TNDOT 2: Trigged Again

This is the second post in our new, Teaching New Dogs Old Trigs/Tricks series. This one, suggested by a reader, is another trig one, with the final part related to our first TNDOT post. It is from the 1913 Victorian Matric Trigonometry Honours Exam. Here it is:

\color{blue}\boldsymbol{\aligned  &\mbox{\bf \textit{Sum}}\\[3\jot]&\mbox{\bf \quad \bf{(i) }}  \sin(k+1)a\,\cos a \, + \, \sin(k+2)a\,\cos 2a \, + \, \cdots \, + \, \sin(k+n)a\,\cos na\, ;\\[3\jot] &\mbox{\bf \quad \bf{(ii) }} \tan x \,+\, \frac12\tan\frac{x}2 \, +\,\frac1{2^2}\tan\frac{x}{2^2}\,+\,\cdots \,+\, \frac1{2^n}\tan\frac{x}{2^n}\,.\\[3\jot] &\mbox{\bf \textit {\quad Express}} \\[3\jot] &\qquad \frac{\sin x -\sin 3x + \sin 5x \mbox{\bf \textit { to n terms}}}{\cos x -\cos 3x + \cos 5x \mbox{\bf \textit { to n terms}} }\\[3\jot] &\mbox{\bf \textit {as the cotangent of an angle.}} \endaligned}

 

UPDATE (25/01/24)

Here is a computation of (i), a minor rephrasing of Simon‘s (second) computation. For the computation, we need the “product to sum” and “sum to product” identities, which were also key to dealing with TNDOT 1:

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

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

\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} \,.}

(I’ve written (2) in a form so as to duck a few minus signs.)

To begin, we apply (2) to each term of the SUM in (i), with \boldsymbol{A = (k+j)a, B = ja} (and so \boldsymbol{A - B= ka, A + B = (k+2)ja}). This gives a bunch of \boldsymbol{\sin ka} terms, plus a simpler sum:

\color{Plum}\qquad\boldsymbol{\aligned 2(\mbox{\bf SUM}) &= n\sin ka + \left[\sin (k+2)a + \sin(k+4)a +\cdots + \sin (k+2n)a\right]\\[3\jot] &= n\sin ka + [\mbox{\bf SECOND SUM}]\,.\endaligned}

This SECOND SUM is more familiar, at least in the world of Fourier series. It can be evaluated using complex number tricks (see Simon’s first proof), or, with hindsight, by induction. A straightforward trig approach is to multiply by \boldsymbol{2\sin a}, expand and apply (1) to each term (with \boldsymbol{A = (k+2j)a, B = a}). This gives a sum that telescopes:

\color{Plum}\qquad\boldsymbol{\aligned&2[\mbox{\bf SECOND SUM}]\sin a\\[3\jot] = \ &[\cos (k\! +\!1)a \! -\! \cos (k\! +\! 3)a]\! +\! [\cos (k\! +\! 3)a \! - \! \cos (k\! +\! 5)a] \! +\! \cdots \! +\! [\cos (k\! +\! 2n\! -\! 1)a \! -\! \cos (k\! +\! 2n\! +\! 1)a]\\[3\jot] =\ & \cos (k +1)a - \cos (k + 2n + 1)a\,. \endaligned}

Now we can apply (3) to this final expression and divide, giving,

\color{Plum}\qquad\boldsymbol{\mbox{\bf SECOND SUM} = \dfrac{\sin(k+n+1)a \, \sin na}{\sin a}\,.}

Plugging back in, we finally get,

\color{Plum}\qquad\boldsymbol{\mbox{\bf SUM} = \dfrac {n\sin ka}2 +  \dfrac{\sin(k+n+1)a \, \sin na}{2\sin a}\,.}

 

UPDATE (31/01/24)

Here is a computation of (ii), again following Simon. This one is easy if you see the trick, but it doesn’t seems easy to see the trick. (I didn’t). Of course, students at the time were clearly practised on this kind of problem. But anyway, the two aspects to seeing the trick seem to be: first, understanding/praying that a telescoping sum must some be part of the solution; secondly, that we’ll have to use the double angle formula for tan:

\color{RawSienna}\boldsymbol{(5)\qquad\tan 2A = \dfrac{2\tan A}{1-\tan^2A}\,.}

Now, simply using (5) directly on (ii) just creates a mess. So, the big trick is to flip both sides, double and split up the RHS. This gives the double angle formula for cot (which was very possibly well known to students at the time):

