I have a question on $$\sin^{-1}(\cos x)$$

Since $\cos(x)=\sin(\frac \pi 2 \pm x)$, the above expression can simplify to either $\frac \pi 2 + x$ or $\frac \pi 2 - x$. This seems like a contradiction.

What's the problem here?

  • $\begingroup$ The $\arcsin$ function is defined (usually) to map $[-1, 1]$ into $[-\frac{\pi}{2},\ \frac{\pi}{2}]$, to make it a function. $\endgroup$ – AdditIdent Oct 19 '18 at 8:41

There is no contradiction. Keep in mind that the $\arcsin$ function is a map from $[-1,1]$ to $\left[-\frac\pi2,\frac\pi2\right]$, in order to avoid ambiguities. So:

  • if $x\in[0,\pi]$, then $\arcsin\bigl(\cos(x)\bigr)=\frac\pi2-x$;
  • if $x\in[\pi,2\pi]$, then $\arcsin\bigl(\cos(x)\bigr)=-\frac{3\pi}2+x$.

Outside the interval $[0,2\pi]$, use the fact that your function is periodic with period $2\pi$.

  • $\begingroup$ should not the second be $\color{red}{+}\frac{\pi}{2}+x$? $\endgroup$ – farruhota Oct 19 '18 at 9:38
  • $\begingroup$ @farruhota No, because then we would be outside the range $\left[-\frac\pi2,\frac\pi2\right]$. $\endgroup$ – José Carlos Santos Oct 19 '18 at 9:40
  • $\begingroup$ but now it is $[\pi/2,3\pi/2]$? $\endgroup$ – farruhota Oct 19 '18 at 9:42
  • $\begingroup$ @farruhota You're right! I've edited my answer. Thank you. $\endgroup$ – José Carlos Santos Oct 19 '18 at 9:45
  • $\begingroup$ +1. For showing how to get different expressions of $x$ for different intervals of $x$. $\endgroup$ – farruhota Oct 19 '18 at 11:03

We have that

$$\sin^{-1}(\cos x)=\sin^{-1}\left(\sin \left(\frac{\pi}2-x\right)\right)$$

and then recall that

$$\sin^{-1}(\sin \theta) =\theta \iff \theta\in \left[-\frac{\pi}2,\frac{\pi}2\right] $$

otherwise we need to adjust the result adding a suitable $k\pi$ term.

We can also use

$$\sin^{-1}(\cos x)=\sin^{-1}\left(\sin \left(\frac{\pi}2+x\right)\right)$$

and in that case we obtain different range for $x$.



You have to play with these relations/definitions:

  • $y=\arcsin x\iff \sin y=x\;\text{ AND }\;-\frac\pi2\le y\le \frac\pi2$,
  • $y=\arccos x\iff \cos y=x\;\text{ AND }\;0\le y\le \pi$,
  • $\arcsin x+\arccos x=\frac\pi2$,
  • $\sin(\arcsin x)=x$ BUT only $\;\arcsin (\sin x)\equiv \begin{cases}x\\[0.5ex]\frac\pi 2-x\end{cases}\mod 2\pi$.

Let $x=\pm\frac{\pi}{2}\mp(y+2\pi n),y\in [-\frac{\pi}{2},\frac{\pi}{2}]$.

Then: $$\begin{align}\arcsin(\cos x)&=\arcsin(\cos [\pm\frac{\pi}{2}\mp(y+2\pi n)])=\\ &=\arcsin(\sin (y+2\pi n))=\\ &=\arcsin(\sin y)=\\ &=y.\end{align}$$ Example 1: If $x=300^\circ$, then $300^\circ=-90^\circ+(30^\circ+360^\circ)$ and: $$\arcsin(\cos 300^\circ)=30^\circ.$$ Example 2: If $x=-300^\circ$, then $-300^\circ=90^\circ-(30^\circ+360^\circ)$ and: $$\arcsin(\cos (-300^\circ))=30^\circ.$$ Example 3: If $x=180^\circ$, then $180^\circ=90^\circ-(-90^\circ+0^\circ)$ and: $$\arcsin(\cos (180^\circ))=-90^\circ.$$


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