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I tried to solve it by using compound angle formulas but in the end I could not leave $\theta$ alone.

It goes like this:

\begin{align} & \frac{2.5}{3}=\sin(3\theta+x)\sin\theta \\[10pt] & \sin(3\theta+x)=\sin(2\theta+\theta)\cos(x)+\sin(x)\cos(2\theta+\theta) \\[10pt] = {} & [(\sin(2\theta)\cos(\theta)+\sin(\theta)\cos(2\theta))]\cos(x)+[(\cos(2\theta)\cos(\theta)-\sin(2\theta)\sin(\theta))]\sin(x) \end{align}

Then I did a couple more steps but I couldn't solve it this way, is there any other way to solve it algebraically?

Michael Hardy
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Sayra
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2 Answers2

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Here's an approach that at least reduces the problem to solving a quartic polynomial.

Let $x=3u$ and $\phi=\theta+u$, so that the equation becomes $3\sin3\phi=2.5/\sin(\phi-u)$, or

$$6\sin3\phi(\sin\phi\cos u-\cos\phi\sin u)=5$$

If you can solve this for $\phi$ in terms of $u$, you can easily convert that solution to one for $\theta$ in terms of $x$.

Now $\sin3\phi=3\sin\phi-4\sin^3\phi$, hence, letting $s=\sin\phi$, we have

$$6(3s-4s^3)\left(s\cos u-\sqrt{1-s^2}\sin u\right)=5$$

or

$$6(3-4s^2)s^2\cos u-5=6(3-4s^2)\sqrt{s^2-s^4}\sin u$$

Squaring both sides produces a quartic polynomial in $s^2$, with coefficients that can be expressed in terms of $\cos u$ (since $\sin^2u=1-\cos^2u$). I.e., letting $S=s^2=\sin^2\phi$ and $C=\cos u=\cos(x/3)$, we have

$$(6C(3-4S)S-5)^2=36(1-C^2)(3-4S)^2(S-S^2)$$

which expands out to a quartic in $S$. Because it's a quartic, this can, in principle, be solved for $S=s^2$, after which you can pick out solutions (if any) to the non-squared equation, but unless a miracle occurs (or there's a simplification I don't see), it looks like a mess. This leads me to wonder where this problem came from; it certainly doesn't strike me as a routine homework exercise.

Barry Cipra
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  • You're right, this isn't a homework exercise. I'm making a model with polar coordinates. $r=\frac{2.5}{\sin\theta}$ meets with $r=3\sin(3\theta+x)$ , I need to observe the angle where they intersect depending on $x$, so I need to express $\theta$ in terms of $x$ and go on with rest of the model. But I didn't think this part would be complex... This is a very good way of approaching it. But I didn't get the part where you went from $6(3s-4s^3)\left(s\cos u-\sqrt{1-s^2}\sin u\right)=5$ to $6(3-4s^2)s^2\cos u-5=6(3-4s^2)\sqrt{s^2-s^4}\sin u$ , what exactly did you do there? Thank you! – Sayra Mar 03 '18 at 17:45
  • @Sayra, I compressed several individual changes there: I moved the $5$ to the left hand side and the stuff with the square root to the right, I factored an $s$ out of the $(3s-4s^2)$ terms, and I pushed it inside the square root as an $s^2$. – Barry Cipra Mar 03 '18 at 18:38
  • @Sayra, thank you for the explanation of where the problem comes from. I see now that you are essentially rotating the propellor-shaped curve $r=3\sin3\theta$ and looking for intersections with the horizontal line $r\sin\theta=2.5$. If you draw those curves separately, you can see there is no intersection for certain values of $x$ (e.g., $x=0$) and two intersections for other values. It still looks like a mess to write down any exact algebraic formulas for the intersections, though. – Barry Cipra Mar 03 '18 at 18:59
  • Thank you for explaining, that part is clear now. Now, I expanded the last equation in your post, and you said this can be solved for S since it is quartic? I'm not very sure about how to do that, is it possible to leave S alone in $36C^2(9S^2-24S^3+16S^4)-60C(3S-4S^2)+25=(36-36C^2)(9S-33S^2+40S^3-16S^4)$ ? – Sayra Mar 03 '18 at 19:16
  • You're definitely right about the polar curves. However, the $x$ values that I'm interested in vary between 7.57 and 8.11 due to this model I'm trying to obtain, so there won't be an issue about the other $x$ values where there is no intersection. I was planning to express $\theta$ in terms of $x$ and then evaluate the rest of the equations according to the $x$ domain I stated; I will happily get to that part if I'm able to solve this problem. – Sayra Mar 03 '18 at 19:25
  • @Sayra, note that as a function of $x$, $\sin(3\theta+x)$ is periodic with period $2\pi\approx6.28$, so you might as well restrict $x$ to vary between $1.29$ and $1.83$. But you may have other reasons for using the interval $[7.57,8.11]$. – Barry Cipra Mar 03 '18 at 19:41
  • Yes, I realized that after I wrote the comment, odd that I didn't think of it before. Thank you :) – Sayra Mar 04 '18 at 10:36
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your equation is equivalent to $$4\,\cos \left( x \right) \left( \sin \left( \theta \right) \right) ^ {2} \left( \cos \left( \theta \right) \right) ^{2}-\cos \left( x \right) \left( \sin \left( \theta \right) \right) ^{2}+4\,\sin \left( \theta \right) \sin \left( x \right) \left( \cos \left( \theta \right) \right) ^{3}-3\,\sin \left( \theta \right) \sin \left( x \right) \cos \left( \theta \right) =2.5$$ you can Substitute $$\sin(\theta)=\frac{2\tan(\frac{\theta}{2})}{1+\tan(\frac{\theta}{2})^2}$$ and $$\cos(\theta)=\frac{1-\tan(\frac{\theta}{2})^2}{1+\tan(\frac{\theta}{2})^2}$$ and then you can Substitute $$\tan(\frac{\theta}{2})=t$$

Dr. Sonnhard Graubner
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  • That makes sense, thank you! – Sayra Mar 03 '18 at 16:13
  • Can you solve your Problem now? – Dr. Sonnhard Graubner Mar 03 '18 at 16:21
  • I worked on it, and I again got stuck at simplifying it. With substituting t, I now have $ \frac{cos(x)(16-40t^4+12t^6-4t^2)+sin(x)(36t^4-30t^3+2t-4t^5-2t^7)}{1+4t^2+6t^4+4t^6+t^8}=\frac{2.5}{3}$ and I don't know how to simplify it. Might there be a way other than substituting $ \tan(\theta/2)$ ? Or can I simplify this equation? – Sayra Mar 03 '18 at 16:57