Methods of Last Resort 6: Right-angled Trigonometry

I must admit to having reservations as I write this blog post. Not because I am unsure as to the approaches I will outline, but rather to do with the categorisation as a 'Method of Last Resort'. Before now I have typically suggested that methods of last resort should be the things we do when our understanding of a situation doesn't allow us to take a more efficient approach - for example considering order of operations a 'Method of Last Resort' as it is the sort of thing we consider when we can't simply work left to right (as in 5 x 6 ÷ 10 for example) or when we can't simplify a calculation (as in 23 x 6 + 7 x 6 = 30 x 6, or 17² – 3²). I am not completely sure that what I am going to outline falls into that category, but nonetheless here goes...

I am going to propose that SOHCAHTOA is a method of last resort. By SOHCAHTOA I don't mean the mnemonic, I mean the idea of treating the trigonometric ratios as formulae:

So what is the alternative? Well the obvious one is the unit circle, but that might be a bit much for the first introduction of trigonometry. Instead I wanted to outline an approach around similar triangles.

Let us first take sine. Sine of an angle between 0 and 90 relates the opposite to the hypotenuse in the following way:

This is all the basis we need to find missing sides in any right triangle with the angle θ. Consider now the triangle below:

This triangle is an enlargement of the first triangle, using a scale factor of 13. This implies that the opposite side is simple 13 × sin θ. Even looking at the triangle below:

This triangle is still an enlargement of the first triangle, but with a scale factor of 13/sin θ. So the hypotenuse must by 13/sin θ.

This approach also be used to find angles. Consider the triangle below:

This triangle is still an enlargement of the original triangle, again by scale factor 13. This would mean that the opposite side of the smaller triangle is 5/13. But remember, in the smaller triangle the opposite side is sin θ. So we have that sin θ = 5/13. This leads of course to θ = sin^-1 (5/13).

Virtually identical approaches can be used with reference to the adjacent and hypotenuse sides, with the cosine function and the tangent function with the opposite and adjacent sides. Importantly, this approach arguably requires a deeper understanding of how trigonometric functions relate sides of a triangle than the formulae provided at the beginning of this blog post, and it is for this reason why I wonder if the formulae couldn't be considered a 'Method of Last Resort.'

Methods of Last Resort 5: Median and Mean.

Lets see if these seem familiar:

Median = middle number in a data set when the set is ordered.
Mean = total of the data set shared equally between the number of data points (or possibly "add them all up and divide by how many there are", but if you still use this, then see my blog here).

In the main, perfectly acceptable approaches to finding median and mean. Note I don't use the term average here: I think a lot more work needs to be done to separate the finding of mode, median and mean with the concept of average, and will blog about that at some future point. For now I want to concentrate on the process of finding median and mean rather than any link they have to the concept of average. Now consider the following:

1) Find the median of the list 3, 5, 6, 7, 8, 13, 10.
2) Find the median of the list 3, 2, 1, 6, 10, 9, 8.
3) Find the mean of the list 7, 9, 10, 11, 13.
4) Find the mean of the list 106, 104, 108, 107, 108.

To anyone that understands the ideas of median and mean,  these questions are a bit different, in that they don't require the definitions provided above. Let us tackle them in pairs.

Firstly the median. In both of the cases above the middle value is the median, and the fact that the lists are not in order makes precisely 0 difference. Now I can hear the arguments already, "yes but these are very contrived data-sets", "yes but that won't work a lot of the time" and I understand where they are coming from. But the point is, as a competent mathematician I get that in these cases there is no need to order. If our goal is to produce competent mathematicians in our pupils, to have pupils that understand these concepts properly, then surely they should understand this as well? And it can't be blamed on my education beyond GCSE - I did precisely no study of statistics beyond GCSE. I had choices for my modules at A-Level and so did all Core and Mechanics, and then my Degree was all in either pure maths or maths modules that linked to classical mechanics and physics. There was no statistics content at all.

A possible solution to this is to re-define the median as something like "the value in the middle position of a data set if all positions below are numerically smaller and all positions above are numerically bigger". Honestly though this definition seems overly convoluted for such a simple concept. There are plenty of times when re-ordering the list is the best strategy, even if it wouldn't be completely necessary (for example 3, 2, 1, 8, 6 only requires the switching of the first and third digit). The point I think is that pupils need to understand what the ordering is trying to achieve, and are shown explicit examples of when this isn't necessary. The ordering of the list can then be treated as a 'Method of Last Resort', something you do when the median is not already in the correct position or very close to the correct position.

