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I am solving this inequality for $\theta$ $$\cos \theta > -\frac{\rho}{\sqrt{1-\rho^2}} \sin \theta$$ with $\rho \in (-1,1)$ given, when trying to integrate a function under polar coordinates and determine the integral region.

My idea is to consider when $\sin \theta < 0, >0$, and $=0$ three cases. But I feel it too complicated to proceed, and my trigonometric knowledge is rusty. Any idea how to solve it clearly? Thanks!

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  • $\begingroup$ try to put $\theta$ to one side and $\rho $ to the other side. $\endgroup$ – Ma Ming May 2 '13 at 16:48
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Since $\rho \in (-1,1)$, set $\rho = \cos(t)$. We then get, since $\sqrt{1- \rho^2} > 0$, $$\cos(\theta) \sin(t) + \cos(t) \sin(\theta) > 0 \implies \sin(t+\theta) > 0 \implies t+\theta \in (2n\pi, 2n\pi + \pi)$$ Hence, $$\theta \in (2n\pi - t, 2n\pi + \pi - t) = (2 n \pi -\arccos(\rho),2 n \pi + \pi -\arccos(\rho))$$

EDIT

If we want $\theta \in [0, \pi]$, since $\arccos(\rho) \in [0, \pi]$, the only possible intersection is $$\theta \in [0, \pi - \arccos(\rho))$$

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  • $\begingroup$ Thanks, but I am solving for $\theta$ not for $\rho$ $\endgroup$ – Mary May 2 '13 at 16:54
  • $\begingroup$ @Mary Updated accordingly. $\endgroup$ – user17762 May 2 '13 at 16:57
  • $\begingroup$ Thanks, how to combine your result with additional requirement that $\theta \in [0, \pi)$? $\endgroup$ – Mary May 2 '13 at 16:57

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