# What is the equation of the tangent line of r=cos(theta+(5pi)/4) * sin(theta-(17pi)/12) at theta=(-5pi)/12?

Mar 13, 2017

Equation of tangent is $y + \frac{3 + \sqrt{3}}{8 \sqrt{2}} = - \left(\frac{38}{37} + \frac{10}{37} \sqrt{3}\right) \left(x - \left(\frac{3 - \sqrt{3}}{8 \sqrt{2}}\right)\right)$

#### Explanation:

In polar coordinates

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\sin \theta \mathrm{dr} + r \cos \theta d \theta}{\cos \theta \mathrm{dr} - r \sin \theta d \theta}$

= $\frac{\sin \theta \frac{\mathrm{dr}}{d \theta} + r \cos \theta}{\cos \theta \frac{\mathrm{dr}}{d \theta} - r \sin \theta}$

As $r = \cos \left(\theta + \frac{5 \pi}{4}\right) \sin \left(\theta - \frac{17 \pi}{12}\right)$

$\frac{\mathrm{dr}}{d \theta} = - \sin \left(\theta + \frac{5 \pi}{4}\right) \sin \left(\theta - \frac{17 \pi}{12}\right) + \cos \left(\theta + \frac{5 \pi}{4}\right) \cos \left(\theta - \frac{17 \pi}{12}\right)$

and when $\theta = \frac{- 5 \pi}{12}$ and

$r = \cos \left(\frac{- 5 \pi}{12} + \frac{5 \pi}{4}\right) \sin \left(\frac{- 5 \pi}{12} - \frac{17 \pi}{12}\right)$

= $- \cos \left(\frac{5 \pi}{6}\right) \sin \left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \times \frac{1}{2} = \frac{\sqrt{3}}{4}$

$\frac{\mathrm{dr}}{d \theta} = - \sin \left(\frac{- 5 \pi}{12} + \frac{5 \pi}{4}\right) \sin \left(\frac{- 5 \pi}{12} - \frac{17 \pi}{12}\right) + \cos \left(\frac{- 5 \pi}{12} + \frac{5 \pi}{4}\right) \cos \left(\frac{- 5 \pi}{12} - \frac{17 \pi}{12}\right)$

= $\sin \left(\frac{10 \pi}{12}\right) \sin \left(\frac{22 \pi}{12}\right) + \cos \left(\frac{10 \pi}{12}\right) \cos \left(\frac{22 \pi}{12}\right)$

= $\cos \left(\frac{22 \pi}{12} - \frac{10 \pi}{12}\right) = \cos \pi = - 1$

and $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- \sin \left(\frac{5 \pi}{12}\right) + \frac{\sqrt{3}}{4} \cos \left(\frac{5 \pi}{12}\right)}{\cos \left(\frac{5 \pi}{12}\right) + \frac{\sqrt{3}}{4} \sin \left(\frac{5 \pi}{12}\right)}$

= (-(sqrt3+1)/(2sqrt2)+sqrt3/4((sqrt3-1)/(2sqrt2)))/((sqrt3-1)/(2sqrt2)+sqrt3/4((sqrt3+1)/(2sqrt2))

= $\frac{- 4 \sqrt{3} - 4 + 3 - \sqrt{3}}{4 \sqrt{3} - 4 + 3 = \sqrt{3}}$

= $\frac{- 5 \sqrt{3} - 1}{5 \sqrt{3} - 1} = - \frac{38}{37} - \frac{10}{37} \sqrt{3}$

and equation of tangent is $y - r \sin \theta = \frac{\mathrm{dy}}{\mathrm{dx}} \left(x - r \cos \theta\right)$ or

$y - \frac{\sqrt{3}}{4} \sin \left(\frac{- 5 \pi}{12}\right) = - \left(\frac{38}{37} + \frac{10}{37} \sqrt{3}\right) \left(x - \frac{\sqrt{3}}{4} \cos \left(\frac{- 5 \pi}{12}\right)\right)$

or $y + \frac{\sqrt{3}}{4} \left(\frac{\sqrt{3} + 1}{2 \sqrt{2}}\right) = - \left(\frac{38}{37} + \frac{10}{37} \sqrt{3}\right) \left(x - \frac{\sqrt{3}}{4} \left(\frac{\sqrt{3} - 1}{2 \sqrt{2}}\right)\right)$

or $y + \frac{3 + \sqrt{3}}{8 \sqrt{2}} = - \left(\frac{38}{37} + \frac{10}{37} \sqrt{3}\right) \left(x - \left(\frac{3 - \sqrt{3}}{8 \sqrt{2}}\right)\right)$