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

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?

• What do you mean by the ray $\pi/4$? Commented Apr 6, 2015 at 21:00
• $\theta = \pi/4$ sorry
– Sean
Commented Apr 6, 2015 at 21:00
• I guess... There isn't a much simpler formula for the tangent in terms of polar coordinates. See here too: tutorial.math.lamar.edu/Classes/CalcII/PolarTangents.aspx Commented Apr 6, 2015 at 21:03

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 $\theta=\dfrac{5\pi}{8}$
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 $x$..