Finding general formula for $\cos^{-1}({\cos{x})}$

Question

My teacher teacher told me that for a general angle $x$, $\cos^{-1}({\cos{x})}$ does't represent $x$ but different straight lines depending upon the intervals in which it lies. For ex:

$$\cos^{-1}{\cos{x}}=$$ $$x,0\leq x \leq \pi \\ 2\pi-x,\pi\leq x \leq 2\pi\\…$$ making the graph look like :- From wolfram alpha

He told us that if we have to find the value of $\cos^{-1}({\cos{x})}$ for a particular $x$ we will have to first find the range in which $x$ lies and then judge with the help of graph but I wondered if there is a direct formula for that. I tried with $\tan^-1({\tan{x}})$ and got it as :- from wolfram alpha

I even verified this with wolfram alpha and got it right but the problem with $\cos$ is that when I try to solve it similarly like I did with the $\tan$ one, and get the interval in which $n$ lies, the extremities of the interval differ by $0.5$ because of which for some values their floor and ceiling match but for some values there isn't an integer value lying in that interval like this :- from wolfram alpha

so what to do in that case and what does no value of $n$ lying in the interval signify?

Thanks for help :)

Answer 1

By definition we have that for $x\in[0,2\pi]$

• for $0\le x\le \pi\quad $ $\cos^{-1}{\cos{x}}=x$
• for $\pi<x\le 2\pi\quad$ $\cos^{-1}{\cos{x}}=2\pi-x$

and this is periodic with period $T=2\pi$.

Thus it is a kind of triangle function and we always need to divide into two parts dependind upon the range in which x lies.

Answer 2

I'm writing $\arccos$ instead of $\cos^{-1}$. By definition, $$\arccos(\cos y)=y\qquad(0\leq y\leq\pi)\ .$$ For arbitrary $x\in{\mathbb R}$ define $$d(x):=\min\bigl\{|x-2k\pi|\,\bigm|\,k\in{\mathbb Z}\bigr\}$$ to be the distance of $x$ from the nearest integer multiple of $2\pi$. Then $$0\leq d(x)\leq\pi,\quad \cos x=\cos\bigl(d(x)\bigr)\qquad\forall x\in{\mathbb R}\ .$$ It follows that $$\arccos(\cos x)=\arccos\bigl(\cos\bigr(d(x)\bigr)\bigr)=d(x)\qquad(x\in{\mathbb R})\ ,$$ which reveals $\arccos\circ\cos$ to be a sawtooth function.

Answer 3

Claim. Let $x\in\left[k\pi,\left(k+1\right)\pi\right]$, where $k\in\mathbb{Z}$. Then:

\begin{alignat*}{1} \cos^{-1}\left(\cos x\right) & =\begin{cases} x-k\pi & \textit{if }k\textit{ is even},\\ \left(k+1\right)\pi-x & \textit{if }k\textit{ is odd}. \end{cases} \end{alignat*}

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Proof. First, note that $x-k\pi$ and $\left(k+1\right)\pi-x$ are both in $\left[0,\pi\right]$.

Moreover, if $y\in\left[0,\pi\right]$, then $\cos^{-1}\left(\cos y\right)\overset{1}{=}y$.

Now, suppose $k$ is even. Then $\cos\left(x-k\pi\right) \overset{2}{=}\cos x$ and so:

$$\cos^{-1}\left(\cos x\right)=\cos^{-1}\left[\cos\left(x-k\pi\right)\right]\overset{1}{=}x-k\pi.$$

Next, suppose $k$ is odd. Then $\cos\left[\left(k+1\right)\pi-x\right]\overset{3}{=}\cos x$ and so:

$$\cos^{-1}\left(\cos x\right)=\cos^{-1}\left[\cos\left(\left(k+1\right)\pi-x\right)\right]\overset{1}{=}\left(k+1\right)\pi-x. \tag*{∎}$$

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(The reader can verify $\overset{2}{=}$ and $\overset{3}{=}$ using the Subtraction Formulae for Cosine.)