Building blocks

## Solution

Can you explain why the points given in the diagram below have coordinates $(1, \tan \theta)$ and $(\cot \theta, 1)$?

Now move the slider for $\delta \theta$.

• What is the relationship between the blue angle and the red angle?

• What can you say about the blue triangles and the red triangle as $\delta \theta$ decreases to $0$?

You may find it helpful to look at the circles section of Going round in circles to see how $\tan \theta$ and $\cot \theta$ relate to the unit circle.

Move the slider so that the blue triangles are visible. Note that the blue angle is $\theta + \delta \theta$ and the red angle is $\theta$. Therefore the blue triangles are both similar to the triangle with angle $\theta + \delta \theta$ and right-angle at $(1,0).$ This triangle tends to the red triangle as $\delta \theta$ tends to $0$ (written $\delta \theta \rightarrow 0$), so the blue triangles get closer and closer to being similar to the red triangle as $\delta \theta \rightarrow 0$.

Consider the blue triangle with one vertex at $(1, \tan \theta)$.

• What is the length of the circular arc drawn through $(1, \tan \theta)$ and subtended by $\delta \theta$?

In triangle $OQP$, Pythagoras’ theorem tells us that $OP=\sec \theta$ (note that $\theta$ is acute in this case, so $\sec \theta >0$).

Alternatively, $\cos \theta = \dfrac{1}{OP}$ so $OP=\dfrac{1}{\cos \theta}= \sec \theta$.

$OP$ is the radius of the circular arc drawn through $P$ and subtended by $\delta \theta$, so the length of this arc is $\sec \theta \,\delta \theta.$ For small $\delta \theta$, we can therefore approximate $PS$ by $PS \approx \sec \theta \,\delta \theta.$

Why is it important that $\theta$ and $\delta \theta$ are measured in radians?

• Which side length shows the increase in $\tan \theta$ as $\theta$ increases by $\delta \theta$? This can be denoted by $\delta (\tan \theta)$.

$PT=\tan (\theta+ \delta \theta) -\tan \theta$, so $PT$ shows the increase in $\tan \theta$ as $\theta$ increases by $\delta \theta$. We’ll write this as $\delta (\tan \theta)$ as it is the change in $\tan \theta$.

How can you use these ideas to find the derivative of $\tan \theta$?

The derivative of $\tan \theta$ is given by $\lim_{\delta \theta \rightarrow 0} \dfrac{\tan (\theta+ \delta \theta) -\tan \theta}{\delta \theta}$ so using our notation, we are interested in $\dfrac{\delta (\tan \theta)}{\delta \theta}.$

We have approximations for the lengths of two sides of the blue triangle when $\delta \theta$ is small. As $\delta \theta \rightarrow 0$, the blue and red triangles get closer to being similar, so it may help to consider corresponding sides and angles in the two triangles.

For small $\delta \theta$, the corresponding ratios of sides gives us $\dfrac{PT}{PS}\approx\dfrac{OP}{OQ}$, so

$\dfrac{\delta (\tan \theta)}{\sec \theta \,\delta \theta}\approx \dfrac{\sec \theta }{1} \quad \text{or equivalently} \quad \dfrac{\delta (\tan \theta)}{\delta \theta}\approx \sec^2 \theta.$

Taking the limit as $\delta \theta \rightarrow 0$ then gives the derivative of $\tan \theta$,

$\dfrac{d}{d\theta}(\tan \theta) = \sec^2 \theta.$

What does this say about the sign of the gradient function of $\tan \theta$? Can you connect this with the graph of $\tan \theta$?

What about the derivative of $\sec \theta$?

We’ll refer back to our first labelled diagram. The length of the hypotenuse of triangle $OQP$ is $\sec \theta$, so as $\theta$ increases by $\delta \theta,$ $ST$ shows the increase in $\sec \theta.$

How can you use these triangles to show that $\dfrac{d}{d \theta}(\sec \theta) = \sec \theta \tan \theta$?

Can you take a similar approach to find the derivatives of $\cot \theta$ and $\cosec \theta$? You may be able to do this in more than one way.

The blue triangle with a vertex at $(\cot \theta, 1)$ can be used to find derivatives of $\cot \theta$ and $\cosec \theta$, so we’ll label the vertices of this triangle as well.

You could use a similar approach to before, by considering ratios of sides in

• triangles $UVM$ and $MNO$ (since $MNO$ is similar to $OQP$ and $OM =\cosec \theta$), or
• triangles $UVM$ and $OQP$ (since the sides of $OQP$ don’t depend on $\delta \theta$), or
• triangles $UVM$ and $PST$ (these are similar even if $\delta \theta$ isn’t small).

We’ll use triangles $UVM$ and $MNO$, but what happens if you use different triangles?

We have worked with a diagram in which $\theta$ is acute. Do these results hold for any value of $\theta$?