What is the slope of the tangent line of r=(2theta+sin2theta)/cos^2theta at theta=(-3pi)/8?

Apr 16, 2017

$\text{m"_"tan} = 16 + 8 \sqrt{2} + \left(12 + 9 \sqrt{2}\right) \pi$

Explanation:

$\frac{\mathrm{dr}}{d \theta} = \frac{\left({\cos}^{2} \theta\right) \left(2 + 2 \cos 2 \theta\right) + \left(2 \cos \theta \sin \theta\right) \left(2 \theta + \sin 2 \theta\right)}{\cos} ^ 4 \left(\theta\right)$

$r ' \left(- \frac{3}{8} \pi\right) =$
$\frac{\left({\cos}^{2} \left(- \frac{3}{8} \pi\right)\right) \left(2 + 2 \cos 2 \left(- \frac{3}{8} \pi\right)\right) + \left(2 \cos \left(- \frac{3}{8} \pi\right) \sin \left(- \frac{3}{8} \pi\right)\right) \left(2 \left(- \frac{3}{8} \pi\right) + \sin 2 \left(- \frac{3}{8} \pi\right)\right)}{\cos} ^ 4 \left(\left(- \frac{3}{8} \pi\right)\right)$

$= 16 + 8 \sqrt{2} + \left(12 + 9 \sqrt{2}\right) \pi$