Inverse Trigonometric Functions

Try this

1. Drag $a$: the horizontal line $y=a$ meets the whole sine curve in infinitely many points (the hollow dots). Knowing the sine, you cannot tell the angle.
2. On the shaded piece $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$, sine is increasing and takes every value in $[-1,1]$, so each $a$ corresponds to exactly one point.
3. Click “mirror in $y=x$”: flip this piece over and you get the graph of $y=\arcsin x$, with domain $[-1,1]$ and range $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$.
4. Switch to “branch 1”: it is monotone too and can also be flipped, but what you get is $\pi -\arcsin x$, not $\arcsin x$.
5. Switch to $\cos$ and $\tan$: their “branch 0” is $[0,\pi ]$ and $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$ respectively.

So $\arcsin a$ means: the angle in $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$ whose sine is $a$. In the same way

Why we cut out a piece

The symmetry of the graphs also gives some useful relations: $\arcsin (-x)=-\arcsin x$ and $\arctan (-x)=-\arctan x$ (odd functions), $\arccos (-x)=\pi -\arccos x$, and $\arcsin x+\arccos x=\frac{\pi }{2}$.

Remove the principal interval

Practice