Polar tangents of $x = cos(3 \theta)$, $y = 5sin(\theta)$ Find the points on the curve where the tangent is horizontal or vertical.
$x = \cos(3 \theta)$ , $y = 5 \sin(\theta)$ 
I have done quite a few of these problems and this was the first one that I keep getting wrong and coming back with same answer. 
Horizontal tangents: $(0,-5) , (0,5)$
Vertical tangent $(-1,0), (1,0)$ 
Any ideas? 
 A: In the $xy$ Cartesian coordinate system, horizontal tangents occur when $\frac{dy}{dx} = 0$ and vertical tangents occur when $\frac{dx}{dy} = 0$. Since $x$ and $y$ are given separately as functions of $\theta$, it'll likely be simplest and easiest to determine these derivatives implicitly by using, for $\frac{dy}{dx}$, the fraction of $\frac{dy}{d\theta}$ divided by $\frac{dx}{d\theta}$, with $\frac{dx}{dy}$ being the reciprocal. We thus get that
$$\frac{dy}{d\theta} = \frac{d\left(5\sin\left(\theta\right)\right)}{d\theta} = 5\cos\left(\theta\right) \tag{1}\label{eq1}$$
$$\frac{dx}{d\theta} = \frac{d\left(\cos\left(3\theta\right)\right)}{d\theta} = -3\sin\left(3\theta\right) \tag{2}\label{eq2}$$
Thus, for example,
$$\frac{dy}{dx} = \frac{5\cos\left(\theta\right)}{-3\sin\left(3\theta\right)} \tag{3}\label{eq3}$$
so $\frac{dx}{dy} = \frac{-3\sin\left(3\theta\right)}{5\cos\left(\theta\right)}$. Because you're looking for values where the derivative is $0$, you only need to check the numerator being $0$ and ensure the denominator is not $0$. However, note you need to consider all valid values where the numerator is $0$, i.e., using that the $0$ values for $\sin$ and $\cos$ are periodic with a period of $\pi$ radians. After doing this, you should have more results than what you've indicated. I trust you can finish the rest yourself.
