Teaching New Dogs Old Trigs

I’ll get back to the anti-festive posts in the near future: there’s plenty of whacking on the to-do-soon list. But for now, while still digesting the Christmas feast as well as recovering from a cold or the plague or whatever, here’s a quick puzzle post.

Recently, I wrote on a 1903 Victorian textbook, and on the snarky comments Ken Clements made about the textbook in his history of school mathematics. I’ve continued hunting for related papers by Clements, and I recently tracked down a paper he had published in the 1975 MAV Proceedings.  Clements’ paper is on the 1904 Royal Commission into the University of Melbourne, and the subsequent reform of senior school exams. The paper is interesting enough, at least for me, but not so interesting as to warrant a post.* Clements’ paper, does however, provide a pleasingly annoying double-puzzle.

In writing about the turn of the century exams, Clements notes that the trigonometry questions on the Geometry and Trigonometry honours exam were “invariably difficult”. Clements elaborates on this point in a footnote:

The five questions on Trigonometry on the December 1902 paper, for example, would have all been too difficult, in the opinion of the writer, for the vast majority of form 6 [i.e. Year 12] mathematics students in Victoria in the 1970s. 

Clements then reproduces in the footnote what he regards as the easiest trig question on the exam (Q8). Here it is:

\color{blue}\boldsymbol{\aligned  &\mbox{\bf \textit{Simplify}}\\[3\jot]&\mbox{\bf \bf{(a) }} \qquad \frac{\sin \theta \,\sin2\theta \, + \, \sin 3\theta\, \sin 6\theta \, + \,\sin 4\theta\, \sin 13\theta}{\sin \theta\, \cos2\theta \, +\, \sin 3\theta \,\cos 6\theta \, + \,\sin 4\theta\, \cos 13\theta} \\[3\jot] &\mbox{\bf \bf{(b) }}\qquad \cot \tfrac{\pi + B}4 \,\cot \tfrac{\pi + C}4 \ + \ \cot \tfrac{\pi + C}4\, \cot \tfrac{\pi + A}4 \ +\ \cot \tfrac{\pi + A}4\, \cot \tfrac{\pi + B}4\\[3\jot] &\quad \mbox{\bf \textit{where }} A + B + C = \pi. \endaligned}

Yes, that was my reaction as well.**

Of course a major reason why 1970s students (and, much more so, 2020s students) would have had little hope of solving such problems is because they were less familiar with relevant trig identities and were unpractised in their application. But of course there’s also much more to it than simply “the curriculum has changed”.

In any case, feel free to give the above problems a go, and to discuss related issues. To begin, perhaps it’s preferable to not suggest full solutions in the comments, although noting relevant formulas and providing similar hints is fine.*** Have fun.


*) Interested readers can try to talk me into it, but I’m not sure how much luck you’ll have.

**) I know something of tackling such problems, but haven’t yet given these ones a go.

***) TeX preferred. Also, please do not suggest further problems in the comments, unless the problems are properly related in form or in methods of solution. I’m happy to, and plan to, write more such posts, on exam problems requiring now-forgotten techniques. But I’d like to avoid it becoming tangled as one big old-stuff mess, and I don’t want this post so derailed.


UPDATE (31/12/23)

Here is a method for the simplification of (a), as suggested by various commenters. First, to attack the numerator, we begin with the “product to sum” identity

\color{RawSienna}\boldsymbol{(1)\qquad2\sin A \, \sin B = \cos(A-B) - \cos(A+B)\,.}

(This identity is easily established by applying the more familiar identities to the RHS.)

Applying (1) to each of the products in the numerator (\boldsymbol{A =2\theta, B=\theta}, etc), and then cancelling, we obtain

    \[\color{Plum} \boldsymbol{\mbox{\bf NUMERATOR} = \cos\theta - \cos17\theta}\]

(Note that the multiples of \boldsymbol{\theta} in the exam question have been chosen exactly so that this cancelling will occur.)

Similarly, applying the identity

\color{RawSienna}\boldsymbol{(2)\qquad 2\sin B\, \cos A = \sin(A+B) - \sin(A-B)\,,}

we obtain

    \[\color{Plum}\qquad\boldsymbol{\mbox{\bf DENOMINATOR} = \sin17\theta - \sin\theta\,.}\]

To go further, we need companion identities to (1) and (2), the “sum to product” identities:

\color{RawSienna}\boldsymbol{(3)\qquad \cos D - \cos C = 2\sin\dfrac{C+D}{2}\,\sin\dfrac{C-D}{2} \,,}

\color{RawSienna}\boldsymbol{(4)\qquad \sin C - \sin D = 2\sin\dfrac{C-D}{2}\,\cos\dfrac{C+D}{2} \,.}

(These follow immediately from (1) and (2), by setting \boldsymbol{A=\frac{C+D}2} and \boldsymbol{B=\frac{C-D}2}.)

Applying (3) to the numerator and (4) to the denominator (\boldsymbol{C=17\theta, D =\theta}), we finally obtain

    \[\color{Plum}\qquad\boldsymbol{\mbox{\bf FRACTION} = \frac{\cos\theta - \cos17\theta}{\sin17\theta - \sin\theta} = \frac{2\sin9\theta\,\sin8\theta}{2\sin8\theta\,\cos9\theta}=\frac{\sin9\theta}{\cos9\theta}=\tan9\theta\,.}\]

Note that the identities (1)-(4) would have been very familiar to 1902 students, and to students long after. I’ve phrased and applied these identities so as to duck the minus signs as much as possible. Without such ducking, one would need to apply the symmetries \boldsymbol{\sin (-\theta) = -\sin\theta} and \boldsymbol{\cos (-\theta) = \cos\theta} at various stages.


