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

Question

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.

Answer 1

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.)

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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.

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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.

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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.

Answer 2

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

Answer 3

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 $y=f_2(\theta)$.

Answer 4

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:

Answer 5

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 $$\boxed{27 (x - y)^2 = (1 - x - y)(x + y + 8)^2} .$$

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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.