Now questions 3 and 4 on the mean. Again as a competent mathematician I understand that I don't need to find the totals in these questions. In the first I can see that 7 and 13 are equally spaced from 10, as are 9 and 11, so these differences are going to even out and make the mean 10. Interestingly, I am not sure I would make the same argument if the list was 13, 9, 11, 10, 7 - I think if presented with this list I would begin to total it and then probably see that the 13 and 7 will combine nicely along with the 9 and 11. In question 4 I can see that I only need to total the 6, 4, 8, 7, and 8 and then find the mean of these 5 numbers before just adding the mean to 100 (to be fair this is something I came across when teaching myself the MEI S1 and S2 units so I could teach my Further Maths A-Level groups - it is called linear coding). Whilst this might mean we could choose to avoid highlighting this particular property of mean at GCSE (although I can't see a good argument for doing so really) it still illustrates that there are other ways of calculating the mean. Again we could solve this by re-defining what we mean by "mean" to better capture the 'evening out' idea, but this would see to again be a bit of overkill. I think the point here is that we should aim to secure understanding of mean to the point where pupils are able to identify whether the total needs to be found or not - totalling becomes a method of last resort to be used if other more efficient methods are not easily identifiable.

As I have been writing this blog, this has highlighted to me what appears to be a subtle difference between the ideas of median and mean and the accepted process for finding them. The idea of median is this idea of centrality, and an accepted process for finding it is ordering. The idea of mean is the idea of evening out the distribution, and totalling then dividing is one way of accomplishing this. I need to consider more what this means for my teaching practice. In the meantime what I will say is that I definitely think we need to be trying to secure the understanding necessary in pupils so that they can discriminate between times when the accepted process is the best, and when it isn't

For those that may not have followed this blog sequence from when I started it following my session at mathsconf, I will reiterate what I have said before - I am not saying whether you should lead with this, or lead with the standard approach before pointing out these special cases. That judgement needs to be made for classes by the teachers that work with them week in and week out. What I am saying is that I passionately believe that our pupils deserve to see these sorts of examples at some point rather than not at all. If we are truly going to teach to develop understanding in our pupils then we need to include this as part of the understanding of median and mean.

Methods of Last Resort 4 - Comparing/Adding/Subtracting Fractions

Working with fractions is notoriously something that teachers complain about when it comes to pupils' understanding and ability to manipulate. As a result it often seems to me that working with fractions is a place where even the best maths teachers can often fall back into what Skemp would call 'instrumental understanding'; pupils mechanically following procedures rather than applying any understanding of the relationships between the different parts of the process or between the question and the result.

This was brought to mind for me recently when I saw the question below mixed into a group of questions about comparing fractions:

From the rest of the questions listed it was quite clear that the intention would be that pupils write the second fraction as a fraction of 30 so that the comparison between the numerators would yield clearly that the first fractions is bigger than the second. Which of course is completely apparent because the first is more than ½ and the second less than ½. Any halfway competent mathematician wouldn't even bother equating the denominators, and this is the sort of thing I would want to highlight to pupils in order to try and develop their relational understanding.

The process of finding common denominators for comparing, adding and subtracting fractions is one that can easily become automatic for pupils, and I would argue that if pupils are to really understand fractions then they need to be able to take a more discriminatory approach. The following are all examples of questions that pupils could tackle without finding common denominators:

I would argue that the first and second points are more easily done by converting to decimals than fractions (which people may or may not agree with), and that the last one certainly doesn't require a common denominator; the first is greater than ½ whilst the second is equal to ½.

So if you are truly committed to developing pupils' relational understanding of fractions then the next time you look at the sorts of comparisons or calculations that often benefit from converting into equivalent fractions with common denominators, it might be worth throwing in some examples and questions of calculations where this is a method of last resort.

Methods of last resort 3 - Straight line graphs

The linear relationship is probably one of the most fundamental relationships in all of mathematics. Functions that have a constant rate of change are the basis of our most rudimentary geometrical transformations, conversions and correlations. It should be fair to say that ensuring pupils have a proper grasp of linear relationships should be an important part of any mathematics curriculum; and yet many pupils are only given a very narrow view of these key mathematical constructs.