UPDATE (31/12/23)

Part (b) is a little more confusing to discuss, since it is not clear which “triangle identities” (identities assuming \boldsymbol{A+B+C=\pi}) the 1902 students would have been expected to know and/or be permitted to assume on the exam. Here, I’ll give a direct simplification of (b) (which is what first occurred to me). In a further update I’ll look at the triangle identity approach.

For ease of writing, let \boldsymbol{\alpha=\frac{\pi +A}{4}}, and similarly define \boldsymbol{\beta} and \boldsymbol{\gamma}. Notice that the assumption \boldsymbol{A+B+C=\pi} implies that also \boldsymbol{\alpha+\beta+\gamma=\pi}. Then writing our mess with a common denominator, we have

    \[\color{Plum}\boldsymbol{\aligned\mbox{\bf MESS} &= \frac{\sin\alpha\,\cos\beta\,\cos\gamma + \sin\beta\,\cos\alpha\,\cos\gamma + \sin\gamma\,\cos\alpha\,\cos\beta}{\sin\alpha\, \sin\beta \,\sin\gamma}\\[3\jot]&= \frac{\sin(\alpha+\beta)\,\cos\gamma +  \sin\gamma\,\cos\alpha\,\cos\beta}{\sin\alpha \,\sin\beta\, \sin\gamma\,.}\endaligned}\]

Subtracting and adding \boldsymbol{\sin\gamma\,\sin\alpha\,\sin\beta} to the numerator, and doing the compound angle thing some more, we have

    \[\color{Plum}\boldsymbol{\aligned \mbox{\bf MESS} &= \frac{\sin(\alpha+\beta)\,\cos\gamma + \cos(\alpha+\beta)\,\sin\gamma + \sin\alpha \,\sin\beta\, \sin\gamma}{\sin\alpha\, \sin\beta\, \sin\gamma}\\[3\jot] &= \frac{\sin(\alpha+\beta+\gamma)+\sin\alpha \,\sin\beta \,\sin\gamma}{\sin\alpha \,\sin\beta\, \sin\gamma}\\[3\jot] &=1\,,\endaligned} }\]

where the last line follows from \boldsymbol{\sin(\alpha+\beta+\gamma)=\sin(\pi) = 0}.


UPDATE (31/12/23)

Here is a second method of simplifying (b), perhaps more along the lines of what would have been expected of the 1902 students. Setting \boldsymbol{\alpha=\frac{\pi +A}{4}} and so on, as above, and recalling \boldsymbol{\alpha+\beta+\gamma=\pi}, we begin with the identity

\color{RawSienna}\boldsymbol{(5)\qquad \tan\alpha + \tan\beta + \tan\gamma= \tan\alpha \, \tan\beta\, \tan\gamma \,.}

(I’ll give a proof this below.) Then, dividing both sides by \boldsymbol{\tan\alpha  \,\tan\beta\, \tan\gamma}, and using \boldsymbol {\cot = 1/\tan}, we have

    \[\color{Plum}\boldsymbol{\qquad \cot\beta\,\cot\gamma + \cot\alpha\,\cot\gamma + \cot\alpha\,\cot\beta= 1 \,.}\]

This is exactly the simplification we were looking for.

It is unclear which triangle identities, if any, could be assumed in the 1902 exam. In any case, (5) is easy to prove. Noting \boldsymbol{\tan(\alpha+\beta+\gamma) = \tan(\pi) = 0}, and doing the compound angle thing, we have

    \[\color{Plum}\boldsymbol{\qquad 0 = \frac{\tan(\alpha+\beta)+\tan\gamma}{\mbox{\bf DON'T CARE}} \,.}\]

Ignoring the denominator and doing  the compound angle thing again, we have

    \[\color{Plum}\boldsymbol{\qquad 0 = \frac{\tan\alpha+\tan\beta}{1-\tan\alpha\,\tan\beta}+\tan\gamma} \,.}\]

Multiplying through by the denominator and tidying up gives (5).

103 Replies to “Teaching New Dogs Old Trigs”

  1. I do think the class of 2023 would have fared better than the class of 2022 were this on a Specialist paper.

    The sum-to-product (and reverse) formulas have (re)appeared in Cambridge texts.

    Of course, that does not always mean much.

      1. I like the second part more than the first, and I suspect that I’m missing a shortcut somewhere (as seems to happen in trig proofs involving tan more than sin, cos).

        I do wonder now why these algebraic techniques fell out of favor with exam setters. Given the recent discussion about Complex Numbers, I feel trigonometry very relevant.

        And all this happened before the CAS was dropped on the unsuspecting (or am I being too generous with my word choice there?)

        1. Exams of yore asked students to solve equations of the form \displaystyle \sin(A) = \sin(B) etc. Simple applications of the complementary angle formulae (or just symmetry of the unit circle). Such theory remains in Maths Methods, such questions do not. You can add that to your wonder list.

          (For fun, you can watch what happens when a typically taught Specialist Maths student is asked to solve \displaystyle \sin(2x) = \cos(3x) over the domain \displaystyle 0 < x < \pi, say. You can even let them use a CAS once ‘by hand’ proves a bridge to far. A little knowledge is a dangerous thing).

            1. I wouldn’t worry. If the exam setters have been listening to this blog there’s no evidence that they’ve comprehended anything.

        2. RF, if you’ve done either/both, feel free to indicate the key identities or ideas or tricks, so others wondering where the hell to start might see where the hell to start.

          1. The first one has the (not quite so) obvious “rearrange to make it look like a known identity” approach, but I’m playing around with a few other ideas as well that are all (but one) about multiplying by \frac{thing}{thing} and seeing if the expansion produces known identities in both numerator and denominator that can then be cancelled.

