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 \(\dfrac{\delta (\tan \theta)}{\delta \theta}.\)
What can the blue and red triangles tell us about this fraction?
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 \(\theta\)?
