)$

$f\left(x\right)=\frac{\mathrm{\pi }}{2}-2{\cos }^{-1}\left(x\right)\mathbf{[}\mathbf{\because }{\mathbf{sin}}^{\mathbf{-}\mathbf{1}}\left(x\right)\mathbf{+}{\mathbf{cos}}^{\mathbf{-}\mathbf{1}}\left(x\right)\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{\pi }}{\mathbf{2}}\mathbf{]}$

Range of $f\left(x\right)\in \left[\frac{\mathrm{\pi }}{2}-2\left(\mathrm{\pi }\right),\frac{\mathrm{\pi }}{2}-0\right]$

Range of $f\left(x\right)=\left[\frac{-3\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]$

Hence the range of the given function is $\left[\frac{-3\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]$