Fluency exercise

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Solution

In the tables below, $0\leq \theta \leq 2\pi$ and any missing functions are from the following list. $\sin \theta \quad \cos \theta \quad \tan \theta \quad \sec \theta \quad \cosec \theta \quad \cot \theta$

Some of the row and column headings are missing. Without using a calculator, try to work out what they could be and complete the table. A function does not appear twice in the same table.

$\theta= \ldots$ $\theta= \ldots$ $\theta= \ldots$
$-1$ $\dfrac{\sqrt{3}}{2}$ $\dfrac{1}{2}$
$\tan \theta$ undefined $\sqrt{3}$
$\sec \theta$ $-\dfrac{2}{\sqrt{3}}$

We have lots of tools that could be used to tackle this problem without relying on a calculator. For example, we could

• use sketch graphs of the functions

• use special values of the functions, such as $\cos \tfrac{\pi}{3} = \tfrac{1}{2}$ and $\tan \tfrac{\pi}{4}=1$

• use triangles or trigonometric identities to work out values of some functions from the values of others

• use the unit circle diagram and symmetry to work out values of functions for angles bigger than $\tfrac{\pi}{2}.$

We will use a mixture of these tools as we describe two approaches to completing the first table.

Approach 1

If $\tan \theta$ is undefined then $\cos \theta=0.$ Therefore $\sec \theta$ is also undefined and the only possibilities for $\theta$ in the first column are $\tfrac{\pi}{2}$ and $\tfrac{3\pi}{2}.$

As $\cot \tfrac{\pi}{2}=\cot \tfrac{3\pi}{2}=0$ the missing function must be $\sin \theta$ or $\cosec \theta$, both of which are $-1$ only when $\theta = \tfrac{3\pi}{2}$, so in the first column $\theta = \tfrac{3\pi}{2}.$

We need more information to decide whether the missing function is $\sin \theta$ or $\cosec \theta.$

From the top row of the table we see that the value of the missing function can be $\tfrac{1}{2}$, so it must be $\sin \theta.$

Now, $\tan \theta = \dfrac{\sin \theta}{\cos \theta} = \sin \theta \sec \theta.$ This means that we can complete the remaining cells in the table without knowing the values of $\theta.$

To find the values of $\theta$, we can recognise certain special values in the table, such as $\tfrac{\sqrt{3}}{2} =\sin \tfrac{\pi}{3}$ and $\tfrac{1}{2}=\sin \tfrac{\pi}{6}$, but we need to decide whether this means that the values of $\theta$ are $\tfrac{\pi}{3}$ and $\tfrac{\pi}{6}$ or angles related to them, such as $\tfrac{\pi}{3}+\pi.$

From the graphs of these functions, we can check where the functions are positive and negative to decide on the values of $\theta.$

Approach 2

An alternative starting point would be to note that in the third column, $\cos \theta = -\tfrac{\sqrt{3}}{2}$, so $\theta$ must be $\tfrac{5\pi}{6}$ or $\tfrac{7\pi}{6}.$ The only function which has value $\tfrac{1}{2}$ at either of these values of $\theta$ is $\sin \theta$, and it only takes this value when $\theta = \tfrac{5\pi}{6}.$ The table can now be completed using a similar argument to the one given above.

How might the answers change if $\theta$ could be any value?

What if you could use functions like $- \sin \theta$ to complete the table?

We were limited to $0\leq \theta \leq 2\pi.$ Otherwise $\theta$ could take any value of the form $\dfrac{\pi}{3}+2n\pi$ and $\dfrac{5\pi}{6}+2n\pi$ where $n$ is an integer. If we were allowed to use functions such as $- \sin \theta$ then the problem does not have a unique solution.

This time we have given some more information about $\theta.$ Try to identify the missing functions and complete the table. Remember not to use a calculator!

$\theta$ is reflex $\theta= \ldots$ $\theta$ is obtuse
$0$ $-\dfrac{3}{5}$
$\cosec \theta$ $1$ $\dfrac{5}{4}$
$\dfrac{5}{12}$ $0$

Let’s start with the second column because we know where certain functions are $1$ or $0.$ Again, $\theta$ is limited to $0\leq \theta \leq 2\pi$, so $\cosec \theta = 1$ is only true when $\theta$ is $\tfrac{\pi}{2}.$ We can now think about which functions are $0$ when $\theta = \tfrac{\pi}{2}.$ Since $\tan \tfrac{\pi}{2}$ and $\sec \tfrac{\pi}{2}$ are undefined, the only possibilities for the missing functions are $\cos \theta$ and $\cot \theta.$

As both $\cos \theta$ and $\cot \theta$ could be $\tfrac{5}{12}$ and $-\tfrac{3}{5}$, we need more information to decide which function goes in the top row and which goes in the bottom row (remember that a function cannot appear twice in the table). Both $\cos \theta$ and $\cot \theta$ are negative when $\theta$ is obtuse and both can be negative or positive when $\theta$ is reflex, so the signs don’t help us here.

Instead, we could use the value of $\cosec \theta$ in the third column. One of the Pythagorean identities for trigonometric functions tells us that $\cosec^2 \theta \equiv 1 + \cot^2 \theta.$

If $\cot \theta$ were the correct function for the top row, then substituting $-\tfrac{3}{5}$ for $\cot \theta$ and $\tfrac{5}{4}$ for $\cosec \theta$ should satisfy this identity. If these values don’t satisfy the identity, then we will know that $\cot \theta$ can’t be the function in the top row.

Once we know where $\cot \theta$ and $\cosec \theta$ go, we can use trigonometric identities to complete the table, taking into account the signs of these functions for reflex angles.

• How can you also state the values of $\theta$ in the $1$st and $3$rd columns?

We can express $\theta$ in the third column as $\arccos \left(-\tfrac{3}{5}\right)$, but in the first column $\theta$ cannot be written $\arccos \left(-\tfrac{5}{13}\right)$ because of the principal value range of $\arccos.$ How is this value of $\theta$ related to $\arccos \left(-\tfrac{5}{13}\right)$?

We could also use $\mathrm{arccot}$ or $\mathrm{arccosec}$ and it could be interesting to think about what their principal value ranges must be.