Simplifying $\sin^{-1}(\cos x)$ 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?
 A: 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 $2\pi$.
A: 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 $x$.
A: Hints:
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 $\;\arcsin (\sin x)\equiv \begin{cases}x\\[0.5ex]\frac\pi 2-x\end{cases}\mod 2\pi$. 

A: 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:
$$\arcsin(\cos (180^\circ))=-90^\circ.$$
