Vtyjkyuk

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

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

He told us that if we have to find the value of $cos^-1(cosx)$ 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 $tan^-1(tanx)$ 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 :)

By definition we have that for $xin[0,2pi]$

• for $0le xle piquad $ $cos^-1cosx=x$


• for $pi<xle 2piquad$ $cos^-1cosx=2pi-x$

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

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.

I'm writing $arccos$ instead of $cos^-1$. By definition,
$$arccos(cos y)=yqquad(0leq yleqpi) .$$
For arbitrary $xinmathbb R$ define
$$d(x):=minbigl$$
to be the distance of $x$ from the nearest integer multiple of $2pi$. Then
$$0leq d(x)leqpi,quad cos x=cosbigl(d(x)bigr)qquadforall xinmathbb R .$$
It follows that
$$arccos(cos x)=arccosbigl(cosbigr(d(x)bigr)bigr)=d(x)qquad(xinmathbb R) ,$$
which reveals $arccoscirccos$ to be a sawtooth function.

Claim. Let $xinleft[kpi,left(k+1right)piright]$, where $kinmathbbZ$. Then:

beginalignat*1
cos^-1left(cos xright) & =begincases
x-kpi & textitif ktextit is even,\
left(k+1right)pi-x & textitif ktextit is odd.
endcases
endalignat*

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

Moreover, if $yinleft[0,piright]$, then $cos^-1left(cos yright)overset1=y$.

Now, suppose $k$ is even. Then $cosleft(x-kpiright) overset2=cos x$ and so:

$$cos^-1left(cos xright)=cos^-1left[cosleft(x-kpiright)right]overset1=x-kpi.$$

Next, suppose $k$ is odd. Then $cosleft[left(k+1right)pi-xright]overset3=cos x$ and so:

$$cos^-1left(cos xright)=cos^-1left[cosleft(left(k+1right)pi-xright)right]overset1=left(k+1right)pi-x. tag*∎$$

(The reader can verify $overset2=$ and $overset3=$ using the Subtraction Formulae for Cosine.)