Deragz

2021-12-26

What is the domain and range of $y=\mathrm{arcsin}x$

Mary Herrera

Expert

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

lenkiklisg7

Expert

user_27qwe

Expert

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 $y=\mathrm{sin}x$ an interval of monotonic behavior is usually chosen as $\left[-\frac{\pi }{2},\frac{\pi }{2}\right],$ 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 $y=\mathrm{arcsin}\left(x\right)$
It has range $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$ and domain from -1 to 1

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