            Part of me hopes that there is a really nice trick like this hiding in the problem. The sum-to-product formulas just feel a bit… gut-slog approach. Nothing wrong with this, of course, I just get a real buzz when there is a neat trick to be found.

        3. RF, for the second one the short cut you’re intuiting might be:

          Consider a triangle with interior angles \displaystyle \alpha, \displaystyle \beta and \displaystyle \gamma and prove \displaystyle \cot(\beta) \cot(\gamma) + \cot(\gamma) \cot(\alpha) + \cot(\alpha) \cot(\beta)= 1.

          The connection to the question is \displaystyle \alpha = \frac{\pi + A}{4} etc.

          For the first one, (no short cut here that I can see off-hand) I assume you multiplied numerator and denominator by 2 and then repeatedly used

          \displaystyle  2 \sin\left ( \frac{A-B}{2}\right) \cos\left ( \frac{A+B}{2}\right) = \sin(A) - \sin(B) …. (1)

          from left to right on the denominator, and

          \displaystyle  2 \sin\left ( \frac{A+B}{2}\right) \sin\left ( \frac{A-B}{2}\right) = \cos(B) - \cos(A) …. (2)

          from left to right on the numerator,

          simplified by cancelling terms, and then used (1) and (2) from right to left on the result.

          I’ve rushed both (in between doing several other things) but I think I’ve used the right ideas.

          1. That was basically my approach the first time, although with problem 2 I did think of using the triple identity for tan:

            tan(A)+tan(B)+tan(C)=tan(A)tan(B)tan(C) whenever A+B+C=\pi

            But couldn’t find a way to neatly work the algebra.

          2. Once either/both of you have what you regard as a nice solution to either/both (a) and (b), I’m happy for you to give details. I haven’t tried to work through (a), but it wouldn’t surprise me if the intended/best solution required a fair degree of computation.

            1. (Please delete if this is not what you meant)

              Part (a) I can simplify to \frac{cos(17x)-cos(x)}{sin(x)-sin(17x)} and then I stopped for the day.

              simplify is a loaded phrase though, because there are varying degrees of simple.

              1. Yes, and then identities (1) and (2) (see my earlier comment) are used from right to left to get products in numerator and denominator, cancel the common factor and you have what I assume is the writer’s intended answer.

                I think ST and RF have travelled the same road.

          3. I will prove

            \displaystyle \cot(\beta) \cot(\gamma) + \cot(\gamma) \cot(\alpha) + \cot(\alpha) \cot(\beta)= 1

            when \displaystyle \alpha + \beta + \gamma = \pi.

            The interested reader can satisfy for themselves that it trivially follows that (b) simplifies to 1.

            (Thinking of a triangle helped me get started).

            \displaystyle \alpha + \beta + \gamma = \pi

            \displaystyle \Rightarrow  \alpha + \beta = \pi - \gamma

            \displaystyle \Rightarrow  \cot(\alpha + \beta) = \cot(\pi - \gamma) = -\cot(\gamma)

            \displaystyle \Rightarrow  \frac{\cot(\alpha) \cot(\beta) - 1}{\cot(\alpha) +  \cot(\beta)} = -\cot(\gamma)

            using the standard identity

            \displaystyle  \tan(\alpha + \beta)  =\frac{\tan(\alpha) + \tan(\beta)}{1 - \tan(\alpha) \tan(\beta)}

            and the definition \displaystyle \cot(\theta) = \frac{1}{\tan(\theta)}

            \displaystyle \Rightarrow  \cot(\alpha) \cot(\beta) - 1 =  -\cot(\gamma) \left( \cot(\alpha) +  \cot(\beta) \right)

            and some simple re-arranging yields the required identity.

            In keeping with the Christmas spirit I say to Clements: Bah humbug! (I hope the The Ghost of Mathematics Future (*) visits him).

            * Or maybe the Ghost of Mathematics Present is sufficient.

          4. Complete solution for (a):

            The following alternative identities to (1) and (2) (*) are initially more convenient:

            \displaystyle  2 \sin(\alpha) \cos(\beta) = \sin(\alpha + \beta) - \sin(\beta - \alpha) …. (1′)

            \displaystyle  2 \sin(\alpha) \sin(\beta) =  \cos(\beta - \alpha) - \cos(\alpha + \beta) …. (2′)

            * Substitute \displaystyle \frac{A-B}{2} = \alpha and \displaystyle \frac{A+B}{2} = \beta.

            Multiply the numerator and denominator of (a) by \displaystyle \frac{1}{2}:


            \displaystyle =  \frac{1}{2} \left( \sin(\theta) \sin(2 \theta) + \sin(3 \theta) \sin(6 \theta) + \sin(4 \theta) \sin(13 \theta) \right)

            Use identity (2′):

            \displaystyle =  \left (\cos(\theta) - \cos(3 \theta) \right) + \left (\cos(3 \theta) - \cos(9 \theta) \right) + \left (\cos(9 \theta) - \cos(17 \theta) \right)

            \displaystyle =  \cos(\theta) - \cos(17 \theta)

            (reminiscent of a ‘telescoping series’).


            \displaystyle = \frac{1}{2} \left( \sin(\theta) \cos(2 \theta) + \sin(3 \theta) \cos(6 \theta) + \sin(4 \theta) \cos(13 \theta) \right)

            Use identity (1′):

            \displaystyle =  \left (\sin(3 \theta) - \sin(\theta) \right) + \left (\sin(9 \theta) - \sin(3 \theta) \right) + \left (\sin(17 \theta) - \sin(9 \theta) \right)

            \displaystyle =  -\sin(\theta) + \sin(17 \theta) = \sin(17 \theta) - \sin(\theta).

