Eliminating $\theta$ from $x^2+y^2=\frac{x\cos3\theta+y\sin3\theta}{\cos^3\theta}$ and $x^2+y^2=\frac{y\cos3\theta-x\sin3\theta}{\cos^3\theta}$

An interesting problem from a 1913 university entrance examination (Melbourne, Australia):

Eliminate $$\theta$$ from the expressions

$$x^{2}+y^{2}=\frac{x \cos{3\theta}+y \sin{3\theta}}{\cos^{3}\theta} \tag1$$ $$x^{2}+y^{2}=\frac{y \cos{3\theta}-x \sin{3\theta}}{\cos^{3}\theta} \tag2$$ Find expressions for $$x$$ and $$y$$ in terms of $$\theta$$

As with many of these historic problems, I'm sure it has been discussed somewhere at length...

Solutions involving complex numbers would have been acceptable at the time.

(EDIT 20/Feb Australian time) I thought I had a solution but upon review I don't think it works (I found an expression for y, then substituted. It was very messy and I don't think the algebra was totally correct.) so editing the post to say: any solutions or suggestions appreciated. The wording of similar questions on the paper suggests that a "simplified" expression is possible.

(Edit 21/Feb) Thanks everyone for the excellent suggestions. Really appreciated. I am wondering if the substitution $$t=\tan{(\frac{\theta}{2})}$$ works here. If $$t=\tan{(\frac{\theta}{2})}$$ then $$\sin{\theta}=\frac{2t}{t^{2}+1}$$ and $$\cos{\theta}=\frac{1-t^{2}}{1+t^{2}}$$.

After some basic manipulation, this seems to have promise, but I'm yet to finish the problem using this approach.

• Welcome to Math.SE! ... Please include your solution. (We can't tell what "neater" means if we have no basis for comparison.) This may help others find a "neater" path to it without having to do all the work from scratch, and it will help them avoid duplicating your work. (It's no fun devising, composing, formatting, and perhaps even illustrating a comprehensive and comprehensible answer to a question only to have the asker respond "That's a lot like what I did; I wanted something better.") If nothing else, they can check their answers.
– Blue
Commented Feb 20 at 9:58
• BTW, this may be "neater": Square the equations, and add them together.
– Blue
Commented Feb 20 at 10:21
• Fair point. Coming soon (I'm a $LaTeX$ novice, so it takes a while...) Commented Feb 20 at 10:31
• Both the equations (1) and (2) have same LHS. It seems inappropriate Commented Feb 20 at 10:32
• Thanks @Blue. I must have made an error in my algebra (it happens a lot). Commented Feb 20 at 11:05

Revising and extending my comment to @Maverick's answer ...

(For completeness, I'll derive the result of that answer. Since I suggested the approach in a comment to the question, I don't believe this steps on anyone's toes.)

First, to arrive at the parameterization, square the equations and add. (For notational convenience, define $$c:=\cos3\theta$$ and $$s:=\sin3\theta$$.) \begin{align} 2(x^2+y^2)^2 &= \left(\frac{xc+ys}{\cos^3\theta}\right)^2 + \left(\frac{yc-xs}{\cos^3\theta}\right)^2 \tag1\\[4pt] &=\frac{(x^2c^2+2xycs+y^2s^2)+(y^2c^2-2yxcs+x^2s^2)}{\cos^6\theta} \tag2\\[4pt] &=\frac{(x^2+y^2)(\cos^33\theta+\sin^23\theta)}{\cos^6\theta)} \tag3\\[4pt] &=\frac{x^2+y^2}{\cos^6\theta} \tag4 \end{align} Ignoring the case $$x^2+y^2=0$$ (but see the Note later), we divide-through by $$2(x^2+y^2)$$ to get $$x^2+y^2=\frac{1}{2\cos^6\theta} \tag5$$ This allows us to write the left-hand sides of the original system without $$x$$ and $$y$$, and we can write $$\frac{1}{2\cos^3\theta} =xc+ys \qquad \frac{1}{2\cos^3\theta} = yc-xs \tag6$$ We readily solve the simple linear system to get $$(x,y) = \left(\frac{\cos3\theta-\sin3\theta}{\cos^3\theta}, \frac{\cos3\theta+\sin3\theta}{\cos^3\theta}\right) \tag7$$ which is our parameterization.

