Find points on curve $r=2\sin\theta$ where the tangent line is parallel to the ray $\theta = \pi/4$

Question

I was thinking to convert to cartesian coordinates and then find when the slope of the tangent line is $1$, but I get a messy equation $2\cos^2\theta -2\sin^2\theta=4\sin^2\theta\cos\theta$ I was wondering if there was an easy way as it is hard to get values from this.

Edit: The equation ends up simplifying to $\tan(2\theta) = 1$, but for future reference is this the best method?

Answer 1

This is the equation of the circle of center $(0,1)$ and radius $1$.

$$r = 2\sin(\theta) \iff r = 2(\frac{y}{r}) \iff r^2 = 2y \iff x^2 + y^2 = 2y \iff x^2 + (y - 1)^2 = 1$$

Differentiate $x^2 + y^2 = 2y$ with respect to $x$..

Answer 2

The curve is $\alpha(\theta)=(2sin{\theta}\cos{\theta},2\sin{\theta}\sin{\theta})$. The velocity vector is $\alpha'(\theta)=(2\cos{2\theta},2\sin{2\theta})$. Now we have to find all points so that this vector is proportional to $(1,1)$.

$\theta=\dfrac{\pi}{8}$ or $\theta=\dfrac{5\pi}{8}$