What is the domain and range of y=arcsin⁡x

Question

Mary Herrera · Accepted Answer

Range: -pi/2 ≤ y ≤ pi/2  Domain: -1 ≤ x ≤ 1

lenkiklisg7 · Accepted Answer

More explanation please

user_27qwe · Accepted Answer

The original sine function defined for any real argument does not have an inverse function because it does not establish a one-to-one correspondence between its domain and a range.
To be able to define an inverse function, we have to reduce the original definition of a sine function to an interval where this correspondence does take place. Any interval where sine is monotonic and takes all values in its range would fit this purpose.
For a function $y = \sin x$ an interval of monotonic behavior is usually chosen as $[- \frac{π}{2}, \frac{π}{2}],$ where the function is monotonously increasing from -1 to 1.
This variant of a sine function, reduced to an interval where it is monotonous and fills an entire range, has an inverse function called $y = \arcsin (x)$
It has range $[- \frac{π}{2}, \frac{π}{2}]$ and domain from -1 to 1

What is the domain and range of y=\arcsin x

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