For the task of eliminating $$\theta$$, the division by $$\cos^3\theta$$ is key. To see how, we expand the triple-angle trig functions:

$$\cos3\theta = \cos\theta (\cos^2\theta - 3\sin^2\theta) \qquad \sin3\theta = \sin\theta (3 \cos^2\theta - \sin^2\theta) \tag8$$ The terms seem to "want" to divide by $$\cos^3\theta$$ to give simple polynomial expressions in $$t:=\tan\theta$$; namely, $$\frac{\cos3\theta}{\cos^3\theta}= \frac{\cos\theta}{\cos\theta}\left(\frac{\cos^2\theta}{\cos^2\theta}-3\frac{\sin^2\theta}{\cos^2\theta}\right)=1-3t^2 \qquad \frac{\sin3\theta}{\cos^3\theta}=t(3-t^2) \tag9$$ Therefore, $$(x,y)=\tfrac12\left(\;1 - 3t - 3t^2 + t^3 \;,\; 1 + 3t - 3t^2 -t^3 \;\right) \tag{10}$$ (The factored forms of the expressions turn out to be non-helpful.) From here, we can write \begin{align} x+y &= 1 - 3t^2 &&\quad\to\quad &t^2 &= \tfrac13(1-(x+y)) \tag{11}\\[4pt] x-y &= t (t^2 - 3) &&\quad\to\quad& (x-y)^2 &= t^2 (t^2-3)^2 \tag{12} \end{align} Substituting from $$(11)$$ into $$(12)$$, and cleaning-up a little, gives $$27(x-y)^2 = -(x+y-1) (x+y+8)^2 \tag{13}$$ This is our $$\theta$$-less equation.

Note. Geometrically, the system describes the locus of points of intersection of two congruent circles of varying radius (specifically, $$\sec^3\theta$$) spinning about the origin. (The origin itself is one of those points of intersection, corresponding to the $$x^2+y^2=0$$ case we ignored earlier.)

The reader is invited to ponder how the appearance of $$x-y$$ and $$x+y$$ in $$(13)$$ relates to the evident $$45^\circ$$ rotation of the curve.

Squaring and adding gives us $$2(x^2+y^2)^2=\frac{x^2+y^2}{\cos^6\theta}\Rightarrow x^2+y^2=\frac{1}{2}\sec^6\theta$$ Putting in original equations we obtain

$$x\cos3\theta+y\sin3\theta-\frac{1}{2}\sec^3\theta=0$$

$$-x\sin3\theta+y\cos3\theta-\frac{1}{2}\sec^3\theta=0$$

Solving for $$x$$ and $$y$$ we have

$$x=\frac{\cos3\theta-\sin3\theta}{2\cos^3\theta}$$ and $$y=\frac{\cos3\theta+\sin3\theta}{2\cos^3\theta}$$

• It's interesting that the triggy nature of the $\cos^3\theta$ has no bearing on the problem; it might just as well be a constant. (One could expand the final result and divide-through to get an expression in tangents, but that's uglier.) ... The problem likely originates as an investigation of a locus —specifically, of the "other" pt of intersection of two congruent, orthogonal circles of varying radius rotating about the origin— as that stuff was popular historically. But, without that context, the $\cos^3\theta$ seems like an unnecessary complication .. maybe to make the exam seem scarier.
– Blue
Commented Feb 20 at 20:36
• Ah! For the task of eliminating $\theta$, the division by $\cos^3\theta$ is key. Defining $t:=\tan\theta$, we have $$(x,y)=\tfrac12\left(\;(1+t)(1-4t+t^2)\;,\;(1-t) (1+4t+t^2)\;\right)$$ Using, eg, the method of resultants (which is a little tricky to do by hand, but not impossible), one can eliminate $t$ to yield $$(x+y)^3+42(x^2+y^2)-24xy+48(x+y)-64=0$$ There may be a "neater" path to this result.
– Blue
Commented Feb 20 at 22:43
• I wonder... would defining $t=\tan{\frac{\theta}{2}}$ be a quicker option? Commented Feb 20 at 23:32

Substituting $$\theta = \arctan t$$ transforms the system into a rational (in fact polynomial) system: \begin{align*} y t^3 + 3 x t^2 - 3 y t - x + (x^2 + y^2) &= 0 \qquad (1) \\ -x t^3 + 3 y t^2 + 3 x t - y + (x^2 + y^2) &= 0 \qquad (2) \end{align*} (The tangent half-angle substitution, $$\theta = 2 \arctan \tau$$, mentioned in the question statement, works, too, but the resulting polynomials have higher degree.)