            Then (a) becomes

            \displaystyle \frac{\cos( \theta) - \cos(17 \theta)}{\sin(17 \theta) - \sin(\theta)}.

            Now apply identities (1) and (2) from right to left:

            \displaystyle = \frac{2 \sin(9 \theta) \sin(-8 \theta)}{2 \sin(-8 \theta) \cos(9 \theta)}

            \displaystyle = \tan(9 \theta).

      2. It [reappearance of sum product formula in the textbook] means that it has reappeared in the study design. Not sure what the intent is though…

  2. A modern source of trigonometric identities and their proofs is Adrian Yeo (2007) “Trig or treat: An encyclopedia of trigonometric identity proofs”. (World Scientific). Problems are organised by degree of difficulty (Easy, Less-easy, Not-so-easy).

    1. Thanks Terry – I’ll look it up.

      Wikipedia has a rather nice collection too, but they are organised by function, not by difficulty.

      I remember loving these way back when…

      They took me forever and many, many scrunched up pages, but I loved them.

    2. Thanks for sharing this Terry – just had a flick through it now and it looks like a great resource for a high school teacher/student. It intentionally has nothing particularly deep and no TNDOT type trig problems!

      It starts with a simple description of all of the key basic functions and identities, then has “The Concordance of Trigonometric Identities” which is an ordered list that looks perfect for finding modern high school trig practice / challenges – all with page referenced proofs.

  3. It is often worth asking – and trying to answer – the unstated question: “What on earth is the point of such exercises?”
    (When learning one’s tables, or mastering fraction arithmetic, came under threat, many knew that they mattered, but could not – and often still cannot – quite say why, So it was hard to “stand firm”.)

    The starting point is Red Five’s “I loved these way back when” – which tells us something about the adolescent learning mind. But there is more to it than that.

    I suggest four aspects that are essential for everyone (to differing degrees):
    (i) the unreasonable effectiveness of algebraic manipulation (where a few simple “re-writing” principles – such as factorising and cancelling – pay massive dividends);
    (ii) the central importance of accuracy (and back-checking for errors as one goes along);
    (iii) the first hand experience of fluency/mastery, and the implicit discovery of what it makes possible for ordinary students;
    (iv) the gradual discovery of what (i), (ii) and (iii) can deliver when combined with a kind of optimism [“I have no idea what is going on here; but if I respect the structure/symmetry of what is given, and do the obvious things accurately, something magical often happens.”]

      1. If we accept Tony’s suggestions as essential, it seems worth pondering:
        a) whether or not i) to iv) are outcomes that current students can successfully achieve, and
        b) how/why things (analogous to this example, shared by Marty) come to change.

    1. The question what the point of these exercises is has been asked and answered by education scientists in the last 25 years. And the answer was entirely predictable.

      For what it’s worth, here’s a problem from the Russian DWI (additional entrance examination for top universities) in 2017 that took me a while to figure out: solve the equation sin(7x) + sin(6x) = sin(x). And solving means finding all solutions, of course without calculator. I struggled with that one for quite a while before I realized that the obvious approach actually works.

      1. What’s “the obvious approach”?
        I could only think of using identity (1) (see my much earlier comment), which is perhaps obvious in the context of the blog. But I first had to re-arrange into

        \displaystyle \sin(7x) - \sin(x) = \sin(-6x)

        (less obvious(?) unless you know exactly where you want to end up).

        Then apply identity (1) to LHS and apply double angle formula to RHS. Then factorise and solve to get all the solutions (not just the ones obvious by inspection).

        So I’ve probably used an elephant gun to kill a fly. What was your fly swatter?

        Footnote: Using the sum-to-product formulae is probably more obvious to students who have studied physics and met interference of waves. It certainly provides the ‘practical application’ that many people seem to demand of the mathematics they study (or teach).

        1. Here’s my elephant gun:

          \displaystyle \sin(7x) - \sin(x) = -\sin(6x)

          (it is left to the interested reader to trivially ponder why this re-arrangement is necessary since there is a sum-to-product identity for \displaystyle \sin(A) + \sin(B))

          \displaystyle \Rightarrow 2\sin(3x) \cos(4x) = -2\sin(3x) \cos(3x)

          using identity (1) (see earlier comment) and the double angle formula

          \displaystyle \Rightarrow \sin(3x) \left( \cos(4x) +\cos(3x) \right)= 0.

          Case 1: \displaystyle \sin(3x) = 0 \Rightarrow x = \frac{n \pi}{3}, \displaystyle n \in Z.

          Case 2: \displaystyle \cos(4x) = -\cos(3x) = \cos(3x + \pi).

          Use symmetry properties:

          Case 2(a):

          \displaystyle 4x = (3x + \pi) + 2m \pi, \displaystyle m \in Z

          \displaystyle \Rightarrow x = (2m + 1) \pi.

          The solutions are a subset of Case 1 solutions (\displaystyle n = 3(2m + 1)).

          Case 2(b):

          \displaystyle 4x = -(3x + \pi) + 2m \pi, \displaystyle m \in Z

          \displaystyle \Rightarrow 7x = (2m - 1) \pi

          \displaystyle \Rightarrow x =\frac{ (2m - 1) \pi}{7}.

          1. Note: Both Case 2’s

            \displaystyle \cos(4x) + \cos(3x) = 0

            \displaystyle \displaystyle \sin\left( \frac{13 x}{2} \right) - \sin\left( \frac{x}{2} \right) = 0

            can be solved without using symmetry. The sum-to-product formulas are just used again.

