# What is the equation of the tangent line of r=2cos(theta-pi/2) + sin(2theta-pi) at theta=pi/4?

Feb 16, 2017

$r \sin \left(\theta - {61.325}^{o}\right) = - \left(\sqrt{2} - 1\right) \sin \left({16.325}^{o}\right)$. The Cartesian
form is $0.877 x - 0.480 y - 0.116 = 0$. See Socratic depiction.

#### Explanation:

Use $r \sin \left(\theta - \psi\right) = a \sin \left(\alpha - \psi\right)$, for the equation to the

tangent at $P \left(a , \alpha\right)$, where the slope of the tangent

$m = \tan \psi = \frac{r ' \sin \theta + r \cos \theta}{r ' \cos \theta - r \sin \theta}$.

Here,

$r = 2 \sin \theta - \sin 2 \theta$

At $\theta = \frac{\pi}{4} = \alpha , r = a = \sqrt{2} - 1$.

So, the point of contact P of the tangent is $\left(\sqrt{2} - 1 , {45}^{o}\right)$.

$r ' = 2 \cos \theta - 2 \cos 2 \theta = \sqrt{2}$, at P.

The slope of the tangent at P is

$m = \tan \psi = \frac{\left(\sqrt{2}\right) \left(\frac{1}{\sqrt{2}}\right) + \left(\sqrt{2} - 1\right) \left(\frac{1}{\sqrt{2}}\right)}{\left(\sqrt{2}\right) \left(\frac{1}{\sqrt{2}}\right) - \left(\sqrt{2} - 1\right) \left(\frac{1}{\sqrt{2}}\right)}$

$= 2 \sqrt{2} - 1$

$\psi = \arctan \left(2 \sqrt{2} - 1\right) = {61.325}^{o}$.

So, the equation to the tangent is

$r \sin \left(\theta - {61.325}^{o}\right) = - \left(\sqrt{2} - 1\right) \sin \left({16.325}^{o}\right)$

Expanding and using $\left(x , y\right) = r \left(\cos \theta , \sin \theta\right)$, the Cartesian

form is

$0.877 x - 0.480 y - 0.116 = 0$
.

graph{(sqrt(x^2+y^2)(x^2+y^2-2y)+2xy)(0.877x-0.48y-0.116)((x-.293)^2+(y-.293)^2-.009)=0 [-5, 5, -2.5, 2.5]}

The Cartesian form is used for the Socratic graph.