Computing $$x \cdot(1) + y \cdot (2)$$ and $$y \cdot (1) - x \cdot (2)$$, discarding the factor $$x^2 + y^2$$ (which eliminates only $$(0, 0)$$ from the solutions of the original system) and rearranging gives the system \begin{align*} x + y &= - 3 t^2 + 1 \\ x - y &= -t^3 + 3 t . \end{align*} Squaring the second equation gives that $$(x - y)^2 = (-t^3 + 3 t)^2 = t^2 (t^2 - 3)^2 ;$$ in particular, $$t$$ only appears in even powers. On the other hand, rearranging the first equation gives $$t^2 = \frac{1 - x - y}3 .$$ and substituting, factoring, and clearing denominators gives $$\boxed{27 (x - y)^2 = (1 - x - y)(x + y + 8)^2} .$$

It's much less practical to do by hand, but one also can proceed algorithmically from the system (1)–(2): Regarding the equations as polynomials in $$t$$ with coefficients in $$\Bbb F[x, y]$$, any solutions $$(x, y)$$ must be a root of the resultant of the polynomials $$f, g$$ on the left-hand side of (1), (2) in $$t$$, namely, satisfying \begin{align*} 0 &= \operatorname{res}(f, g) \\ &= (x^2 + y^2)^3 (x^3 + 3 x^2 y + 3 x y^2 + y^3 + 42 x^2 - 24 x y + 42 y^2 + 48 x + 48 y - 64), \end{align*} hence our curve is (again setting apart the exceptional point at $$(0, 0)$$) $$x^3 + 3 x^2 y + 3 x y^2 + y^3 + 42 x^2 - 24 x y + 42 y^2 + 48 x + 48 y - 64 = 0 ,$$ which coincides with the above boxed answer.

HINT.-We have $$(\cos^3 \theta) y^2-(\sin 3\theta) y+[(\cos^3 \theta) x^2-(\cos3\theta) x]=0\\(\cos^3 \theta) y^2-(\cos 3\theta) y+[(\cos^3 \theta) x^2-(\sin3\theta) x]=0$$ then, solving as quadratic equations in $$y$$, $$y=\sin 3\theta\pm\sqrt{\sin^2 3\theta-4\cos^3\theta[(\cos^3\theta)x^2-(\cos(3\theta)x}]\\y=\cos 3\theta\pm\sqrt{\cos^2 3\theta-4\cos^3\theta[(\cos^3\theta)x^2-(\sin(3\theta)x}]$$ Equating the two expressions of $$y$$ we obtain an equation from which we deduce $$x=f_1(\theta)$$. Similarly we can get $$y=f_2(\theta)$$.

• Both excellent. Thanks. These questions were from "honors" papers set by The University of Melbourne and so were meant to have "tricks" in them. The question asks to eliminate \theta from the expression, which none of these answers have completely managed yet (despite them both being excellent for the second part). Commented Feb 20 at 22:17
• Thanks everyone for the ideas. Here is a related question I'm puzzling over: math.stackexchange.com/questions/4867383/… Commented Feb 22 at 6:14

Equating numerators of given RHS, letting $$t= \theta,$$

$$x \cos(3t) +y \sin( 3t) = y \cos(3t) -x \sin(3t)$$

$$\text{ Let} ~x=r \cos \alpha ,~ y= r \sin \alpha ~, r=\sqrt{x^2+y^2}, \alpha = \tan ^{-1}(y/x)$$

Plug in and simplify trig

$$\cos(\alpha -3t) =\sin (\alpha-3t),~ \tan(\alpha-3t)=1$$

$$3\theta= \alpha -(2k+1) \pi/4$$

Let $$\gamma = \tan ^{-1}(y/x)-(2k+1) \pi/4, 3 \theta= \gamma, \theta=\gamma/3 ~~ \cos 3\theta= \cos \gamma,~\sin 3\theta= \sin \gamma;$$

Plug into RHS of the first given equation

$$x^2+y^2=\frac{x \cos \gamma+ y \sin \gamma}{\cos^{3} (\gamma/3)}$$

which contains $$x,y$$ and no $$\theta,$$ but amenable to further simplification. The plot, if I made no mistakes:

• Thanks everyone for the brilliant ideas. I'm about to post a similar question (same structure as this but with one change that makes quite a difference) Commented Feb 22 at 6:09