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The curve $r=5\sinθ$ is given for $0\leθ\le a$
It appears that if $a = \pi$, the graph is complete. Now my question is why? Since I thought it should be $a = 2\pi$ since $2\pi = 360°$, which is a full circle.

enter image description here

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If you plot the points at some spacing of $\theta$ you will see the graph is complete when $\theta=\pi$. Because of the relationship $\sin (\theta+\pi)=-\sin (\theta)$ as $\theta$ rises from $\pi$ to $2\pi$ you have $r$ being negative, which reflects the angle through the center and retraces the graph you have already plotted.

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  • $\begingroup$ But if the sin is negative, then theta is also negative right? Which is not allowed. $\endgroup$ – Stallmp Jul 11 at 14:55
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    $\begingroup$ Usually $\theta$ is allowed to run over $(-\infty, \infty)$ and you plot enough of it to get what you want. That doesn't matter in this case. $\sin \frac {3\pi}2$ is negative even though $\frac {3\pi}2$ is not $\endgroup$ – Ross Millikan Jul 11 at 15:05
  • $\begingroup$ I see, so sin can be negative while theta remains strictly positive? $\endgroup$ – Stallmp Jul 11 at 15:37
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    $\begingroup$ Sure. Just plot $y=\sin x$ and you will see. It is negative on $(\pi, 2\pi)$ and on all intervals $2k\pi$ above and below $\endgroup$ – Ross Millikan Jul 11 at 17:52

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