            For the former, use

            \displaystyle \cos(A) + \cos(B) = 2 \cos\left( \frac{A+B}{2} \right) \cos\left( \frac{A-B}{2} \right) …. (4)

        2. Can you re-write the LHS using a Chebyshev polynomial, factor out the sin(x) and then use the null factor law?

          EDIT: Scrap that idea. Taking sin(x) from both sides first seems to work better.

          1. Yes you can. This is clearly the obvious approach FL was alluding to (the clue is the Russian context. So obvious, I don’t know how I missed it). You have to use Chebyshev polynomials of the second kind. Then:

            \displaystyle U_6(t) \sin(x) + U_5(t) \sin(x) - \sin(x) = 0

            where \displaystyle t = \cos(x)

            \displaystyle \Rightarrow \sin(x) \left( U_6(t) + U_5(t) - 1\right)  = 0.

            Case 1: \displaystyle \sin(x) = 0 \Rightarrow x = n \pi, \displaystyle n \in Z.

            Case 2:

            \displaystyle U_6(t) + U_5(t) - 1 = 0

            \displaystyle \Rightarrow 32t^6 + 16t^5 - 40t^4 - 16t^3 + 12 t^2 + 3t - 1 = 0

            using the usual recurrence relation definition of a Chebyshev polynomial of the second kind (remember, it has to be done ‘by hand’).

            Case 2(a): By inspection and the rational root theorem I see the following solutions:

            (i) \displaystyle t = -1 \Rightarrow \cos(x) = -1

            and the solutions are obviously a subset of the Case 1 solutions.

            (ii) \displaystyle t = -\frac{1}{2} \Rightarrow \cos(x) = -\frac{1}{2} \Rightarrow

            (iii) \displaystyle t = \frac{1}{2} \Rightarrow \cos(x) = \frac{1}{2} \Rightarrow

            The solutions from (ii) and (iii) can be written as \displaystyle x = \frac{n \pi}{3}, \displaystyle n \in Z.

            (The Case 1 solutions and the solutions from (i) are a subset of these solutions).

            Case 2(b): Using the roots from Case 2(a) you can construct a cubic factor of \displaystyle 32t^6 + 16t^5 - 40t^4 - 16t^3 + 12 t^2 + 3t - 1 and hence factorise it as a product of two cubics (remember, it has to be done ‘by hand’). So you’re left with solving

            \displaystyle 8t^3 - 4t^2 - 4t + 1 = 0 …. (A)

            Equation (A) has no rational solutions (they were all found in Case 2(a)) so you must solve it using the usual cubic formula. There are three real solutions (this follows from Rolle’s Theorem and the Intermediate Value Theorem). Then you equate these solutions to \displaystyle \cos(x) and solve for x. I sense from the EDIT in your comment that you’ve done all the calculations ‘by hand’ without too much trouble. The interested reader is invited to fill in the trivial details.

                1. FL’s Russian equation can be modified so that (without the losing sum-to-product methods) if you prefer using the obvious method of Chebyshev polynomials, needing the cubic formula is obviated (you still need the rational root theorem but you get all the roots):

                  \displaystyle \sin(7x) + \sin(3x) = \sin(x).

                  Case 1: \displaystyle \sin(x) = 0.

                  Case 2: \displaystyle 64t^6 - 80t^4 + 28t^2 - 3 = 0
                  (where \displaystyle t = \cos(x)).

                    1. You’re welcome.
                      I’ve obviously had every single one of these thoughts all at once but just wanted to get my comment count up by drip feeding them.

                      I’m not sitting here spending hours and hours figuring all this stuff simply to post one final clean epic comment. I was commenting as I write this stuff up for a SAC (whose final form will be weeks in the making). I thought it would be good to post my on-going thoughts as part of a potential groupthink effort and a repository of ideas that each reader could pursue and perhaps even post their own thoughts on. Clearly a fanciful notion.

                      You’re right, it is better to figure this stuff out cleanly and not just post comment after comment.

  4. A book of problems by Mark Skanavi full of similar questions. Looking at this problem sends me back to my childhood. School didn’t offer such questions, but those who worked with Skanavi book, trying to achieve more than school can offer, definitely would have a dejavu seeing this problem. Those who are interested can easily find pdf of Skanavi book.

  5. One way to think about these:

    What is the thing(s) you have got? (Describe it, is it a sum, a product, a …)
    What is the thing you want?
    What is the link? (What else do you notice?)

    The link is likely an identity (or two), a technique (or two), … , when selected well, and applied accurately, some magic takes place, where simplicity came out of complexity. (Thanks Tony …)

    Re part a)

    I have products being summed.
    I need something simple(r), given it is not a prove that …, and so maybe the products can become sums and some will cancel.

    Maybe the product to sums identities might help.
    Do the numbers work nicely? Yes they do. I see a cascade-y thing coming.
    And out pops something simple(r).

      1. Further to above, turning products into sums:


        Applying the same idea to each of the other products in the numerator provides some terms that reduce to zero by subtraction, leaving a difference, which can then be turned backed into a helpful product (using the reverse of the identity first applied).

        Attack the denominator in a similar manner and …

              1. Very.

                Do you think this level of puzzle should be in the scope of Year 12 students who study Specialist Mathematics?
                Or is the time it would take to prepare students for such better spent on …?

                1. If you got a function of a single argument, well done, your result is simpler than mine and (in my opinion) a superior result.

                  I went down the route of looking at the biggest angle, expanding, and seeing what could be cancelled. Something I used to call telescoping in my first year of undergraduate mathematics.

                  I’m not sure my result is valid or the intended answer, as the checking process has not occurred.

                  I mostly posted my (perhaps intermediate) result in the hope of provoking discussion which I see has happened, so all is good.

