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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 ?

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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...)

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Just as I thought. My students had a problem from a test which clearly had two solutions but did not get credit for that. Thank you very much. – hyg17 Apr 17 '13 at 20:17

You can look at graphs of all three functions here:

Do they look the same to you?

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Yes, they do. Thanks for confirming. – hyg17 Apr 17 '13 at 20:16

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