<s>[INST] 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?
 [/INST] 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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