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

Feb 8, 2018

$f ' \left(\frac{- 5 \pi}{3}\right) = \frac{3 \sqrt{3}}{2}$

#### Explanation:

Now, remember the chain rule and the fact that $\frac{d}{\mathrm{dx}} \left(\cos x\right) = - \sin x$

The chain rule states that:
If $f \left(x\right) = g \left(h \left(x\right)\right)$, then $f ' \left(x\right) = g ' \left(h \left(x\right)\right) \cdot h ' \left(x\right)$

Since $r = \cos \left(- 3 \theta - \frac{2 \pi}{3}\right)$, we have:

$f \left(x\right) = \cos \left(- 3 x - \frac{2 \pi}{3}\right)$

=>$f ' \left(x\right) = - \sin \left(- 3 x - \frac{2 \pi}{3}\right) \cdot \frac{d}{\mathrm{dx}} \left(- 3 x - 2 \frac{\pi}{3}\right)$

=>$f ' \left(x\right) = - \sin \left(- 3 x - \frac{2 \pi}{3}\right) \cdot - 3$

=>$f ' \left(x\right) = 3 \sin \left(- 3 x - \frac{2 \pi}{3}\right)$

=>$f ' \left(\frac{- 5 \pi}{3}\right) = 3 \sin \left(- 3 \cdot \frac{- 5 \pi}{3} - \frac{2 \pi}{3}\right)$

=>$f ' \left(\frac{- 5 \pi}{3}\right) = 3 \sin \left(- 1 \cdot \frac{- 5 \pi}{1} - \frac{2 \pi}{3}\right)$

=>$f ' \left(\frac{- 5 \pi}{3}\right) = 3 \sin \left(- \left(- 5 \pi\right) - \frac{2 \pi}{3}\right)$

=>$f ' \left(\frac{- 5 \pi}{3}\right) = 3 \sin \left(5 \pi - \frac{2 \pi}{3}\right)$

=>$f ' \left(\frac{- 5 \pi}{3}\right) = 3 \sin \left(\frac{13 \pi}{3}\right)$

=>$f ' \left(\frac{- 5 \pi}{3}\right) = 3 \sin \left(4 \frac{\pi}{3}\right)$

Using our special angles, we know that $\sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$

We have:

=>$f ' \left(\frac{- 5 \pi}{3}\right) = 3 \cdot \frac{\sqrt{3}}{2}$

=>$f ' \left(\frac{- 5 \pi}{3}\right) = \frac{3 \sqrt{3}}{2}$