How to find third horizontal tangent to $r=1-\sin\theta$

Question

Given the curve $r=1-\sin\theta, $ I'm supposed to find the horizontal tangents. I took the derivative of x and y to find that...

$$dy/dx = \frac{\cos\theta-2\cos\theta \sin\theta}{\sin^2\theta-\cos^2\theta-\sin\theta}$$

I set the derivative equal to zero and get

$$0=\cos\theta-2\cos\theta \sin\theta$$

$$2\cos\theta \sin\theta = \cos \theta$$

Dividing by $\cos$ gives

$$2\sin\theta=1$$ which has two solutions between $0$ and $2\pi$.

These appear correct when I graph it, too, as $\theta = \pi/6, 5\pi/6$ both point to horizontal tangents.

But it's obvious that $3\pi/2$ is also a horizontal tangent. How would I have known this from the math, without looking at the graph?

Answer 1

Dividing by $\cos(\theta)$: Note the $\cos(\theta)$ could be zero, in which case it doesn't make sense to straight away divide $\cos(\theta)$.

There so there should be some case analysis:

$\text{Case } 1: \cos{\theta} \neq 0$

Then $$2\sin(\theta) = 1 \\ \theta = \dfrac{\pi}{6}, \dfrac{5\pi}{6}$$

$\text{Case } 2: \cos{\theta} = 0$

Then $$\theta = \dfrac{\pi}{2}, \dfrac{3\pi}{2}$$ However, by substituting into the equation for the derivative, the denominator, $\sin^2\theta−\cos^2\theta−\sin\theta$ becomes $0$ when $\theta = \dfrac{\pi}{2}$. Hence, the derivative does not exist at $\theta = \dfrac{\pi}{2}$.

Therefore we dismiss the root $\dfrac{\pi}{2}$, so the roots to the derivative, and hence the horizontal tangents to the curve are: $$\theta = \dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{3\pi}{2}$$.