Most pupils first view of the graphs of linear relationships between two variables are through algebra in the form y = mx + c. Pupils will be given equations of this form, and asked to substitute to find coordinates and then plot coordinates to draw lines. Some pupils may be given the opportunity to draw parallels between the equation and the relationship between the variables x and y but not all. Eventually concepts like gradients and intercepts will be taught, and here is where the narrowing will begin. Most pupils will be given an algebraic definition of gradient, such as "change in y over change in x" or similar. Can we first be very clear from the start please that this is not what gradient is, this is just one way to find the gradient if you happen to know the horizontal and vertical distance travelled (for those people who think I am being picky, another way to find the gradient is to take the tangent of the angle the line makes with the horizontal, which is seldom taught in this way).

What gradient actually is is the vertical distance travelled for a unit increase in horizontal distance. Dividing a given vertical by a given horizontal will calculate the the value, as will applying the tangent function to the angle made with the horizontal, but neither tell you what it actually is. Pupils should have a proper understanding of what gradient is, before they begin calculating it (in my opinion). But this is not actually the point of this blog post so I will get back on track...

Once gradient is 'taught' the link between its value and the value of m in the formula given above is very quickly highlighted, often either explicitly or through some form of 'discovery'. Here comes the second narrowing - from this point onward virtually every attempt to ascertain the value of the gradient of a particular line when given any form of linear algebraic relationship invariably leads back to writing the equation in the form y = mx + c. Remember lines are very often defined in a different form; x + y = 5, 3x + 2y + 4 = 0 etc. Ask any competent school age pupil to find the value of the gradient of these lines, and I will guarantee that the vast majority of the time a rearrangement into the form y = mx + c is attempted if the pupil is even able to attempt the problem at all. And while this approach is perfectly correct and if done well will reveal the value of the gradient, it isn't the only approach; many pupils labour in ignorance when better methods may be applied.

Take the line x + y = 5 for example. Now for most mathematicians it would be straightforward to rearrange this to give y = -x + 5, and hence find the value of the gradient of -1, and the y-intercept of (0,5). However I would argue at least equally straightforward would be to say "the points (0,5) and (5,0) are on the line, and so the value of the gradient = -5/5 = -1 and the y-intercept is (0,5) [and, by the way, the x intercept is (5,0) - which is not nearly so often asked about]. To be fair, there is probably not a huge difference in the mechanics, but as Anne Watson highlights in her blog (see postscript below) there is perhaps a difference in pupils understanding of what this line actually looks like, as well as providing more of an opportunity to reinforce the idea of vertical distance travelled for unit horizontal distance.

If we then take the line 3x + 2y + 4 = 0, the rearrangement is a bit messier - I know plenty of pupils that wouldn't be able to rearrange successfully. However it is still a rearrangement that you would want pupils to be able to do and expect that they could if they had the proper grounding in inverse operations etc. The other side of this though is that I can quite quickly see that the point (0,-2) is on this line, and that the point (-1⅓, 0) is on this line. So I can also calculate the gradient as -2/1⅓ = 1½, as well as tell you about the x-intercept and y-intercept. Perhaps even more straightforwardly I could have told you that the point (1, -3½) is on the line, and so arrived at the value of the gradient immediately, I have gone 1½  units down when x increased by 1 (from 0 to 1).

Whether you want to consider rearrangement to the form y = mx + c as a 'method of last resort' or not is up to you; clearly it is an important mathematical idea that relationships can be expressed in different forms. However I would suggest that it is not the only idea that pupils should be able to draw upon when talking and thinking about finding gradient values, and that we should be aiming to give pupils a range of strategies linked to a deeper understanding of what gradients, and lines of constant gradient, are.

Postscript: Emeritus Professor of Education at Oxford University Anne Watson recently released a blog about a similar topic (and actually using one of the same equations!) here. I have actually been writing this blog post since late January and was just trying to find time to finish it off, so wanted to go ahead and publish it anyway!

