# What is the equation of the tangent line of r=2cos(theta-(3pi)/4) +2sin(-theta+(3pi)/2) at theta=(-5pi)/12?

Jun 3, 2016

$y = 2.17303 - 0.767327 \left(x + 0.582262\right)$

#### Explanation:

Given $r \left(\theta\right)$ we have

{ (x(theta)=r(theta)cos(theta)), (y(theta)=r(theta)sin(theta)) :}

 { (dx/(d theta) = -r(theta)sin(theta)+cos(theta)((dr)/(d theta))), (dy/(d theta) = r(theta)cos(theta)+sin(theta)((dr)/(d theta))) :}

but

$r \left(\theta\right) = 2 \cos \left(\theta - 3 \frac{\pi}{4}\right) + 2 \sin \left(- \theta + 3 \frac{\pi}{2}\right)$

then

 { (dx/(d theta) = 2 (sin(2 theta) + sin(pi/4 + 2 theta))), (dy/(d theta)=-2 (cos(2 theta) + cos(pi/4+ 2 theta))) :}

but

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(\frac{\mathrm{dy}}{d \theta}\right)}{\left(\frac{\mathrm{dx}}{d \theta}\right)}$

at point ${\theta}_{0} = - \frac{5 \pi}{12}$ we have

${p}_{0} = \left\{r \left({\theta}_{0}\right) \cos \left({\theta}_{0}\right) , r \left({\theta}_{0}\right) \sin \left({\theta}_{0}\right)\right\} = \left\{- 0.582262 , 2.17303\right\}$

and

${m}_{0} = {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{{\theta}_{0}} = - 0.767327$

$y = 2.17303 - 0.767327 \left(x + 0.582262\right)$