                  To answer ST’s question, I found the scope of this definitely OK for Specialist 3&4 (ditto NSW Ext 2 and IBHL) but doubt VCAA would ever ask such a question in the current climate.

              1. Nice. I didn’t think of going into negative angles, but it so nicely works.

                Really nice problem, as much for it being an algebraic workout as anything else.

                1. Of course, it has been cooked to be nice. Which takes skill. It plays the identities like an accordion.
                  The question is generic (or used to be). A challenge for the interested reader is to write a similar one for use in their 2024 assessment (or as a gift to a colleague for \displaystyle their assessment).
                  (Maybe 2024 is the year that VCAA exam setters listen to this blog and comprehend something. Anything).

                  1. As if VCAA’s/MAV’s/etc’s “open-ended” swill isn’t cooked. The only difference is that they’re lousy cooks.

  6. A good exam question is one that tests routines rather than requiring ingenuity (though part (b) can go a little further).

    Here “sinA.sinB” should have a standard response: “multiply top and bottom by “2”, and then replace each term in the numerator by “-[cos (A+B) – cos(A-B)]”, which leads to lots of cancellation.
    Then do the obvious with each “2.sinA.cosB” in the denominator, with more cancellation.

    Then do the reverse with numerator (cosX – cosY) and denominator (-sinX + sinY) to get the final “tan[(X+Y)/2]”.

    1. Thanks, Tony. Yep, there’s really only one thing to try (and at least one thing to try if you know the identities), and then you work carefully and cross your fingers.

      I think when I was learning trig in Victoria in the mid 70s, this kind of problem was just dying out. I had a teacher who would give us a list of such identities to prove as a challenge, but I don’t remember anything as involved on exams. The product-to-sum identities for (a) were expected to be known and used, but I think the applications were always more straight-forward.

      1. I do wonder… (being serious, in a way) if the fact that these questions do not have the end-point specified may allow them to be considered “open-ended” for the purposes of, say, a Specialist SAC…

        (Having recently been through an audit I seriously doubt it, unfortunately. VCAA’s definition of “open-ended” seems to differ greatly from my own).

          1. It depends on what simplify means. For (a) I strongly doubt that the writer had any answer other than \displaystyle \tan(9 \theta) in mind. I dislike the notion that multiple answers such as RF’s would be acceptable (sorry RF). I also dislike like the idea of specifying the \displaystyle form of the final answer in order to ‘define’ the intent of ‘simplify’.

            A work-around, which I also dislike, is to have a second part along the lines of “Hence solve …… = 0”

            which would effectively force the student to continue from RF’s answer to \displaystyle \tan(9 \theta).

            Given that the intended answer is actually \displaystyle \tan(9 \theta), how do you remove the ‘open-endedness’? Apart from already having an established convention or precedent for the meaning of “simplify” (an exam report could do this, but that only helps \displaystyle after the first time the question appears and can get forgotten over time unless that type of question appears more regularly than Halley’s comet).

            1. I’m getting lost. RF wants (a) to be open-ended, and I’m querying how it could possibly be thought of as anything else. I don’t think anyone is advocating that (a) be thought of or rejigged to not be open-ended.

              1. What did you mean by

                “That is a really excellent point. How could one possibly argue that (a) and (b) are anything other than open-ended?”

                in your earlier comment?

                1. 1) It is an excellent point that very good problems such as the trig problems here *should* satisfy VCAA’s “open-ended” fetish.

                  2) As does RF, I can imagine that VCAA *would* object, but I cannot imagine the substance of any such argument for any such objection.

                  1. OK. Not trolling:
                    How is (a) open-ended? Is it in the meaning of “Simplify”? (See earlier comment). What do \displaystyle you mean by “open-ended”?

                    VCAA would object on the basis of \displaystyle it's definition of open-ended (which can be found somewhere. Also, there is the VCE Mathematics 2023-2027 – Mathematical Investigations implementation advice).
                    It wants students to be mathematical explorers (*) in a name you own adventure. You get to choose your own values of n. You get to choose your own functions. You get to embark on a journey of exploration and discovery (but take care because you only have a few hours in this mathematical wonderland).

                    I quote from an old audit report shown to me:

                    “For example, in the Algebra and Calculus task, students could be asked to model the structure of the bridge using their own appropriate functions, rather than be given the general form for a quartic, and could then analyse each function as to their suitability.”

                    Neither (a) or (b) come remotely close to this (thank goodness). The roads to the answer are limited and they all lead to one answer.

                    * Usually Burke and Wills rather than Lewis and Clarke.

        1. Re: Audits.
          VCAA’s definition of mathematical correctness also seems to differ greatly from your own (and mine, and Marty, and nearly 70 university mathematicians in a signed open letter). We need to remember the ‘Amber Gambit’:

          You tell VCAA that all your procedures were followed and that you commissioned an independent external review which confirmed this and concluded that there were no errors on your SAC (but that some wording could be improved). You also offer to provide a summary of the review.

          1. If anyone from VCAA is reading this, please stop now or take what follows to be a work of pure fiction.

            I “failed” a SAC audit recently because my SAC 2 and 3 were not open-ended and/or balanced.

            I thought they were both, but there were other issues to worry about, so I went to the VCAA implementation videos and copied without editing a bunch of their questions and added them to my SACs.

            I then passed the audit.

            (VCAA if you are reading, none of this actually happened, OK…)

            Marty – I sent you a copy of the (for VCAA readers, non-existent) audit report but please do not quote from it. I may no longer work at the school in question, but feel some level of duty to protect those who still do.

            1. Again, this is not where to begin. The beginning is VCAA’s precise guidance on how a SAC is to be “open-ended and/or balanced”.

