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If $$x=2\cos\theta-\cos\theta\cos 2\theta$$ $$y=2\sin\theta-\sin\theta\sin 2\theta$$ find a relation between $x$ and $y$ (not involving $\theta$).

Another trig elimination that has me stumped. One approach is to express as homogeneous functions,

$$x=3\sin^2\theta \cos\theta+\cos^3\theta$$ $$y=2\sin^3\theta+2\sin\theta\cos^2\theta-2\sin^2\theta\cos \theta$$

and by forming linear combinations of $x,y$ find some third power trig polynomials eg$(\cos \theta+\sin\theta)^3$. But I am not finding any success with this problem.

Note that this problem is from Hobson, Treatise on Plane Trigonometry 2 ed pg.97 #47. Slightly altered in the form I ask it. And may contain a typo, see discussion.

Rene Schipperus
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3 Answers3

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Let $u=\cos\theta$ so $x=3u-2u^3$ and $y=-2u\pm2\sqrt{1-u^2}+2u^3$.

Let $z=x+y$ so $4(1-u^2)=(y+2u-2u^3)^2=(z-u)^2$ giving $$5u^2-2zu+z^2-4=0\implies5u=z\pm2\sqrt{5-z^2}.$$ Apply the quadratic equality repeatedly to get \begin{align}x&=3u-2u\cdot5^{-1}(2zu-z^2+4)=(3+2\cdot5^{-1}z^2-8\cdot5^{-1})u-4\cdot5^{-1}zu^2\\&=(3+2\cdot5^{-1}z^2-8\cdot5^{-1})u-4\cdot5^{-2}z(2zu-z^2+4)\\&=(3+2\cdot5^{-1}z^2-8\cdot5^{-1}-8\cdot5^{-2}z^2)5^{-1}(z\pm2\sqrt{5-z^2})+4\cdot5^{-2}(z^2-4).\end{align} Simplification leads to $$125x=22(x+y)^3-45(x+y)\pm2(35+2(x+y)^2)\sqrt{5-(x+y)^2}.$$

ə̷̶̸͇̘̜́̍͗̂̄︣͟
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  • I am reviewing trig and would appreciate seeing the steps to derive $X$ and $Y$ as functions of $\cos\theta$ in the simplest manner since it does seem this would be the necessary step to eliminate $\theta$. And are cubes of trig functions the key to solving this type of problem, or is this an unusual problem to be presented in a trig course? – Gwendolyn Anderson Oct 24 '21 at 14:58
  • What I am seeing is that the square root expression is the way to force the lone $\sin\theta$ term so that it is all in terms of $\cos\theta$. But then an expression in Y has to be squared in order to remove that root, in order to arrive at a solution, since $\theta$ will not be removed if it is in an expression under a root. That is all I was looking for in the answer. @SimpliFire – Gwendolyn Anderson Oct 24 '21 at 17:10
  • The graph corresponding to the final equality appears to differ from that obtained from the problem statement and from @Blue's equation in the comments. Blue's equation should be obtained by rationalizing your equation. This is currently not the case. The method is sound, however. – user26872 Oct 25 '21 at 01:21
  • The final equation should read $125x=22(x+y)^3-45(x+y)\pm{\color{red} 2}(35+2(x+y)^2)\sqrt{5-(x+y)^2}$. – user26872 Oct 25 '21 at 15:25
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    @user26872 Thanks a lot for finding the mistake. Here is validation of the result: https://www.desmos.com/calculator/hvuiaxmmdw. – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Oct 25 '21 at 15:28
  • I am glad to help. (+1) for an interesting solution. – user26872 Oct 25 '21 at 15:33
  • Thanks for this solution +1, I still have to study it, but what I find strange is that it is unsymmetrical in $x$ and $y$. – Rene Schipperus Oct 26 '21 at 13:01
  • I think it's asymmetrical because the expansions of $\sin k\theta$ and $\cos k\theta$ are always off by a small factor but I'm not sure exactly what happens geometrically. – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Oct 26 '21 at 13:08
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    @TheSimpliFire: +1. Your equation is equivalent to mine, and has a fairly straightforward derivation. (FYI, clearing your radical, and/or grouping my polynomial, gives $$4z^6-12z^4-44xz^3+33z^2+90xz+125x^2-196=0$$ where $z:=x+y$.) However, I don't think either form is consistent with the spirit of other exercises of this kind. I remain on Team Typo. :) – Blue Oct 26 '21 at 13:22
  • @Blue I guess that's true, the coefficients don't seem to have a generalisable pattern. The system $(k\cos\theta-\cos\theta\cos k\theta,k\sin\theta-\sin\theta\sin k\theta)$ is unlikely to have a closed form solution in terms of $k$ as we may have to invert Chebyshev polynomials at some point. – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Oct 26 '21 at 13:27
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    @Blue My remark to Team Typo (to which I belong) is that it must be a typo in the original formulation of the question. – Rene Schipperus Oct 26 '21 at 13:40
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Indulging the possibility/likelihood of a typographic error in the source material (which was the source of the problem in OP's previous question), consider the system $$\begin{align} x &= \cos\theta\, ( 2 - \cos2\theta ) \\ y &= \sin\theta\, ( 2 - \color{red}{\cos2\theta} ) \quad\color{red}{\leftarrow\text{instead of $\sin2\theta$}} \end{align}$$ From here, we easily have, defining $r:=2-\cos2\theta$, $$\begin{align} x^2+y^2 &= (2-\cos2\theta)^2\phantom{\,\cos2\theta} = r^2 \\ x^2-y^2 &= (2-\cos2\theta)^2\cos2\theta = r^2(2-r) = (x^2+y^2)(2-r) \end{align}$$ whence $r(x^2+y^2) = x^2+3y^2$, so that