\color{RawSienna}\boldsymbol{(6)\qquad2\cot 2A =\cot A - \tan A\,.}

Rearranging, we have

\color{Plum}\boldsymbol{\tan A=\cot A - 2\cot 2A\,,}

and that is exactly what we need to transform (ii) into a telescoping sum of cots. Setting \boldsymbol{A = \frac{x}{2^j}}, the SUM in (ii) becomes,

\color{Plum}\qquad\boldsymbol{\aligned\mbox{\bf SUM} &= (\cot x - 2\cot 2x) + \tfrac12(\cot \tfrac{x}{2} - 2\cot x)+\cdots + \tfrac1{2^n} (\cot \tfrac{x}{2^{n}} - 2\cot\tfrac{x}{2^{n-1}})\\[3\jot] &= -2\cot 2x + \frac1{2^n} \cot \frac{x}{2^{n}\,.}\endaligned }

 

UPDATE (02/02/24)

Here is the last “Express” bit, following Red Five and Simon. The trick is to multiply the top and bottom by \boldsymbol{2\cos x}. Then, for the NUMERATOR we can apply (2) to each term, giving,

\color{Plum}\qquad\boldsymbol{\aligned \mbox{\bf NUM} &= 2[\sin x -\sin 3x + \sin 5x+\cdots+(-1)^{n-1}\sin(2n\!-\!1)x]\cos x\\[3\jot] &=(\sin 2x +\sin 0x) - (\sin 4x +\sin 2x) +\cdots + (-1)^{n-1}(\sin 2nx +\sin (2n\!-\!1)x)\\[3\jot] &=(-1)^{n-1}\sin 2nx\endaligned}

The denominator can be handled similarly, using the third “product to sum” formula:

\color{RawSienna}\boldsymbol{(7)\qquad 2\cos A\, \cos B = \cos(A-B) + \cos(A+B)\,.}

Applying (7) to each term in the denominator, we get

\color{Plum}\qquad\boldsymbol{\aligned \mbox{\bf DENOM} &= 2[\cos x -\cos 3x + \cos 5x+\cdots+(-1)^{n-1}\cos(2n\!-\!1)x]\cos x\\[3\jot] &=(\cos 2x +\cos 0x) - (\cos 4x +\cos 2x) +\cdots + (-1)^{n-1}(\cos 2nx +\cos (2n\!-\!1)x)\\[3\jot] &= 1 +(-1)^{n-1}\cos 2nx\endaligned}

To finish, we apply the double angle formulas to \boldsymbol{\sin 2nx} and \boldsymbol{\cos 2nx}. The only wrinkle is, which formula we apply to \boldsymbol{\cos 2nx} depends upon whether n is even or odd (so that the 1 always gets cancelled out). Finally, we get,

\color{Plum}\boldsymbol{\mbox{\bf MESS} = \left\{\aligned &\frac{-2\sin nx\,\cos nx}{2\sin^2 nx} = -\frac{\cos nx}{\sin nx}=-\cot nx\qquad &&n \mbox{ \bf even,} \\[3\jot] &\frac{2\sin nx\,\cos nx}{2\cos^2 nx} =\frac{\sin nx}{\cos nx}=\tan nx =\frac1{\cot nx}\qquad&& n \mbox{ \bf odd.}\right.\endaligned}

29 Replies to “TNDOT 2: Trigged Again”

  1. It is rather curious that the final part of this specifies the COTANGENT of an angle when the presence of \sin{x} terms in the numerator and \cos{x} terms in the denominator would lead one, presumably, to assume the answer would be of the form \tan{something(x)}

  2. Spoiler alert: I *think* I have a solution (that is not too involved).

    If anyone wants to compare answers, I’m happy to post.

    I used the same four rules/formulas quoted in the previous TNDOT post plus some thinking about the graph of y=\cot{x}.

    Took me three attempts to get started (with #1 and #2 both going nowhere after a few lines of working, so abandoned in search of a better approach) but then seemed to all fall out nicely. Hopefully correctly!

    1. Hi RF, I suppose the answer would be cot(2x)?
      Just quickly worked through the last part and it was an interesting question.
      Will be curious to see how others approach this part.

      Still pondering the first two parts.

        1. Yep, works well.

          I took the (perhaps less efficient, but worked for me) approach of multiplying the numerator by \frac{\cos{x}}{\cos{x}} and the denominator by \frac{\sin{x}}{\sin{x}} and then applying the product-to-sum identities. The \cot{x} then dropped out nicely.

  3. These seem rather wonderful – and not at all easy. I’m not sure how I would fare.

    For the cotangent question, I agree with Sentinel that:
    when n=2, the simplified form cancels to cot(2x).
    But when n=1, the answer is clearly 1/[cotx].
    So it seems that we should not expect a single form.

