Things you might have noticed

Take a look at the diagram below.

Half circle divided into sectors

We’ve used inequalities involving \(\sin \theta\), \(\cos \theta\) and \(\tan \theta\) to divide the semicircle into sectors. Each sector in the diagram is defined by a different inequality. For example, one sector is defined by the angles \(\theta\) between \(0\) and \(\pi\) for which \(\cos \theta < \sin \theta < \tan \theta.\) Another sector is defined by \(\cos \theta< \tan \theta < \sin \theta.\)

  • Can you work out which inequality has been used to define each sector?

There are many ways to approach this problem, and we will illustrate a few of them here. If you have used different approaches, or used similar approaches but for different parts of the problem, you might find it interesting to compare the methods.

However, we may need to do a bit more work to complete the sketch graphs. For example, do the graphs \(y=\tan \theta\) and \(y=\sin \theta\) cross between \(\theta = 0\) and \(\theta =\dfrac{\pi}{2}\)? Here are a couple of ways to think about this.

Can you explain why the identity \(\tan \theta \equiv \dfrac{\sin \theta}{\cos \theta}\) tells us the following?

  • For \(\theta\) between \(0\) and \(\dfrac{\pi}{2}\), \(\sin \theta < \tan \theta\).

  • The graphs of \(y=\tan \theta\) and \(y=\sin \theta\) only meet when \(\theta\) is a multiple of \(\pi\).

Thinking more generally, we can also use the diagram to show the following.

  • For small values of \(\theta\), \(\sin \theta < \tan \theta <\cos \theta\). This defines the red sector.

  • \(\cos \dfrac{\pi}{4} = \sin \dfrac{\pi}{4}\)

  • For \(\theta\) between \(\dfrac{\pi}{4}\) and \(\dfrac{\pi}{2}\), \(\cos \theta < \sin \theta < \tan \theta\), which defines the green sector, as we also saw before from the graph.

We now only need to decide which inequality defines the orange sector. You may find it helpful to look at the graphs or to visualise a point moving round the unit circle.

If we think about the behaviour of the functions as \(\theta\) increases from \(0\), we first have \(\sin \theta < \tan \theta < \cos \theta\) (which defines the red sector). Another inequality must define the orange sector, and then we have \(\cos \theta < \sin \theta < \tan \theta\) for \(\theta\) between \(\dfrac{\pi}{4}\) and \(\dfrac{\pi}{2}\), which defines the green sector.

As \(\tan \theta\) and \(\cos \theta\) are continuous between \(0\) and \(\dfrac{\pi}{4}\), there must be some value of \(\theta\) between \(0\) and \(\dfrac{\pi}{4}\) where the graphs \(y=\tan \theta\) and \(y=\cos \theta\) cross over.

Therefore the inequality that defines the orange sector is \(\sin \theta < \cos \theta < \tan \theta\). Why is this the only option?

Why is the boundary between the green and yellow sectors different from the boundaries between other sectors?

  • Which is the biggest sector?

The boundary of the orange and green sectors is at \(\theta=\dfrac{\pi}{4}\), and therefore the green sector is larger than the red sector. However, the yellow or blue sectors may be bigger than the green sector. Here are a few ways to tackle this problem.

What is the value of the different functions at \(\dfrac{3\pi}{4}\)?

What are the solutions of \(\tan \theta = \cos \theta\)? Identities may help to make this into a simpler equation to solve.

Do you need to solve this equation completely to decide which sector is the largest?

It may be of interest that solutions of \(\tan \theta = \cos \theta\) are related to the golden ratio.

From the symmetry of the graphs, the angle between the boundary of the yellow and blue sectors and \(\theta=\pi\) must be the same as the angle between \(\theta=0\) and the boundary of the red and orange sectors. In other words, the blue and red sectors are the same size. What does this tell us about the sizes of the other sectors?

If you used more than one approach, can you connect them? For example, can you connect the symmetry argument with solving \(\tan \theta = \cos \theta\)?

  • If you extended the diagram to make a complete circle, how many extra sectors would you need?

To complete the circle we can use similar methods to those discussed above. Another approach might be to consider the possible orderings of \(\sin \theta\), \(\cos \theta\) and \(\tan \theta\).

How many ways can the functions \(\sin \theta\), \(\cos \theta\) and \(\tan \theta\) be ordered?

  • Which of these ways of ordering occur between \(\theta =0\) and \(\theta=\pi\)?

  • Which ways might be repeated?

  • Will all these ways of ordering occur between \(\theta=0\) and \(\theta =2\pi\)?

Can you explain how the behaviour of the functions gives us these sectors and inequalities?

complete circle with all the sectors drawn and labelled by their corresponding inequalities as follows:  sin theta < tan theta < cos theta, sin theta < cos theta < tan theta, cos theta < sin theta < tan theta, tan theta < cos theta < sin theta, cos theta < tan theta < sin theta, cos theta< sin theta < tan theta, sin theta < cos theta < tan theta, tan theta < sin theta < cos theta

There were many ways to approach this problem. Instead of considering all three functions at the same time, we could have considered pairs of functions.

  • When is \(\sin \theta < \cos \theta\)?

  • When is \(\sin \theta < \tan \theta\)?

  • When is \(\tan \theta < \cos \theta\)?

Can you see how these inequalities feature in the diagram above?