# What is the slope of the tangent line of r=5theta+cos(-theta/3-(pi)/2) at theta=(-pi)/6?

$r \left(\setminus \theta\right) = 5 \setminus \theta + \cos \left(- \setminus \frac{\theta}{3} - \setminus \frac{\pi}{2}\right)$
$r ' \left(\setminus \theta\right) = \frac{\mathrm{dr} \left(\setminus \theta\right)}{d \setminus \theta} = 5 + \frac{1}{3} \sin \left(- \setminus \frac{\theta}{3} - \setminus \frac{\pi}{2}\right)$
Assessing at $\setminus \theta = - \frac{\pi}{6}$:
$r ' \left(- \frac{\pi}{6}\right) = 5 + \frac{1}{3} \sin \left(- \setminus \frac{\pi}{18} - \setminus \frac{\pi}{2}\right)$
$r ' \left(- \frac{\pi}{6}\right) = 5 + \frac{1}{3} \sin \left(- \frac{5 \setminus \pi}{9}\right) \setminus \approx 4.67$