    When n=3, the simplified form cancels to 1/[cot(3x)];
    when n=4, the simplified form (combining the first and fourth terms and the second and third terms separately) cancels to cot(4x).

    1. Think about the expression \cot{(A-n x)}. Where A is either a fixed angle or dependent on n.

      Can a general rule be found that works whether n is odd or even?

      (I think it can be done, but my rule may be wrong!)

      1. I’m not sure if you have a typo in your TeX, but I wouldn’t interpret the question as requiring the one expression to work for both n even and n odd. (One can always hammer two expressions into one, but if it is not natural to do so, then I wouldn’t think to do it here.)

        1. My solution doesn’t discriminate for odd/even but I thought it worth considering while working through the problem.

    2. Hi Tony,
      Thanks for pointing it out – yep I overlooked case-by-case discussion. I will look back and try to generalize the result.

      I think this question itself can form a very nice investigation question for VCE Spesh kids. Even at year 11 level.

      1. Think about this:

        \frac{1}{\cot{x}}=\tan{x}=\cot({\frac{\pi}{2}-x})

        cot{2x}=-cot{(\pi-2x)}

        There is a pattern here…

        …maybe.

        (I’m still not 100% sure my solution is totally valid for all positive integers n, but will share when I’ve checked it through)

  4. Thanks to everyone, for your attempts so far. As with the first, 1902 TNDOT, the 1913 students would have presumably been familiar with such questions and with the formulas and techniques to solve them.

    The forms of the expressions suggest natural approaches but I haven’t pondered the questions beyond that. Eventually, once you slaves have done the hard work, I’ll update the post with tidied answers.

    1. I think there are multiple correct answers to this one, which in a way is nice.

      My earlier suggestions did work, but I’ve since found another approach which makes the answer drop out much more efficiently, albeit to an answer in a different form (an equivalent form if I may use the word, probably incorrectly!)

  5. My approach was as follows:

    Try to find a rule for n=2. Easy enough.

    Try to find a rule for n=3. Didn’t look anything like the rule for n=2 so decided to attack the general case.

    Assume (for the moment) n is even. Found a rule. Now, assume n is odd…

    And yes, my original answer involved \tan{x} rather than \cot{x} which took a few moments of pondering, but it made sense in the end.

    It is really enjoyable (yes, I know I’m a bit different…) to watch all those terms cancel once the identities are applied.

  6. Hi Marty – that was fun! Lots of telescoping sums, but definitely ones I’m not familiar with. Also lots of practice with the newly reintroduced (to SM) trig sum ↔ product formulas. There are skills that fall in and out of favour in maths and other parts of education. Not sure why these types of sums were previously common in the Victorian Matric – I definitely wouldn’t like to sit this paper in exam conditions without more practice!

    The first question fell out fairly naturally. But I needed to Google a hint for the second one. Then from my experience with the 1st question the final one fell out ok – in a similar way to Red Five.

    My solutions are in https://www.mathcha.io/editor/0EJ5dI0VtjPidDCNm0MnjHM3WgEuNYXlJH21mz57
    I’ve checked them for the first few n (using CAS 😁), so I have hopefully caught dumb errors. But maybe they could be more efficient…

    1. Thanks, Simon. I’m glad they were fun, and thanks for your solutions. Once I get a chance to breathe I’ll ponder the questions, take a look at yours and other’s solutions, and update the post with tidy versions.

    1. Thanks, RF. I thought you came to this forum to be told off, like everyone else. I’ll look at your and other’s solutions soon.

      1. I can do both…

        Plus, yelling at clouds.

        (I try to learn from my mistakes. There are plenty of new ones I should be making)

  7. I’ve updated with a computation for the last “express” bit, following Red Five and Simon. Thanks to you all for playing. I have a post I *really* have to do, but I will try to to get to updating TNDOT 3 and posting TNDOT 4 soon.

    1. If anyone has PDFs scans of any textbooks that students may have used around the time to study for these exams, I’m sure a few commenters here would be rather interested…

      Hopefully any copyright issues have long expired.

      1. Please don’t post books or exams without clearing with me. I’m absolutely snowed right now, but I’ll try to sort out the copyright situation in the near future.

  8. I was going through some trig with my year 12s yesterday and noticed that the extended response questions in the review section includes the trig identities from / similar / used in this TNDOT

    See Q11, 16-19 on pages 141-142 of chapter 3 of the Cambridge Specialist 3&4 textbook attached (definitely within fair dealing rules).
    The questions are more structured than the identities required in this matric paper, which speaks to students’ expected familiarity with such things.

    Q14 is also quite cute – ties in nicely to the x + 1/x expression that is common in some maths competition questions – but also is secretly hiding some unit length complex numbers!

    Pg141-142

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