            2. If VCAA worried less about SAC questions not being open-ended and/or balanced and more about being mathematically correct on its exams, we would all be very much better off. Glass houses and stone throwing come to mind.

              RF, you’re leaving a lot to the imagination but I’ll have a lash. I assume in your work of fiction that:
              1) you failed Stage 1 (*) and were progressed to stage 2, where you were required to revise the SACs to meet the requirements of the study design and submit the revised tasks as further evidence, and
              2) further evidence should be submitted no later than Friday 20 October, and
              3) students had already sat your SACs (**) and were, in fact, on swot vac when you received the 2023 school-based assessment audit: unit outcome. So Stage 2 was simply serving a pointless, senseless bureaucratic purpose, and
              4) you were jesting that there were other issues to worry about (surely not report writing, internal exam writing and marking, helping students prepare for VCAA exams (***), teaching Yr 7 – 10 classes etc), and
              5) You checked that it was not actually April 1, there were no hidden cameras etc

              You must be attending the same fiction writing classes as other teachers I know, and are trying to get out of teaching and into a writing gig on Utopia. (It’s lucky you’re an experienced teacher rather than a recent graduate). Personally, I would applied the Amber Gambit.

              * there is a skill to writing when completing stage 1. You have to use the right words, the right phrases, the right language. It helps if you can embellish and exaggerate. And don’t ever feed them with assessments.

              ** which had also been marked and results submitted.

              *** which possibly included explaining various mistakes on past exams and reports that were confusing your students.

              1. BiB, I am more than happy to discuss this off-blog (we have met before, btw…). Marty has my details.

                But yes, I am deliberately leaving out a lot of details.

                Your guesses have been remarkably accurate though, save a few key (and in some ways, direction-altering) oversights.

  7. Thank you all for your comments so far. I’ve only been peeking at the discussion, so may I ask where we are?

    I gather that for (a) people agree on the natural approach and the natural final form.

    Is there agreement on (b)? (I’m not sure my first thought was the same as BiB’s.)

    For Franz’s bonus question, I gather BiB has a nice and natural approach. So it’s just a question of whether Fran’s “obvious” approach is the same?

    I’ll update the post soon, at least with a tidied version of people’s solution to (a).

    1. A few comments on the original question:

      1. I like the question. It is a good question and a really good exercise in algebra, accuracy and fundamental trigonometry.

      2. I think there are different levels of “simplify”, but the answer suggested by ST is superior to my own because it is more “simple”.

      3. There are a few different ways to approach the problem, some lead to longer working than others. This only enhances my enjoyment of the problem.

      4. I now wonder, what integer values of \theta can lead to similar, simple results and will likely spend a lot of the remainder of 2023 playing around with these numbers to see where it all goes.

      5. Unfortunately, I cannot imagine a way by which this can be made into an “acceptable” SAC. I think there is the possibility of something at SM1&2 level though, so I will ponder further.

      1. I would really like to pursue the question of 5, if anyone can be bothered to dig up VCAA’s damn SAC rules.

        You write that you “cannot imagine” that such questions are acceptable. I wouldn’t be surprised. But if not, I’d like to know exactly why the hell not.

        1. VCAA gives no definitive definition or statement. It’s definition seems almost self-referential. But you can get the general flavour if you’re able to read through stuff like what is here:

          Component 3 of old CATs provide many examples of what VCAA has in mind. I have attached an old (2004) VCAA Bulletin Supplement that gives further insight (I have others, and the old CATs can be found on-line somewhere).

          VCAA Bulletin Supplement 2 December 2004 – Starting Points

                1. RF said he thought (a) and (b) might be open-ended but did not say why. You gave total agreement but did not say why. I’ve said plenty in response. The way I see it, it’s the job of either you or RF to justify your original statements.

      2. There are many ways these sorts of questions can be incorporated into a SAC (note that not all questions on a SAC must be -open-ended’, some can be ‘closed’). But as posed they would not be considered by VCAA to be ‘open-ended’ questions. However, consider including the following ‘question on a SAC:

        Now construct your own question of this generic type and provide a detailed solution.

        By its own definitions, I don’t see how VCAA could NOT consider this as an ‘open-ended’ question. However, I would predict a number of students saying “I am just going outside and may be some time.”

        1. For fuck’s sake.

          “But as posed they would not be considered by VCAA to be ‘open-ended’ questions.”

          I keep asking you guys to justify this claim by proper reference to VCAA’s SAC guidance and you simply won’t do it.

          1. See my DECEMBER 28, 2023 AT 8:26 PM comment. That’s all I’ve got and it’s all that VCAA’s got – examples. Definition by examples.

            Closed: Find the value of -4 + 3. Open-ended: Find two numbers whose sum is -1. Now find two numbers whose sum is a.

            If that’s still unsatisfying, you should contact the VCAA and ask for a definition. Or ask the MAV, they tout their commercial SACs as containing open-ended questions (that I would assume meet VCAA requirements) so they must know the definition.

            PS – There’s a difference between “won’t” and “can’t”.

          2. VCAA’s SAC guidance is not helpful. As a teacher, the only way I know what is and what is not acceptable is by going through the audit process.

            I find this wholly unsatisfactory.

            There is a lot more to say, but this is not the forum.

            Short version: SAC writing requires a lot of guess-work and a lot of teachers do not have the time so modify a SAC that has already passed audit or was written by MAV.

            Writing a good SAC is difficult and in the end, probably not worth the effort.

  8. A quick reply to BiB, on “open-ended”.

    It doesn’t really matter for this discussion what “open-ended” means: what matters is whether VCAA would or would not accept (a) and (b) for a SAC, in the context of however VCAA characterises this required open-endedness. (And, if not, what are VCAA’s presumed arguments for this objection.)

