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

Jan 27, 2017

$r \sin \left({\theta}^{o} - {20.3}^{o}\right) = 8.74 \sin \left({99.7}^{o}\right)$

#### Explanation:

I recall that, I have given here the formula, for the polar equation io

the tangent at $P \left(a , \alpha\right)$, with slope

$m = \tan \psi = \frac{r ' \sin \theta + r \cos \theta}{r ' \cos \theta - r \sin \theta}$,

evaluated at $\left(a , \alpha\right)$, as

$r \sin \left(\theta - \psi\right) = a \sin \left(\alpha - \psi\right)$

Here,

$r = 5 \theta + 2 \sin \left(4 \theta + \frac{2}{3} \pi\right)$

$\alpha = \frac{2}{3} \pi = 2.0944 r a \mathrm{di} a n = {120}^{o}$

$a = \frac{10}{3} \pi - \sqrt{3} = 8.74$

$r ' = 5 = 8 \cos \left(4 \theta + \frac{2}{3} \pi\right) = 1 ,$ at P

$m = 0.3400 \mathmr{and} \psi = {20.3}^{o}$.

Now, the equation to the tangent at $P \left(8.74 , {120}^{o}\right)$ is

$r \sin \left({\theta}^{o} - {20.3}^{o}\right) = 8.74 \sin \left({120}^{o} - {20.3}^{o}\right) = 8.74 \sin \left({99.7}^{o}\right)$