# 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?

• The $\arcsin$ function is defined (usually) to map $[-1, 1]$ into $[-\frac{\pi}{2},\ \frac{\pi}{2}]$, to make it a function. – 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 $$2\pi$$.

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

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 $$\;\arcsin (\sin x)\equiv \begin{cases}x\\[0.5ex]\frac\pi 2-x\end{cases}\mod 2\pi$$.

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