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Posts Tagged ‘Trigonometry’

Thoughts on the understanding paradox and introducing trigonometry

July 14, 2017 4 comments

Recently I read a blog entitled “The understanding paradox” (William, 2017) which discussed the idea of maths teaching and put forward the idea that actually, it is better to bypass understanding when first teaching a topic and then fill that understanding in later. This was then applied to the teaching of right angled triangle trigonometry in an example that I found confusing to say the least.

The author, Rufus William, suggested that when teaching trig for the first time we should be solely teaching procedurally using SOHCAHTOA as a mnemonic, but then went on to say we shouldn’t be discussing ratio or similarity and how that links until later on. This confused me as the mnemonic SOHCAHTOA is designed to help you remember the trig ratios. I.e. Sine is the ratio of the opposite side over the hypotenuse. Just by teaching that you ARE teaching the trig ratios and purely by the fact that you are teaching the students that this will work for all right angled triangles you are telling the students that the ratios are the same for any triangle with the same angle no matter what the length of the sides are. THIS IS THE VERY DEFINITION OF SIMILAR TRIANGLES.

This perplexed me a lot and I spent a lot of time thinking about it and asking the author to elaborate on what he meant. The only way I can fathom to teach this without reference to ratio and similarity would be to say: ” “SOHCAHTOA” it gives you 3 triangles. Label the sides circle them to see which triangle you use. Put numbers in, cover the missing one, its either a divide or a times”. To me this seems like a backwards way to go about things. It feels like you are teaching them unnecessary procedures to avoid discussing the underlying concepts of trigonometry, and it doesn’t really make sense to me.

I find that by the time students reach right angled triangle trigonometry they have already met the concept of similarity, I like to use this a way in to discussing the topic and to show that ratio of two sides that are the same in relation to an angle will be the same for all similar triangles. Students will have always encountered simplifying fractions before they meet trig and as such can see why this is. This is when I specifically discuss the sine, cosine and tangent ratios and introduce the procedural manner in which they can solve the problems, although I do avoid the dreaded formula triangles (for many reasons which I have blogged about here). I will show them some common mnemonics, and SOHCAHTOA is one of them. I’m not a fan of mnemonics personally, I’ve never found them that useful except for musical ones, but I know a lot of people do.

Rufus does make some salient points in his post about teachers who refuse to allow students to memorise things and the dangers this will have on learning. Although I’m not entirely sure that they exist, and if they do I certainly don’t think there are many of them. I’ve certainly never met any.

He also suggests that students cannot have a full understanding of the ins and outs of trigonometry when they first meet it. I would very much agree with him in that respect, I know many people who have taught trigonometry for decades and still don’t, but I don’t think that means we have to bypass all information.

Reference List:

Cavadino, S.R. 2014. Formula Triangles. 12th October. Cavmaths. [online] accessed 14th July 2017. available: https://cavmaths.wordpress.com/2014/10/12/formula-triangles/

Cavadino, S.R. 2016. Catchy Mnemonics. 16th September. Cavmaths. [Online] accessed 14th July. Available: https://cavmaths.wordpress.com/2016/09/16/catchy-mnemonics/

William, R. 2017. The understanding paradox. 7th July. No easy answers. [online] accessed 14th July 2017. available: https://noeasyanswerseducation.wordpress.com/2017/07/07/the-understanding-paradox/

 

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A lovely old problem

July 11, 2016 Leave a comment

Recently Ed Southall shared this problem from 1976:

I’m not entirely sure if it is from an A level or and O level paper. It covers topics that currently sit on the A level, but I think calculus was on the O level at some point. Edit: it’s O level I saw the question and couldn’t help but have a try at it.

First, I drew the diagram – of course:

I have the coordinates of P, and hence N so I needed to work out the coordinates of Q.  To do this I differentiated to get the gradient of a tangent and followed to get the gradient of a tangent at P.

Next I found the equation, and hence the X intercept.

And then, because I’m am idiot, I decided to work out the Y coordinate I already knew and had used!

The word in brackets is duh…..

Now I had all three point.

It was a simple division to find the tangent ratio of the angle.

The next 2 parts were trivial:

And then I misread the question and assumed I’d been asked to find the shaded region (actually part d). 

Because I decided calculators were probably not widely available in 1976 I did it without one:

I thought it was quite a lot of complicated simplifying, but then I saw part c and the nice answer it gives:

Which makes the simplifying in part d simpler:

I thought this was a lovely question and I found it enjoyable to do. It tests a number of skills together and although it is scaffolded it still requires a little bit of thinking. I hope to see some nice big questions like this on the new specification.

Edit: The front cover of the paper:

Is one solution more elegant?

May 14, 2016 13 comments

Earlier this week I wrote this post on mathematical elegance and whether or not it should have marks awarded to it in A level examinations, then bizarrely the next day in my GCSE class I came across a question that could be answered many ways. In fact it was answered in a few ways by my own students.

Here’s the question – it’s from the November Edexcel Non-calculator higher paper:

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I like this question, and am going to look at the two ways students attempted it and a third way I think I would have gone for. Before you read in I’d love it if you have a think about how you would go about it and let me know.