    I’m not going to hunt through VCAA’s SACs guidance, but I also see no point in guessing games. What matters first is the precise wording of VCAA’s guidance. A “for example” of an audit report may be relevant later, but it is not where to begin.

    1. I would argue that this misses the most important point… VCAA’s SACs rules are crap, I think most readers here agree on that.

      Do questions like this still have a MEANINGFUL place in the Specialist classroom? I would hope so, but I’m not entirely sure. Does it depend on the school/class and their motivation(s) for taking the class? Could such a question POSSIBLY find its way onto a Paper 1, as a nice sting in the tail (as has been known to happen)?

      These are the questions which are perhaps more pressing to the SM3&4 teacher looking ahead to 2024.

      1. Jesus, RF.

        Of course VCAA’s SAC rules are crap. Of course such questions should still have a meaningful place in SM. Yes, these are much more important points, but no one here is likely to debate any of this.

        So let’s debate the SACs, which is still damn important. You brought this up. If you’re not simply whining then put some flesh on your accusation that VCAA would reject such questions on a SAC.

        1. I could, but it would be guessing.

          From speaking to other teachers, VCAA rejects SACs for two main reasons:

          1. Not open-ended enough.

          2. Not “balanced”.

          I believe, but have no evidence, that a SAC on trig identities would fail on point (2) because:

          (a) technology to “explore” is not going to do much.

          (b) there is not much “increasing complexity” which is a phrase VCAA documents have a few times in their SAC guidance.

          But still, guessing. Or maybe simply whining. Each to their own opinion on that.

  9. I should not be saying this (partly because I am not clear on VCAA’s misplaced demand for “open-ended” formulations, and partly because it sounds like thumping my own tub). However, this blog’s recent thread on “problem solving” had a piece of mine which included an analysis suggesting the limitations of “open-ended” as a goal, and focused more strongly on “open-middled”:

    “[open-ended extended tasks’] lack of predictability and control means that extended tasks are not an effective way for most pupils to learn the art of solving multi-step problems. For most teachers, this art is much more effectively addressed through short, easily stated problems in a specific domain (such as number, or counting, or algebra, or Euclidean geometry), where

    * what is given and what is required are both clear,
    * but the route from one to the other requires pupils to identify one or more intermediate stepping-stones (that is, they are “open-middled”)—as with
    – solving a simple number puzzle, or
    – interpreting and solving word problems, or
    – proving a slightly surprising algebraic identity, or
    – angle-chasing (where a more-or-less complicated figure is described and has to be drawn, with some angles given and some sides declared to be equal, and certain other angles are to be found—using the basic repertoire of angles on a straight line, vertically opposite angles, angles in a triangle, and base angles of an isosceles triangle), or [etc. ad nauseam]”

  10. OK, enough of this insane “open-ended SAC” discussion. But one thing I really can’t stand is etherial whining. That’s what the Heads of Maths is for.

    There is a serious point here, of what VCAA will permit as part of their exploratory SAC bullshit. I’ll put it on the list.

    1. Starting with tan 9\theta and trying to find an alternative starting point might be another option.

      The fact that this question was (likely) written by a Mathematician who worked at a university is likely a significant factor in it being an excellent question.

  11. Re: “The five questions on Trigonometry on the December 1902 [Geometry and Trigonometry honours exam] paper, for example, would have all been too difficult, in the opinion of the writer, for the vast majority of form 6 [i.e. Year 12] mathematics students in Victoria in the 1970s.”

    It looks to me that the paper did not contain a lot of what is in today’s syllabus: Calculus, Complex numbers etc. I think if teachers and students had a laser focus purely on Geometry and Trigonometry, students would have met many examples of these sorts of questions. They would be well drilled. This would render (a) and (b) and their ilk a lot more routine in 1902 than in 1972 (or indeed 2022). The historical context is important. Clements is comparing apples with oranges and his assertion must be taken with a large chunk of salt.

    Having said this, I think a decent, well taught and well-drilled student in Specialist Maths would handle (a). But I accept that they might find (b) tougher (even so, I think the stronger students would experience the appropriate epiphany of starting with A + B + C = pi).

    1. One gem in this post I missed at first was the link to the 1904 Royal Commission into University exams; I gave the document a skim-read after reading this post and I think there are some interesting points raised in the report (as well as an awful lot I didn’t understand).

      Basically the report said that Mathematics exams were getting repetitive and good schools such as Xavier (Wesley was also named) were learning how to game the system.

      Taken in that context, maybe this question was not so difficult for a student in 1902.

      The exams were driving the teaching in 1903 apparently.

      120 years later, CAS seems to be the main difference in that regard…

      1. Thanks, RF. Exams will always be “repetitive” and will always be “gamed”. The issue is, what is being repeated and how is that being gamed?

        I didn’t try to read the RC itself. Clements’ discussion of it was confusing and seemed to me untrustworthy. I’m also not sure I’d trust the RC. It seems pretty clear that everyone had an agenda, and the RC was guided by whichever agenda they bought in to. What does seem clear is that there was a mismatch between: (a) what students could reasonably achieve at the time, at either the elite schools or the (few) others; (b) what was examined at Matric; (c) what Unimelb wished or required from the incoming students. I think it’d take a lot of work to determine the nature and the extent of that mismatch, and the causes of it.

        1. Agreed. Difficult enough for the 2023 exams, impossible now (probably) for the 1903 papers.

          I still think parts of it have relevance today, I’m just not sure what that relevance is and whether it can be helpful.

          Thanks for bringing it to my attention either way.

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