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I'm trying to find the angle of intersection between two polar curves:

$$\begin{cases}r= 5 + 3 \sin\theta \\ r' = 3\end{cases}$$

I've set them as $5 + 3\sin\theta = 3$ and got to $\sin\theta = -2/3$ but from here how do I find $\theta$? The unit circle seems useless here. Can someone please teach me the way?


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Do you know the inverse sine function? – Hakim Aug 2 '14 at 14:57
Nope, I don't.. – math_n3b Aug 2 '14 at 15:00
If you have $\sin\theta=\alpha$ then $\arcsin(\sin\theta)=\arcsin\alpha\implies\theta=\arcsin\alpha$, where $\arcsin$ is the inverse sine function. – Hakim Aug 2 '14 at 15:14
up vote 3 down vote accepted

$$\sin \theta = -\frac 23 \iff \arcsin (\sin \theta) = \arcsin\left(-\frac 23\right) \iff \theta =\arcsin\left(-\frac 23\right)$$

You'll need to approximate $\theta$, since there is no "nice" angle $\theta$ such that $\sin(\theta)=-\frac 23$.

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thanks, now I understand :) – math_n3b Aug 2 '14 at 15:16
You're welcome, math_n3b. – amWhy Aug 2 '14 at 15:16
No problem, @Hakim. I've done that on occasion myself. I didn't downvote you, though. I rarely downvote, and even less so on answers. – amWhy Aug 2 '14 at 15:41
@amWhy Oh sorry, my fault... – Hakim Aug 2 '14 at 15:41

This is really just a footnote to amWhy's answer. If you graph the two equations in your system you'll get something like:

$\phantom{XXXXXXXXX}$ xkcd-style-polar-plot ;-)

So there are two points where the two curves meet. The angles at which they meet is given by: $$\arcsin\left(-\dfrac23\right)\quad\color{grey}{\text{ and }}\quad\pi-\arcsin\left(-\dfrac23\right).$$

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