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

Question

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 $\frac{3\pi}2$.

Answer 1

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 $y=-\frac34$.

Is this what you wanted?