Food for thought

## Solution

Take a look at the three graphs below.

One of the graphs shows $y=\arctan(\tan x)$. Another shows $y=\tan (\arctan x)$. Which graphs are these, and why?

As $\tan x$, $\sin x$ and $\cos x$ are periodic functions, there are many values of $x$ that give the same value of $\tan x$, $\sin x$ or $\cos x$. This means that inverse functions such as $\arctan x$ and $\arcsin x$ have to be very carefully defined. You can read more about this in Inverse trigonometric functions, and these ideas are used in this solution.

Looking at the three graphs, I notice that Graph C looks like a graph of $y=x$ for all real values of $x.$ In contrast, Graph B looks like a graph of $y=x$ but only for $x$ in the interval $\left(\frac{-\pi}{2}, \frac{\vphantom{-}\pi}{2}\right)$. Having thought about Graph B, I can now think of Graph A as a repeating version of Graph B, with period $\pi$.

To see which graphs show the functions $\arctan(\tan x)$ and $\tan (\arctan x)$ I can think about domains and ranges.

When composing functions $f(x)$ and $g(x)$ to form $g(f(x))$, I need to think about the domain of $f(x)$. I need to check that any output of $f$ is in the domain of $g$ and what these outputs are, so the range of $g(f(x))$ depends on the domain and range of $f$ as well as the range of $g$.

Starting with the inner function in $\arctan(\tan x)$, I know that $\tan x$ is defined for all real $x$ except where $x=\frac{(2n+1)\pi}{2}$. This is the domain of $\tan x$. The range is the whole of the real numbers. This means the input to the outer $\arctan$ function is all the real numbers, so the output for $\arctan$ is its principal value range, which is the interval $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$. Therefore, the graph of $\arctan(\tan x)$ has a domain which is the whole of the $x$-axis except the points where $x=\frac{(2n+1)\pi}{2}$, and the range is $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$, so Graph A shows $y=\arctan(\tan x)$.

I’ll now use similar thinking to work out which is the graph of $y= \tan (\arctan x)$. This time $\arctan x$ is the inner function. Its domain is the whole of the real numbers, but the range is $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$.

This graph shows $y= \tan x$ for $x$ in the interval $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$. Can you use it to explain why Graph C must show $y=\tan (\arctan x)$?

Match these equations to the graphs below and explain your reasoning.

$y=\arcsin(\sin x)$

$y=\sin(\arcsin x)$

$y=\arccos(\cos x)$

$y=\cos(\arccos x)$

One thing I notice is that all four of the functions map $0$ to $0$, so the graphs must pass through the origin. This eliminates Graph G.

$y=\arcsin(\sin x)$

The domain of $\sin x$ is all real $x$ and the range is the interval $[-1,1]$. The domain of $\arcsin x$ is the interval $[-1,1]$, and the principal value range is $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ (see Inverse trigonometric functions for more explanation).

So the function $\arcsin(\sin x)$ is defined for all real $x$ and has range $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$. Within the domain $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ the function will map $x$ to itself so the graph will look like that of $y=x$. So this one must be Graph D.

Notice that as $x$ increases from $\frac{\pi}{2}$ to $\frac{3\pi}{2}$, $\sin x$ decreases from $1$ to $-1$ and so $\arcsin(\sin x)$ decreases from $\frac{\pi}{2}$ to $-\frac{\pi}{2}$. And then this pattern continues, forming a zig-zag graph.

$y=\sin(\arcsin x)$

The domain of $\arcsin x$ is the interval $[-1,1]$ and it is undefined elsewhere. Within this domain it has range $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$, and these values as input to $\sin x$ produce values in the range $[-1,1]$. So the graph will look like $y=x$ restricted to the domain $-1\le x\le1$, which is Graph E.

$y=\arccos(\cos x)$

$\cos x$ is defined for all real $x$ and has range $[-1,1]$. $\arccos x$ is defined on the domain $[-1,1]$ and its principal value range is $[0,\pi]$. So the domain of this composition is all the real numbers and its range is$[0,\pi]$. The only graph that matches this is J.

By way of a check, as $x$ increases from $0$ to $\pi$, the value of $\cos x$ decreases from $1$ to $-1$ and so $\arccos(\cos x)$ increases from $0$ to $\pi$. On this domain, the graph looks like $y=x$. As $x$ increases further from $\pi$ to $2\pi$, $\cos x$ increases from $-1$ to $1$ and $\arccos(\cos x)$ decreases from $\pi$ to $0$. Hence we get a zig-zag pattern.

$y=\cos(\arccos x)$

$\arccos x$ is defined only for $x$ in the interval $[-1,1]$. It’s range is $[0,\pi]$ and $\cos$ of these values has range $[-1,1]$. So our graph will look like $y=x$ restricted to the domain $[-1,1]$, and it must be Graph E, the same as for equation (2).

It is interesting to note that $\cos x$ and $\arccos x$ are decreasing functions on these intervals but the composition of the functions is increasing. Is this the case for any pair of decreasing functions?

For which values of $x$ is $\tan(\arctan x)=x$? What about $\arctan(\tan x)=x$?

What can you say about the solutions of similar equations, such as $\sin(\arcsin x)=x$ or $\arccos(\cos x)=x$?

As Graph C is $y=\tan(\arctan x)$, it must be that $\tan(\arctan x)=x$ for all real $x$.

Graph A shows $y=\arctan(\tan x)$, so $\arctan(\tan x)=x$ only for $x$ in the interval $\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$ as this is the only part of the graph for which $y=x$ matches the graph of $y=\arctan(\tan x)$.

To illustrate what happens if $x$ is outside this interval, let’s try $x=\tfrac{4\pi}{3}$. I know $\tan \tfrac{4\pi}{3} = \sqrt{3}$, but $\arctan \sqrt{3}= \tfrac{\pi}{3}$. Therefore $\arctan\left(\tan \tfrac{4\pi}{3}\right)= \arctan \sqrt{3}=\tfrac{\pi}{3}$. Generalising from this example, I can see that if $x$ is outside the interval $\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$ and $\tan x$ is defined, then $\arctan(\tan x)$ takes $x$ to the corresponding value inside the interval $\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$.

We can draw corresponding conclusions for the $\sin$ and $\cos$ compositions by looking at graphs D, E and J.