# $r=1-\sin(\theta)$ horizontal and vertical tangents

I am having problems determining the horizontal and vertical tangents for $$r=1-\sin(\theta)$$.

I thought the tangent lines occurred at $$\frac{5\pi}6$$, $$\frac\pi6$$, $$\frac{3\pi}2$$ while the vertical tangent lines occurred at $$\frac{11\pi}6$$ and $$\frac{7\pi}6$$.

I solved for $$x$$ to get $$x=\cos(\theta)-\sin(\theta)\cdot\cos(\theta)$$ which produced $$-\frac{\sqrt3}4$$ for $$\frac{5\pi}6$$, $$\frac{\sqrt3}4$$ for $$\frac\pi6$$, $$0$$, and $$\frac{3\pi}2$$.

• It seems completely right so far. You just need to plug in the rest of the numbers and get all the relevant $x$ and $y$ values. Where do you have a problem? Nov 5, 2019 at 0:09
• At first, I was plugging them into my x/y equations which caused a problem. However, I figured out plugging the values into the original equation generated the correct r values. Now, I can't seem to get the vertical tangents correct. Nov 5, 2019 at 0:20

Vertical tangents: Your values of $$\theta$$ are correct, so we can find the $$x$$ and $$y$$ coordinates of the intersections by just plugging into $$x=\cos\theta(1-\sin\theta)$$ and $$y=\sin\theta(1-\sin\theta)$$. That gives $$(x,y) = (\pm \frac{3\sqrt{3}}{4},-\frac{3}{4})$$ So the tangent lines are given by $$x=\pm \frac{3\sqrt{3}}{4}$$, and they intersect the curve at $$y=-\frac34$$.
• I think they are correct. Except there there is actually also a vertical tangent of sorts at the cusp at $(x,y)=(0,0)$, $(r,\theta)=(0,\frac\pi 2)$. Nov 5, 2019 at 1:29