TITLE: Find the equation of the tangent line to the polar curve at given $(x,y)$.
QUESTION [5 upvotes]: Find the equation(s) of the tangent line(s) to the curve given $(x,y)$ point.
$$r=1-2\sin(\theta )$$ at $(0,0)$.  I am not sure how to go about find the the tangent line. Do I need to convert from polar to rectangular?
Thanks!

REPLY [2 votes]: Let's use the polar form for the slope, which can be derived using the product rule as follows:
$$\frac{dy}{dx}=\frac{d(r\sin\theta)}{d(r\cos\theta)}=\frac{r'(\theta)\sin\theta+r(\theta)\cos\theta}{r'(\theta)\cos\theta-r(\theta)\sin\theta}$$
Now since $r(\theta)=1-2\sin\theta$, we have $r'(\theta)=-2\cos\theta$, so our slope becomes:
$$\frac{dy}{dx}=\frac{\cos\theta-4\sin\theta\cos\theta}{2\sin^2\theta-2\cos^2\theta-\sin\theta}=\frac{\cos\theta-2\sin2\theta}{4\sin^2\theta-\sin\theta-2}$$
Now $(0,0)$ corresponds to angles $\theta=\frac{\pi}{6},\frac{5\pi}{6}$, and plugging these in gives two slopes:
$$m=\frac{\sqrt{3}}{3},\;\frac{-\sqrt{3}}{3}$$
corresponding to the lines
$$y=\pm\frac{\sqrt{3}}{3}x$$