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

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Answer 1 · 2018-08-10T13:02:06

Domain: $abs x <= sqrt 2$
Range: $[ - pi/2, pi/2 ]$

$y = arcsin ( 1 - x^2 )$ is constrained to be in $( -pi/2, pi/2 )$ and,

as $( 1 - x^2 )$ is a sine value, $- 1 <= 1 - x^2 <= 1$

$rArr -1 <= 1 - x^2 rArr x^2 <= 2 rArr -sqrt2 <= x <= sqrt 2$ .

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

$(sin)^(-1)( 1 - x^2 ) = kpi + (-1)^k arcsin( 1 - x^2 )$ , 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] }

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