Question 1142691
the largest domain of y = arcsin(x) appears to be 0 < x <= 1.


y = sin(x) gives you a value of y = -1 to 1.


that's the range of y = sin(x).


y = arcsin(-1) = -90 degrees.
y = arcsin(1) = 90 degrees.


the value of x goes from -1 to 1 and the value of y goes from -90 degrees to 90 degrees.


the domain of y = arcsin(x) is -1 <= x <= 1.


the domain of y = ln(arcsin(x)) is 0 < x <= 1.


that's because the natural log of a number requires the number to be greater than 0.


since arsin(0) = 0, that would be invalid.
as long as arcsin(x) has x > 0, then the value would be valid.


i think this is what you are looking for.


the domain would be 0 <= x <= 1
the range would be -infinity < y <= ln(90) when working with degrees.
since 90 degrees * pi / 180 = pi/2 radians, then the range would be -infinity to pi/2 radians when working with radians.


whether working with degrees or radians, the domain is still 0 < x <= 1.


the following graphs illustrate this.


the first 2 graphs show the value of y in degrees.


the last 2 graphs shosw the value of y in radians.


in both cases, the value of ln(arcsin(x)) has a domain of 0 < x <= 1.


<img src = "http://theo.x10hosting.com/2019/062801.jpg" alt="$$$" >


<img src = "http://theo.x10hosting.com/2019/062802.jpg" alt="$$$" >


<img src = "http://theo.x10hosting.com/2019/062803.jpg" alt="$$$" >


<img src = "http://theo.x10hosting.com/2019/062804.jpg" alt="$$$" >


i'm not exactly sure that this is what you're looking for, but it's what i get based on my interpretation of the problem.