Big trig

Big trig

Excerpt: This provides an opportunity for students to consider how the sine and cosine functions behave in a "which is bigger?" context. Working on this problem could help to reinforce students' understanding of the behaviour of trigonometric functions and the problem can be attempted in a variety of different ways.

Problem

Which is larger: $\sin(\cos x)$ or $\cos(\sin x)$?

Does it depend on $x$?

Suggestion

• What are the minimum and maximum values of $\sin x$ and $\cos x$?

• Where are $\sin x$ and $\cos x$ positive or negative?

• Where are $\sin x$ and $\cos x$ increasing or decreasing?

Things you might consider

Let’s start by thinking about what we know about the behaviour of $\sin x$ and $\cos x.$

• Both $\sin x$ and $\cos x$ have period $2\pi$ so $\sin(\cos x)$ and $\cos(\sin x)$ will be periodic. The period of each function can’t be more than $2\pi$ so we only need to consider what happens for $x$ in the interval $[0,2\pi].$

• $\sin x$ and $\cos x$ vary between $-1$ and $1.$ What does this tell us about the minimum and maximum values of $\sin(\cos x)$ and $\cos(\sin x)$?

• For $x$ in $\big[0,\tfrac{\pi}{2}\big]$ and $\big[\tfrac{3\pi}{2},2\pi\big]$ $\sin x$ increases and for $x$ in $\big[\tfrac{\pi}{2},\tfrac{3\pi}{2}\big]$ it decreases. Where does $\cos x$ decrease or increase between $0$ and $2\pi$?

• We know that $\cos x$ is an even function and $\sin x$ is an odd function, i.e. $\cos (-x) = \cos x$ and $\sin(-x)=-\sin x.$

Special points

We’ll consider some multiples of $\tfrac{\pi}{2}$ to get a sense of how the two functions behave. You may find it helpful to start marking some points on a sketch graph as you read.

• When $x=0, \pi, 2\pi$ the value of $\cos(\sin x)$ is $1$, which must be the maximum value of the function. This means the period of $\cos(\sin x)$ could be $\pi.$

• $\sin(\cos 2\pi)=\sin(\cos 0) = \sin 1.$ The value $\sin 1$ must be positive, but as $\sin x$ does not reach $1$ until $x=\tfrac{\pi}{2}$, $\sin(\cos 0) <1.$ Similarly $\sin (\cos \pi)=\sin (-1) = -\sin 1 >-1.$

• $\sin \tfrac{\pi}{2}=1$, so $\cos (\sin \tfrac{\pi}{2})=\cos 1$, which is positive, but less than $1.$ Also, $\cos (\sin \tfrac{3\pi}{2})=\cos 1$ because of the period of $\cos (\sin x).$

• $\cos x=0$ when $x=\tfrac{\pi}{2}$ and $x=\tfrac{3\pi}{2}$ so $\sin(\cos x)=0$ at these points.

This means that $\cos(\sin x) >\sin(\cos x)$ at $0,\tfrac{\pi}{2},\pi,\tfrac{3\pi}{2}$ and $2\pi.$ The values of $\cos 1$ and $\sin 1$ seem to be important. Without using a calculator or graphing software, how could you decide which of these values is bigger?

Behaviour between special points

Positive and negative

As $\cos x$ is negative between $\tfrac{\pi}{2}$ and $\tfrac{3\pi}{2}$ and $\sin x$ is negative for $x$ between $-1$ and $0$ the value of the function $\sin (\cos x)$ must be negative for $x$ between $\tfrac{\pi}{2}$ and $\tfrac{3\pi}{2}.$

What does the same reasoning tell us about $\cos(\sin x)$?

