# How do you find the domain and range of arcsin(1-x^2)?

Aug 10, 2018

Domain: $\left\mid x \right\mid \le \sqrt{2}$
Range: $\left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$

#### Explanation:

$y = \arcsin \left(1 - {x}^{2}\right)$ is constrained to be in $\left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$ and,

as $\left(1 - {x}^{2}\right)$ is a sine value, $- 1 \le 1 - {x}^{2} \le 1$

$\Rightarrow - 1 \le 1 - {x}^{2} \Rightarrow {x}^{2} \le 2 \Rightarrow - \sqrt{2} \le x \le \sqrt{2}$.

See graph, depicting domain and range.
graph{(y-arcsin(1-x^2))(y-pi/2 +0y)(y+pi/2+0y)=0}

Had it been the piecewise wholesome sine inverse

${\left(\sin\right)}^{- 1} \left(1 - {x}^{2}\right) = k \pi + {\left(- 1\right)}^{k} \arcsin \left(1 - {x}^{2}\right)$, you can use

the common-to-both inverse $1 - {x}^{2} = \sin y$, to get the

wholesome unconstrained graph, for range unlimited.
graph{1-x^2-sin y = 0[ -20 20 -10 10] }