$$(x^2+y^2)^3 = ( x^2 + 3y^2 )^2 \tag{$\star$}$$

Alternatively, if we had $$\begin{align} x &= \cos\theta\, ( 2 - \color{red}{\sin2\theta} ) \quad\color{red}{\leftarrow\text{instead of $\cos2\theta$}}\\ y &= \sin\theta\, ( 2 - \sin2\theta ) \end{align}$$ then, with $r:=2-\sin2\theta$, $$\begin{align} x^2+y^2 &= (2-\sin2\theta)^2 \phantom{\sin2\theta\,} = r^2\\ 2 x y &= ( 2 - \sin2\theta)^2\sin2\theta = (x^2+y^2)(2-r) \end{align}$$ whence

$$(x^2 + y^2)^3 = 4 (x^2 - x y + y^2)^2 \tag{$\star\star$}$$

Of the two, I favor $(\star)$ as the correction of the ostensible typo. Either way, these solutions are more in keeping with the spirit of similar exercises in the source than the (perfectly valid) ones given in other answers to this question as stated.

Blue
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    Thanks for these two possibilities. +1. – Rene Schipperus Oct 26 '21 at 14:35
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    My appreciation for trying to fix the possible typo. But IMO, these refinements make the question very easy and even remove the good look ($x$ in $\cos$ and $y$ in $\sin$). Also note that in this similar question from the same OP, he calls it a "monster of question". So this question also can be such a monster : ) – ACB Oct 26 '21 at 14:35
  • @ACB Well yes. People talk about things like implicit function theorem, but these problems expose the practical nature of those questions. – Rene Schipperus Oct 26 '21 at 14:37
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    I am very pleased with Blues answer, I cannot accept it for obvious reasons, but please up vote it. – Rene Schipperus Oct 26 '21 at 14:39
  • @YvesDaoust: This is true. Either "fixed" version is effectively the Cartesian-ification of a polar eqn ($r=2-\cos2\theta$ or $r=2-\sin2\theta$ ... hence my sly use of $r$ in the solutions. :) The standard brute-force substitutions $\cos\theta\to x/r$, $\sin\theta\to y/r$ also work to get the solution, but it seemed instructive to work through the problem differently. In the context of the source exercises, I don't believe the reader is expected to be thinking in terms of polar eqns (or else the problem would be stated as such), but in terms of manipulating expressions using trig identities. – Blue Oct 26 '21 at 14:55
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Since there is a discussion over the possibility of typos in the source of the question and inspired by @Blue's answer, here's another (possible) typo fixed approach: [Note: I am presenting this answer because I was more concerned about the appearance of the question, not the answer :) ]

$$x=\cos\theta-\cos\theta\cos2\theta\tag1$$ $$y=\sin\theta-\sin\theta\sin2\theta\tag2$$

(Observe $x=\color{red}{\not2}\cos\theta-\cos\theta\cos2\theta$ and $y=\color{red}{\not2}\sin\theta-\sin\theta\sin2\theta$)

Adding two equations, $$x+y=\cos\theta+\sin\theta-\underbrace{(\cos\theta\cos2\theta+\sin\theta\sin2\theta)}_{\cos\theta}=\sin\theta$$

From the first equation we get, $$x=\cos\theta(1-\cos2\theta)=2\sin^2\theta\cos\theta$$

By squaring, $$x^2=4\sin^4\theta\cos^2\theta=4\sin^4\theta(1-\sin^2\theta)$$

Substituting the first result, $$x^2=4(x+y)^4(1-(x+y)^2)$$

ACB
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  • FYI: In the version of Hobson's Treatise on Plane Trigonometry on archive.org ,the equations are $a\cos\theta\cos2\theta=2(a\cos\theta-x)$ and $a\sin\theta\sin2\theta=2(a\sin\theta-y)$ those are slightly different from the OP's equations. So I removed the factor 2 in OP's edition. – ACB Oct 27 '21 at 08:19