Graph of an inverse trig function. Which of the following is equivalent to the graph of $arcsin(x)$ ?
(a) Reflecting $arccos(x)$ about the y-axis, then shift down by $\pi /2$ units.
(b) Reflecting $arccos(x)$ about the x-axis, then shift up by $\pi /2$ units.
I think they are both the same thing. Can someone confirm this ?

The inverse trig functions take numbers and give angles, so the relation between these functions can be written, for instance, as $\sin^{-1}x + \cos^{-1}x  = \frac{\pi}{2}.$ (The two functions represent complementary angles in the first quadrant, and the relationship still holds otherwise.)  So $\sin^{-1}x = \frac{\pi}{2} - \cos^{-1}x $, which indicates that the graph for inverse sine could be obtained by reflecting inverse cosine about the x-axis and shifting upward by $\frac{\pi}{2}$.
Ah, okay, as I look at this again, a reflection about the y-axis on inverse cosine produces $\cos^{-1} (-x)$; since the range of inverse cosine is $[0, \pi]$, this gives $\cos^{-1} (-x) = \pi - \cos^{-1}(x)$, that is, the supplementary angle.  Thus, $$\cos^{-1} (-x) = \pi - [\frac{\pi}{2} - \sin^{-1}(x)] \Rightarrow \sin^{-1} x = \cos^{-1}(-x) - \frac{\pi}{2}.$$  So the graph for inverse sine can also be produced by rotating the graph of inverse cosine about the y-axis and shifting it downward by $\frac{\pi}{2}$.  (There is an error in the specification for one of the WolframAlpha graphs...)
