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

Tangent line $\textcolor{b l u e}{y - \frac{\sqrt{3} \pi}{3} + \frac{9}{4} = \left(\frac{3 \sqrt{3} - 2 \pi}{15 - 2 \pi \sqrt{3}}\right) \cdot \left(x - \frac{3 \sqrt{3}}{4} + \frac{\pi}{3}\right)}$
$\textcolor{b l u e}{y + 0.4362006358 = - 0.2640221212 \left(x - 0.2518405545\right)}$

#### Explanation:

From the given equation $r = \theta - 3 \cdot \sin \left(\theta + \frac{5 \pi}{3}\right)$
solve $r$ when $\theta = \frac{2 \pi}{3}$

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

$\textcolor{red}{r = \frac{2 \pi}{3} - \frac{3 \sqrt{3}}{2}}$

Solve for the slope $m$ with the formula

color(blue)(m=(r' sin theta+r*cos theta)/(r' cos theta-r*sin theta)

$r ' = \frac{\mathrm{dr}}{d \theta} = 1 - 3 \cos \left(\theta + \frac{5 \pi}{3}\right)$

$m =$
(((1-3 cos (theta+(5pi)/3)) sin theta+(theta-3*sin(theta+(5pi)/3))*cos theta))/(((1-3 cos (theta+(5pi)/3) )*cos theta-(theta-3*sin(theta+(5pi)/3))*sin theta)

with the value $\theta = \frac{2 \pi}{3}$

$m =$
(((1-3 cos ((2pi)/3+(5pi)/3)) sin ((2pi)/3)+((2pi)/3-3*sin((2pi)/3+(5pi)/3))*cos ((2pi)/3)))/(((1-3 cos ((2pi)/3+(5pi)/3) )*cos ((2pi)/3)-((2pi)/3-3*sin((2pi)/3+(5pi)/3))*sin ((2pi)/3))

$m =$
(((1-3 cos ((7pi)/3)) sin ((2pi)/3)+((2pi)/3-3*sin((7pi)/3))*cos ((2pi)/3)))/(((1-3 cos ((7pi)/3) )*cos ((2pi)/3)-((2pi)/3-3*sin((7pi)/3))*sin ((2pi)/3))

m=((1-3/2) (sqrt3/2)+((2pi)/3-(3sqrt3)/2)(-1/2))/((1-3/2 )*(-1/2)-((2pi)/3-(3*sqrt3)/2)((sqrt3)/2)

$m = \frac{3 \sqrt{3} - 2 \pi}{15 - 2 \pi \sqrt{3}}$

convert $\left(r , \theta\right)$ to $\left(x , y\right)$

${x}_{1} = r \cos \theta = \left(\frac{2 \pi}{3} - \frac{3 \sqrt{3}}{2}\right) \cos \left(\frac{2 \pi}{3}\right) = \frac{3 \sqrt{3}}{4} - \frac{\pi}{3}$

${x}_{1} = \frac{3 \sqrt{3}}{4} - \frac{\pi}{3}$

${y}_{1} = r \sin \theta = \left(\frac{2 \pi}{3} - \frac{3 \sqrt{3}}{2}\right) \sin \left(\frac{2 \pi}{3}\right) = \frac{\sqrt{3} \cdot \pi}{3} - \frac{9}{4}$

${y}_{1} = \frac{\sqrt{3} \cdot \pi}{3} - \frac{9}{4}$

Solve the tangent line using $m$ and $\left({x}_{1} , {y}_{1}\right)$

$y - {y}_{1} = m \left(x - {x}_{1}\right)$

$y - \frac{\sqrt{3} \cdot \pi}{3} + \frac{9}{4} = \left(\frac{3 \sqrt{3} - 2 \pi}{15 - 2 \pi \sqrt{3}}\right) \left(x - \frac{3 \sqrt{3}}{4} + \frac{\pi}{3}\right)$

Kindly see the graph of $r = \theta - 3 \cdot \sin \left(\theta + \frac{5 \pi}{3}\right)$ and tangent line $y - \frac{\sqrt{3} \cdot \pi}{3} + \frac{9}{4} = \left(\frac{3 \sqrt{3} - 2 \pi}{15 - 2 \pi \sqrt{3}}\right) \left(x - \frac{3 \sqrt{3}}{4} + \frac{\pi}{3}\right)$

God bless...I hope the explanation is useful.