Method 1

Before I go into this method I should state that the students weren’t working through the paper, they were completing some booklets I’d made based on questions taken from towards the end of recent exam papers q’s I wanted them to get some practice working on the harder stuff but still be coming at the quite cold (ie not “here’s a booklet on sine and cosine rule,  here’s one on vectors,” etc). As these books were mixed the students had calculators and this student hadn’t noticed it was marked up as a non calculator question.

He handed me his worked and asked to check he’d got it right.  I looked, first he’d used the equation to find points A (3,0) and D (0,6) by subbing 0 in for y and x respectively. He then used right angled triangle trigonometry to work out the angle OAD, then worked out OAP from 90 – OAD and used trig again to work out OP to be 1.5, thus getting the correct answer of 7.5. I didn’t think about the question too much and I didn’t notice that it was marked as non-calculator either. I just followed his working, saw that it was all correct and all followed itself fine and told him he’d got the correct answer.

Method 2

Literally 2 minutes later another student handed me her working for the same question and asked if it was right, I looked and it was full of algebra. As I looked I had the trigonometry based solution in my head so starter to say “No” but then saw she had the right answer so said “Hang on, maybe”.

I read the question fully then looked at her working. She had recognised D as the y intercept of the equation so written (0,6) for that point then had found A by subbing y=0 in to get (3,0). Next she had used the fact that the product of two perpendicular gradients is -1 to work put the gradient of the line through P and A is 1/2.

She then used y = x/2 + c and point A (3,0) to calculate c to be -1/2, which she recognised as the Y intercept, hence finding 5he point P (0,-1.5) it then followed that the answer was 7.5.

A lovely neat solution I thought, and it got me thinking as to which way was more elegant, and if marks for style would be awarded differently. I also thought about which way I would do it.

Method 3

I’m fairly sure that if I was looking at this for the first time I would have initially thought “Trigonometry”, then realised that I can essential bypass the trigonometry bit using similar triangles. As the axes are perpendicular and PAD is a right angle we can deduce that ODA = OAP and OPA = OAD. This gives us two similar triangles.

Using the equation as in both methods above we get the lengths OD = 6 and OA = 3. The length OD in triangle OAD corresponds to the OA in OAP, and OD on OAD corresponds to OP, this means that OP must be half of OA (as OA is half of OD) and is as such 1.5. Thus the length PD is 7.5.

Method 4

This question had me intrigued, so i considered other avenues and came up with Pythagoras’s Theorem.

Obviously AD^2 = 6^2 + 3^2 = 45 (from the top triangle). Then AP^2 = 3^2 + x^2 (where x = OP). And PD = 6 + x so we get:

(6 + x)^2 = 45 + 9 + x^2

x^2 + 12x + 36 = 54 + x^2

12x = 18

x = 1.5

Leading to a final answer of 7.5 again.

Another nice solution. I don’t know which I like best, to be honest. When I looked at the rest of the class’s work it appears that Pythagoras’s Theorem was the method that was most popular, followed by trigonometry then similar triangles. No other student had used the perpendicular gradients method.

I thought it might be interesting to check the mark scheme:

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All three methods were there (obviously the trig method was missed due to it being a non calculator paper). I wondered if the ordering of the mark scheme suggested the preference of the exam board, and which solution they find more elegant. I love all the solutions, and although I think similar triangles is the way I’d go at it if OD not seen it, I think I prefer the perpendicular gradients method.

Did you consider this? Which way would you do the question? Which way would your students? Do you tuink one is more elegant? Do you think that matters? I’d love to know, and you can tell me in the comments or via social media!

Cross-posted to Betterqs here.

A lovely simple trigonometry puzzle

May 8, 2016 4 comments

Sometimes a puzzle can look complicated,  but be rather simple (see this geometry puzzle). I love puzzles like this and I particularly like to test them out on classes to try and build their problem solving ability.

Just now, I saw the following trig puzzle from brilliant.org and I love it! It’s amazing!

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Have you done it yet?

How long did it take you to spot it?

My initial thought was, it’s got three terms,  it’s bound to be a disguised quadratic that will factorise. A few seconds later I realised that it wasn’t. I saw the – sin^4 and suspected a difference of two squares but then a few seconds later it became clear.

If you haven’t spotted it yet, have a look at the expression rearranged:

Sin^6 + sin^4 cos^2 – sin^4

See it now? What if I rewrite it as:

Sin^4 sin^2 + sin^4 cos^2 – sin^4

I’m sure you have seen it now, but to be complete,  take the common factor of the first two terms:

Sin^4 (sin^2 + cos^2) – sin^4

Obviously sin^2 + cos^2 = 1, so we’re left with:

Sin^4 – sin^4 = 0

A lovely, satisfying, simple answer to a little brain teaser. Hope you liked it as much as I did.

Cross-posted to Betterqs here.

May 8, 2016 Leave a comment

Sometimes a puzzle can look complicated,  but be rather simple (see this geometry puzzle). I love puzzles like this and I particularly like to test them out on classes to try and build their problem solving ability.