Symmetry

As

$\cos x$

is an even function and

$\sin x$

is odd,

$\cos (\sin (-x))=\cos (-\sin x)=\cos (\sin x)$

and

$\sin (\cos (-x))=\sin (\cos x)$

, so both functions are even. As well as telling us that the

$y$

-axis is a line of symmetry of the graphs of

$y=\sin (\cos x)$

and

$y=\cos (\sin x)$

, this also shows that these graphs are symmetric about the line

$x=\pi.$

(For example,

$\sin (\cos (\pi-x))=\sin (\cos (x-\pi)).$

) This helps because we are interested in what happens between

$0$

and

$2\pi.$

Increasing and decreasing behaviour

For $x$ between $0$ and $\tfrac{\pi}{2}$, $\sin x$ increases from $0$ to $1$ and $\cos x$ decreases, so $\cos(\sin x)$ decreases from $1$ to $\cos 1$. On $\big[\tfrac{\pi}{2},\pi\big]$, $\sin x$ decreases from $1$ to $0$ and $\cos x$ decreases from $0$ to $-1$. This means that $\cos(\sin x)$ increases from $\cos 1$ to $1$ and therefore has a minimum when $x=\tfrac{\pi}{2}.$ What happens between $\pi$ and $2\pi$?

These ideas, can be used to show that $\cos(\sin x)$ has a period of $\pi$ and its range is $[\cos 1,1]$ whereas $\sin(\cos x)$ has period $2\pi$ and its range is $[-\sin 1, \sin 1].$

For all the points considered, $\cos(\sin x) >\sin(\cos x)$ and we now know a lot about the graphs $y=\cos (\sin x)$ and $y=\sin (\cos x).$ It seems reasonable to conjecture that $\cos(\sin x) >\sin(\cos x)$ for all $x$, but to convince ourselves that this is true, we need to show that the graphs do not cross anywhere.

Do the graphs of $y=\cos(\sin x)$ and $y=\sin(\cos x)$ ever cross?

The graphs can’t cross between $\tfrac{\pi}{2}$ and $\tfrac{3\pi}{2}$ because $\sin(\cos x)$ is negative here, but what about between $0$ and $\tfrac{\pi}{2}$ or between $\tfrac{3\pi}{2}$ and $2\pi$?

We want to know whether $\cos (\sin x) = \sin (\cos x)$ has any solutions for $x$ in $\left[0, \tfrac{\pi}{2}\right].$ Now, if $\cos a = \sin b$ then $a=\tfrac{\pi}{2}-b$, so $a+b=\tfrac{\pi}{2}$, or in our case $\sin x+ \cos x = \tfrac{\pi}{2}.$

The diagram shows part of a circle of radius $1$ and an angle $x$ between $0$ and $\tfrac{\pi}{2}.$ The length of arc $APB$ is $\tfrac{\pi}{2},$ but $\sin x$ is smaller than the length of arc $AP$ and $\cos x$ is smaller than the length of arc $PB.$

Therefore $\sin x + \cos x < \tfrac{\pi}{2}$ so $\cos (\sin x) = \sin (\cos x)$ does not have any solutions in $[0, \tfrac{\pi}{2}]$. The symmetry of the graphs about $x=\pi$, means that there aren’t any solutions in $[\tfrac{3\pi}{2},2\pi]$ either. Therefore $\cos (\sin x)$ is larger than $\sin (\cos x)$ for all real $x.$

Another approach is to use the inequality $\sin \theta < \theta$ for positive $\theta.$ To see how this inequality arises, consider the diagram below. This shows the chord $AB$ of a circle subtending an angle $\theta$ at the centre of the circle, where $0<\theta < \pi.$

If we compare the areas of the triangle $OAB$ and the minor sector $OAB,$ we get $\tfrac{1}{2}r^2 \sin \theta < \tfrac{1}{2}r^2 \theta$ and therefore $\sin \theta < \theta$.

We can combine this inequality with the increasing and decreasing behaviour of $\sin x$ and $\cos x$ on $\big[ 0, \tfrac{\pi}{2}\big]$ to show that $\sin (\cos x) < \cos x \quad \text{and} \quad \cos (\sin x) > \cos x$

Therefore $\cos x$ is always sandwiched between $\sin (\cos x)$ and $\cos (\sin x)$ for $x$ between $0$ and $\tfrac{\pi}{2}.$ As above, the symmetry of the graphs completes the proof that $\cos (\sin x)$ is larger than $\sin (\cos x)$ for all real $x.$

Graph