Methods of last resort 2 - Order of Operations

Teaching the correct order of operations is possibly one of the most debated topics for maths teachers. In my #mathsconf8 session I was asked 'what is my problem with BIDMAS' and proceeded to outline times when this acronym is redundant (e.g. 4 x 3 ÷ 2) or even downright wrong (4 - 5 + 6 would mistakenly be given as -7 rather than the correct answer as 5). Various diagrams have been mooted as the solution to this, and there are several examples below:

 

I have several issues with these diagrams, which can be summarised as:

(a) It isn't specific enough for all of the possible functions that can be applied to numbers (even those that include square roots don't involve higher roots, and no mention of sin, cos, tan, log etc)

(b) BRACKETS ARE NOT AN OPERATION (please forgive the shouting). This may seem like semantics but for me it is an important distinction - brackets are used to either alter or clarify the order of operations intended, but are not an operation in themselves (just a note on clarify, an example of this is 12 ÷ (3 x 4) needed clarity as without these brackets the answer would be 16 and not 1). If we are going to teach pupils to understand the maths they are doing then we need to be communicating understanding like this, and not allowing pupils to mistakenly believe that brackets are an operation themselves.

But this post is not about teaching correct order of operations (although that segue has outlined my thoughts on it quite nicely); this is about when you wouldn't want pupils teaching using the correct order of operations in the first place. The example I used in my #mathsconf8 session was:

673 x 405 — 672 x 405

Any mathematician is definitely not applying the correct order of operations in this situation; and is quickly writing down that this is just 405. With the advent of 'teaching for mastery' gaining ground in mathematics education pupils are being increasingly exposed to questions like this when looking at distributive laws, or factorisation but I am yet to see it, or anything like it, thrown into a lesson on Order of Operations as a non-example. There is good evidence out there now to back up the idea that non-examples are important in communicating a concept and so if we are trying to communicate the correct order of operations we should be highlighting cases like this as when applying the correct order of operations is not wrong, but is just wildly inefficient compared to use of the distributive laws (in this case the formal statement would be something like 673 x 405 - 672 x 405 = 405 x (673 - 672) = 405 x 1 = 405).

Some other examples of times when correct order of operations are an inefficient way to solve problems (particularly without a calculator) are:

• 12 x 345 ÷ 6
• 18² ÷ 9²
• √128 ÷ √32 (although this one does require some real mathematical understanding)
• 372 + 845 – 369

I would be exploring all of these questions prior to teaching the correct order of operations, and then including questions like it in the deliberate practice on the correct order of operations to ensure that pupils are recognising when not to apply them alongside when they are absolutely necessary.

Methods of Last Resort 1 - Percentages

Following on from my session in Kettering at #mathsconf8 I will be writing a series of blogs about the areas of maths I find or figure out that might be better looked at separate to any problems that might be solved using a standard approach or a 'method of last resort'. The first area I want to look at is percentages.

Because of the multiplicative nature of percentages there are lots of questions that can be solved without having to resort to approaches such as "Find 10% first..." or "What multiplier calculates...." or other standard approaches. The point I made at mathsconf is that I would want pupils to understand why these questions can be solved quickly and straightforwardly, and that actually by exploring the special nature of some of these calculations we can deepen pupils understanding of the topics - in this case percentages.

Find 32% of 75

This is the example I used at mathsconf. There are still plenty of teachers that don't realise that 32% of 75 is the same as 75% of 32, but once they see it they understand why. What I like is that in explaining why this is true really does get at the heart of percentages and how they are calculated and so it is a perfect little 'explain why' to stretch pupils as well as then serving as reinforcement of concepts for others.

Find 32% of 100

Try it; you will be surprised how many pupils don's immediately link the % with the 100 or are unsure when they want to say 'isn't that just 32?' Again this sort of question gets at the heart of percentages as parts of 100.

Find 32% of 50

If you have built up to it these are actually now becoming quite straightforward, but encouraging pupils to talk and explain why is still powerful.

Find 32% of 200, 300, 400 etc

I probably don't need to say much more at this point.

As well as calculating percentages, equally there are similar questions for writing one value as a percentage of another. Again there are standard approaches for this (writing and converting fractions or similar) but there are questions that anyone with a real understanding of percentages would look at and solve. This set of questions comes from a well known worksheet provider; see if you can spot the ones that could be done without requiring the use of a standard approach or 'method of last resort'.

Even if you don't really know your fractions, questions 3, 5, 6, 7, 11 and possibly 12 and/or 17 can be solved using some relatively straightforward multiplication and division. Do we always teach pupils though that if they can see an obvious way to write it as 'a percentage of 100' that this will be much quicker than a standard approach, and more importantly to support them in understanding why this works which would lead to a deeper understanding of percentages as a whole.