Just now, I saw the following trig puzzle from brilliant.org and I love it! It’s amazing!

image

Have you done it yet?

How long did it take you to spot it?

My initial thought was, it’s got three terms,  it’s bound to be a disguised quadratic that will factorise. A few seconds later I realised that it wasn’t. I saw the – sin^4 and suspected a difference of two squares but then a few seconds later it became clear.

If you haven’t spotted it yet, have a look at the expression rearranged:

Sin^6 + sin^4 cos^2 – sin^4

See it now? What if I rewrite it as:

Sin^4 sin^2 + sin^4 cos^2 – sin^4

I’m sure you have seen it now, but to be complete,  take the common factor of the first two terms:

Sin^4 (sin^2 + cos^2) – sin^4

Obviously sin^2 + cos^2 = 1, so we’re left with:

Sin^4 – sin^4 = 0

A lovely, satisfying, simple answer to a little brain teaser. Hope you liked it as much as I did.

Accuracy with Trigonometry

April 11, 2016 Leave a comment

This post was originally posted here on Cavmaths and here on BetterQs, on 5th March 2016, however the original post somehow got deleted so I’m re posting it.

This week I was planning to cover upper and lower bounds with year 11 as on the last mock a lot of them made mistakes so I felt it would be a good topic to revise. As part of the planning process I had a look through the higher textbooks our department has bought for the new specification GCSE (we bought the Pearson ones, the full suite at KS3 and 4. Some great questions in them and the online version, activeteach, is great to take questions and place into your lessons. I’d definitely recommend it, if used correctly, but I will admit to being disappointed to see a formula triangle being advised…) to see if there were any good questions I could pilfer, and I came across the section on using upper and lower bounds in trigonometry.

My first thought was, “that’s a nice topic”, and then the full spectrum of the topic began to unfold.

Initially, I had like the idea that students would be required to think about the fraction, and how minimising the denominator actually maximises it, but thin I remembered the nature of the cosine function! This example shows what excited:

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Not only would students be required to understand the nature of a fraction, they’d also need a deep understanding of the cosine function itself, to understand that the bigger cos x is, the smaller x is, and vice versa  (where x is between 0 and 90 of course). This could be a real deep understanding of the graph, or the unit circle, or just the geometry of a right angled triangle.

The example itself is very procedural based, which is a shame,  but it does give a teacher a good frame to start discussions. I wouldn’t use textbook example as teaching anyhow, just as an additional example to talk through one on one with students who were still struggling.

The textbook goes on to pose this awesome discussion question:

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A real nice prompt to get an in depth discussion around the trig ratios going. I often use similar prompts when looking at maximum values for sine and cosine “what’s the biggest opp/hyp can ever be?” for example. This often gives a nice discussion focus.

I think that this topic shows how different the new specification will be. Students are going to need a much deeper relational understanding if they are to achieve the top grades with questions like this being posed.

What do you think of bounds being questioned in relation trigonometry? Have you used prompts like this before? How have you found them?

Hippocrates’s First Theorem

April 11, 2016 5 comments

Over the half term I was doing some reading for my MA and I happened across Hippocrates’s First Theorem. (Not THAT Hippocrates, THIS Hippocrates!)

Here is the mention in the book I was reading (Simmons 1993):

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It’s not a theorem I’d ever come across before, and it doesn’t seem to have any real applications, however it is still a nice theorem and it made me wonder why it worked, so I set about trying to prove it.

First I drew a diagram and assigned an arbitrary value to the hypotenuse of triangle A.

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I selected 2x, as I figured it would be easier than x later when looking at sectors.

I then decided to work out the area of half of A.

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A nice start – splitting A into two smaller right angled isosceles triangles made it nice and easy.

I then considered the area b. And that to find it I’d need to work out the area the book had shaded, I called this C.

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Then the area of B was just the area of a semi circle with the area of C subtracted from it:

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Which worked out as the area of the triangle (ie half the area of A) as required.

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This made me wonder if it worked for all triangles that are inscribed in semi circles this way – ie the areas of the semicircles on the short legs that fall outside the semicircle on the longest side equal the area of the triangle.

My first thought was that for all three vertices to sit on the edge of a semi circle in this was then the triangle must be right angled (via Thales’s Theorem).

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I called the length eg  (ie the diameter of the large semi circle and the hypotenuse of efg) x and used right angled triangle trigonometry to get expressions for the two shorter sides ef and fg. Then I found the area of the triangle:

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Then the semi circles:
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I then considered the diagram, to see where to go next:

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I could see that the shaded area needed to be found next, and that this was the area left when you subtract the triangle from the semicircle.
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I could now subtract this from the two semi circles to see if it did equal the triangle.

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Which it did. A lovely theorem that I enjoyed playing around with and proving.

I think there could be a use for this when discussing proof with classes, it’s obviously not on the curriculum, but it could add a nice bit of enrichment.

Have you come across the theorem before? Do you like it? Can you see a benefit of using it to enrich the curriculum?

Reference:

Simmons M, 1993, The Effective Teaching of Mathematics, Longman: Harlow

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