TNDOT 4: Two Trigs in One

I’m still trying to deal with the backlog, and the other “topical” posts are requiring a bit more work than envisaged. So, to keep things ticking along, and by request, here is a new TNDOT. As for TNDOT 2, this one is from the 1913 Victorian Matric Trigonometry Honours Exam. Have fun.

\color{blue}\boldsymbol{\aligned  &\mbox{\bf {From the equations}}\\[3\jot] &x^2 + y^2 = \frac{x\cos 3\theta + y\sin 3\theta}{\cos^3\theta}=\frac{y\cos 3\theta - x\sin 3\theta}{\sin^3\theta}\,.\\[3\jot]&\mbox{\bf \quad \bf{(i) Eliminate }} \theta \mbox{ \bf{and express the result in a simple form; }} \\[3\jot] &\mbox{\bf \quad \bf{(ii) Find }} x \mbox{\bf { and }} y \mbox{\bf { in terms of }}\theta. \endaligned}


OK, after a very long delay, here is a tidy solution to the question. I’ve followed the hints of J. D. and the notes of Red Five. (I also suspect, as do Simon and Red Five, that there may be some neat complex numbers approach to the problem, but I didn’t pursue it.)

To simplify the writing a little, we’ll first set

\color{Plum}\boldsymbol{r^2 = x^2 + y^2,\ c = \cos3\theta,\ s =\sin3\theta.}

Then the two equations we are given can be written as

\color{Plum}\boldsymbol{\aligned &(a)\quad  &&\phantom{ioj}cx + sy = r^2\cos^3\!\theta\,, \\[3\jot] &(b)\quad &&-sx+cy = r^2 \sin^3!\theta\,.\endaligned}

Thinking of this as two linear in equations in x and y, this system can then be easily “solved”, either by matrix trickery or, as was probably expected in 1913, by combining the equations trig-naturally (c(a) – s(b) and s(a) + c(b)), and using \boldsymbol{c^2 + s^2 = 1}:

\color{Plum}\boldsymbol{\aligned &(c)\quad  x = r^2\left(c\cos^3\!\theta-s\sin^3\!\theta)\,, \\[3\jot] &(d)\quad y =  r^2\left(s\cos^3\!\theta+c\sin^3\!\theta)\,.\endaligned}

The next step is to write the triple-angle guys and the cubic guys as the same kind of thing. It seems easiest to convert the cubic guys, using the following formulas (which would have been very well known in 1913):

\color{RawSienna}\boldsymbol{(1)\qquad 4\sin^3\!\theta = 3\sin\theta -\sin3\theta\,,}

\color{RawSienna}\boldsymbol{(2)\qquad 4\cos^3\!\theta = 3\cos\theta +\cos3\theta\,.}

Plugging these into (c) and (d) and simplifying, we get,

\color{Plum}\boldsymbol{&(e)\quad  4x/r^2= \cos3\theta(3\cos\theta +\cos3\theta) -\sin3\theta(3\sin\theta -\sin3\theta)= 1+3\cos4\theta\,,}

\color{Plum}\boldsymbol{&(f)\quad  4y/r^2= \sin3\theta(3\cos\theta +\cos3\theta) +\cos3\theta(3\sin\theta -\sin3\theta)= 3\sin4\theta\,.}

Now we’re home. Squaring (e) -1 and (f), and adding, we get

\color{Plum}\boldsymbol{\left(\dfrac{4x-r^2}{r^2}\right)^2 + \left(\dfrac{4y}{r^2}\right)^2= 9\,.}

Multiplying through by \boldsymbol{r^4}, expanding and using \boldsymbol{x^2 + y^2 = r^2}, this easily simplifies to

\color{Plum}\boldsymbol{(g)\qquad 2-x = x^2 + y^2\,,}

which is the desired answer to (i).

Next, to write x in terms of \boldsymbol{\theta}, we can multiply (e) through by \boldsymbol{r^2} and plug in (g):

\color{Plum}\boldsymbol{4x = (2-x)(1+3\cos4\theta)\,.}

This is a linear equation in x, which unravels to give

\color{Plum}\boldsymbol{x = \dfrac{6\cos4\theta + 2}{3\cos4\theta + 5}\,.}

Finally, to write y in terms of \boldsymbol{\theta} we can plug (g) and then this expression for x into (f), giving,

\color{Plum}\boldsymbol{&(f)\quad  4y= (2-x)3\sin4\theta= \left(2 - \dfrac{6\cos4\theta + 2}{3\cos4\theta + 5} \right)3\sin4\theta\,.}

Getting a common denominator, this easily simplifies to

\color{Plum}\boldsymbol{y = \dfrac{6\sin4\theta}{3\cos4\theta + 5}\,,}

and we’re done.

