Given $f(x)=\cos^{-1}(\cos x)$, why is $f(4)\neq 4$?

Question

Let $f(x)=\cos^{-1}(\cos x)$.

• a) What does $f(x)$ mean and what should it be?

  A: $f(x)=\cos^{-1}(\cos x)=x$.

• b) Test your answer in a) for $x=1,\;2,\;3,\;4$.

  A: $f(1)=1,\;f(2)=2\;,f(3)=3,\;f(4)\neq 4$

• c) Explain the result in b).

Translation: If $\cos\;x=k$ and $\cos^{-1} k=x$, $\cos x$ must be limited to an interval where every allowed $k$ gives one $x$. We have chosen $0\leq x\leq\pi$. For $x>\pi$ we get the $x$ in this interval for which $\cos\;x=\cos4$.

I do not understand their explanation. Could someone write/explain it in another way?

Answer 1

Since the cosine function is periodic, it does not have an inverse unless we restrict its domain to an interval in which it is injective, meaning that there is one value of $x$ in the domain for each value of $y$ in the range. By convention, we take the restricted domain to be $[0, \pi]$. Thus, we need the inverse of the function $g: [0, \pi] \to [-1, 1]$ defined by $g(x) = \cos x$.

The inverse of $g$ is called the arccosine function, denoted $h(x) = \arccos x$ or $h(x) = \cos^{-1}x$. It maps the interval $[-1, 1] \to [0, \pi]$ and is defined by $$h(x) = \arccos x = y \iff \cos y = x~\text{and}~y \in [0, \pi]$$

Therefore, the restriction of the function $f: \mathbb{R} \to [0, \pi]$ defined by $f(x) = \arccos(\cos x)$ to the interval $[0, \pi]$ is the identity function on that interval. Thus, \begin{align*} f(1) & = 1\\ f(2) & = 2\\ f(3) & = 3 \end{align*} However, outside the interval $f(x) = t$ if $t$ is the value in $[0, \pi]$ such that $\cos x = \cos t$ since $$f(x) = \arccos(\cos x) = \arccos(\cos t) = t$$
where $t \in [0, \pi]$.

Since $4 > \pi$, we must find the value of $t \in [0, \pi]$ such that $f(4) = \cos t$. Since $\cos(\pi + x) = -\cos x$, $$\cos(4) = \cos(\pi + 4 - \pi) = -\cos(4 - \pi)$$
Since $\cos(\pi - x) = -\cos x$, $$\cos(4) = -\cos(4 - \pi) = \cos(\pi - (4 - \pi)) = \cos(2\pi - 4)$$ Moreover, $2\pi - 4 \in [0, \pi]$. Hence, $$f(4) = 2\pi - 4$$

Answer 2

You know that $\cos(x)$ can take any real value $x$ and return a real value between $[-1,1]$. So you see that $\cos^{-1}(y)$ must admit only values for $y$ between $[-1,1]$ because that is the range of $\cos(x)$.

Now, if we look at the function $\cos^{-1}(\cos(x))$, we can still input any $x$ but the output of $\cos^{-1}(y)$ must lie somewhere between $[0,\pi]$ or $[\pi,2\pi]$ but in no interval that has the same $\cos$-value for a specific $x$.

Answer 3

$\cos(x)$ is not an injective function on $\mathbb{R}$. In other words, if we look at $\cos(x)=1$ (for example) we can find more than $1$ value of $x$ for which $\cos(x)=1$. In fact we can find infinitely many $x$ for which $\cos(x)=1$.

The issue arises. What is the value of $\cos^{-1}(1)$? We can say $x=0$, but we can also put $x=2\pi$, or any $2k\pi$ for any integer $k$. But we can't have a function spewing out multiple outputs for a single input - this simply goes against the definition of a function.

But if we restrict $\cos(x)$ to a domain of $[0, \pi]$ (other restrictions work but this one is pretty standard) then $\cos(x)$ only gives one output per input. Now, we can uniquely define $\cos^{-1}(1)=0$, because no other value in this new restriction works!

Answer 4

Remember the equation $\;\cos x=\cos\alpha$ has solutions $$x\equiv \pm\alpha\mod 2\pi\mathbf Z,$$ and the restriction of $\cos$ to the interval $[0,\pi]$ is a decreasing bijection from this interval to the range $[-1,1]$ of $\cos$.

The arccosine is, conventionally, the inverse bijection of this restriction.

Thus $y=\arccos(\cos x)$ is the unique number in $[0,\pi]$ such that $$\cos y=\cos x.$$ So $\;\arccos(\cos1,2,3)=1,2,3$, but $\;\arccos(\cos 4)=2\pi-4=\arccos(\cos(-4))$.