19 Replies to “TNDOT 4: Two Trigs in One”

  1. I love this question. My solution runs into a few pages though, so I’m very curious to see what others come up with…

    As with others, use the sum-to-product identities at one point (well, I did).

    Other hints on request.

  2. There’s an interesting 90 degree rotation in theta hiding in there. Maybe could be exploited for a nice solution, but I haven’t had time to play yet!

    1. A very interesting observation. I missed that (my method is more brute-force than finesse).

      After eliminating \theta, finding an expression for x was easy enough. The expression for y took a few more lines of scribble though, so again, I’m fairly certain I’ve missed the “nice” solution.

      Would writing \sin{\theta}=\cos{(x-\frac{\pi}{2})} help at all I’m now wondering…


      Thanks for the idea.

    2. UPDATE: While the rotation idea is certainly a nice one, I cannot (yet) use it to simplify my solution at all. Complex numbers don’t seem to help either (I’m sure a solution using the complex exponential forms of sin and cos would work, I’m just not convinced it is an improvement on the solution using product-to-sum identities).

      If no one has posted further ideas by Monday, I might drop a hint or two as to how I went about it. Full solution (remembering that I really don’t think it is the best solution possible) available on request.

  3. If you temporarily consider the x^2 + y^2 on the left to be a constant R^2, then the system becomes a 2 \times 2 linear system in the unknowns x and y. The determinant of the system being 1, it’s not hard to write down the solution using Cramer’s Formula, and after that the expressions for x and y can in fact be simplified considerably (without ever changing R^2 back to x^2 + y^2). Once that’s done, \theta can be eliminated by using \cos^2 4\theta + \sin^2 4\theta = 1. In principle, when \theta is eliminated, there should be an R^2 factor in the equation obtained, because (x,y) = (0,0) is always a solution.

    1. Cramer’s rule. That is genius. Thanks. I’ll have a go to see where that leads.

      I did think of a geometric approach, but didn’t get very far.

      UPDATE: I think (still checking my work) that this leads to x^{2}+y^{2}=\frac{8}{5+3\cos{4\theta}}

      From there x and y should be not too difficult to extract.

      1. Yes, I also have that result. After simplifying the expressions for x and y, I get x = (1 + 3 \cos 4\theta)R^2/4, y = (3 \sin 4\theta)R^2/4. Eliminating \theta leads to R^2(R^2 + x - 2) = 0. After that, on the assumption that (x,y) \ne (0,0), we can substitute 2 - x for R^2 in the first equation to obtain an expression for x in terms of \theta, and then again put R^2 = 2 - x into the second equation to find y.

    2. Very nice, J. D. (Just calling it “the inverse of a 2 x 2” is less scary than “Cramer”.)

      I doubt students at the time (and much less now) would have been expected to have approached the question in such a manner. But probably they would have done the simultaneous equations work that mimicked your approach.

      1. Your comment made me wonder whether Cramer’s Rule was taught at the School Certificate and Higher School Certificate levels in England in the past. It seems determinants are not mentioned in Barnard and Child’s “New Algebra” (1912), but Cramer’s Rule is discussed in Ferrar’s “Higher Algebra for Schools” (1944), which was for mathematics and science specialists at HSC level. The 2 x 2 version is discussed in Part III of Durell’s “New Algebra for Schools” (1931), which was for “Additional Mathematics” at SC level.

        1. Thanks, J.D. I think of matrices and determinants as a more modern thing, particularly as a school topic. I haven’t hunted, but I don’t think these were school topics in Australia at the time of this question.

          1. Matrix determinants were a thing in some of the 1900s exam papers up to 3×3.

            Matrix operations beyond determinants do not seem to have been a thing.

              1. That is the Victorian papers. I have yet to get my hands on the NSW papers of the time.

                If anyone has any though… I am very interested. The NSW state library apparently has paper copies (bound into annual volumes) back to pre-1900 but won’t allow inter-library borrowing.

    1. That’s some good work — your solutions for x and y do solve the problem… but I agree that it seems like too much for an exam question and there must be some other way through it all. I played a bit with this question, but struggled to find a simple way to eliminate θ or solve for x and y. Maybe you path through isn’t so bad… especially now that I review it again, it is fairly logical.

      1. Thanks Simon. I struggled with this question for DAYS, convinced that I was missing something before I decided to just brute-force it.

        I still think there may be a complex numbers solution hiding somewhere (as a few papers around the same time made passing reference to De Moivre’s theorem and Euler’s formula).

  4. OK, I finally got around to updating the post with solutions. Thanks to everyone for their contributions. I’ll try to update TNDOT 3 and to post a TNDOT 5 in the next couple days.

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