# What is the equation of the tangent line of r=4cos(-theta+(3pi)/2) +sin(-theta) at theta=(-5pi)/12?

Tangent line $\textcolor{red}{y + \frac{5}{4} \left(2 + \sqrt{3}\right) = \frac{1}{\sqrt{3}} \left(x - \frac{5}{4}\right) \text{ }}$

#### Explanation:

We have some long solution here. I will try to shorten it.
From the given $r = 4 \cos \left(\frac{3 \pi}{2} - \theta\right) + \sin \left(- \theta\right)$
this equation is reducible to
$r = 4 \left[\cos \left(\frac{3 \pi}{2}\right) \cos \left(\theta\right) + \sin \left(\frac{3 \pi}{2}\right) \sin \theta\right] - \sin \theta$

also because $\sin \left(- \theta\right) = - \sin \theta$

$r = 4 \left[\cos \left(\frac{3 \pi}{2}\right) \cos \left(\theta\right) + \sin \left(\frac{3 \pi}{2}\right) \sin \theta\right] - \sin \theta$

$r = 4 \left[0 \cdot \cos \theta + \left(- 1\right) \sin \theta\right] - \sin \theta$
$r = - 4 \sin \theta - \sin \theta$
$r = - 5 \sin \theta$

Solve for the point of tangency: given $\theta = \frac{- 5 \pi}{12}$
$r = - 5 \sin \theta = - 5 \sin \left(\frac{- 5 \pi}{12}\right) = \frac{5}{4} \left(\sqrt{6} + \sqrt{2}\right)$

The polar coordinate of the point is
$\left(r , \theta\right) = \left(\frac{5}{4} \left(\sqrt{6} + \sqrt{2}\right) , \frac{- 5 \pi}{12}\right)$

We will use rectangular coordinates for convenience
$x = r \cdot \cos \theta$ and $y = r \cdot \sin \theta$

Solve x

$x = r \cdot \cos \theta$
$x = \frac{5}{4} \left(\sqrt{6} + \sqrt{2}\right) \cdot \cos \left(\frac{- 5 \pi}{12}\right)$
$x = \frac{5 \left(\sqrt{6} + \sqrt{2}\right)}{4} \frac{\left(\sqrt{6} - \sqrt{2}\right)}{4}$

$x = \frac{5}{4}$

Solve y

$y = r \cdot \sin \theta$
$x = \frac{5}{4} \left(\sqrt{6} + \sqrt{2}\right) \cdot \sin \left(\frac{- 5 \pi}{12}\right)$
$x = \frac{5}{4} \left(\sqrt{6} + \sqrt{2}\right) \left(- \frac{\sqrt{6} + \sqrt{2}}{4}\right)$
$x = - \frac{5}{16} \left(8 + 4 \sqrt{3}\right)$

$x = - \frac{5}{4} \left(2 + \sqrt{3}\right)$

We now have the rectangular coordinate point
$\left(x , y\right) = \left(\frac{5}{4} , - \frac{5}{4} \left(2 + \sqrt{3}\right)\right)$

We need the slope $m$ on the curve $r = - 5 \sin \theta$
For convenience , we convert this to rectangular coordinate system
$r = - 5 \sin \theta$
$\sqrt{{x}^{2} + {y}^{2}} = - 5 \cdot \frac{y}{\sqrt{{x}^{2} + {y}^{2}}}$
Simplify
${\left(\sqrt{{x}^{2} + {y}^{2}}\right)}^{2} = - 5 \cdot \frac{y}{\cancel{\sqrt{{x}^{2} + {y}^{2}}}} \cdot \cancel{\sqrt{{x}^{2} + {y}^{2}}}$

${x}^{2} + {y}^{2} = - 5 y$

${x}^{2} + {y}^{2} + 5 y = 0$

Solve for m by differentiating implicitly

${x}^{2} + {y}^{2} + 5 y = 0$
$2 x + 2 y y ' + 5 y ' = 0$
$y ' \left(2 y + 5\right) = - 2 x$
$y ' = \frac{- 2 x}{2 y + 5}$

$m = \frac{- 2 x}{2 y + 5} = \frac{- 2 \left(\frac{5}{4}\right)}{2 \left(- \frac{5}{4} \left(2 + \sqrt{3}\right)\right) + 5}$

$m = \frac{- \left(\frac{5}{2}\right)}{\left(- \frac{5}{2} \left(2 + \sqrt{3}\right)\right) + 5}$

$m = \frac{1}{\sqrt{3}}$
Finally, let us determine the tangent line: using Two-Point Form

$y - {y}_{1} = m \left(x - {x}_{1}\right)$
$y - - \frac{5}{4} \left(2 + \sqrt{3}\right) = \frac{1}{\sqrt{3}} \left(x - \frac{5}{4}\right)$

$\textcolor{red}{y + \frac{5}{4} \left(2 + \sqrt{3}\right) = \frac{1}{\sqrt{3}} \left(x - \frac{5}{4}\right) \text{ }}$The Tangent Line

kindly see the graph of $r = 4 \left[\cos \left(\frac{3 \pi}{2}\right) \cos \left(\theta\right) + \sin \left(\frac{3 \pi}{2}\right) \sin \theta\right] - \sin \theta$ and $y + \frac{5}{4} \left(2 + \sqrt{3}\right) = \frac{1}{\sqrt{3}} \left(x - \frac{5}{4}\right)$the Tangent Line

graph{(y+5/4(2+sqrt3)-1/sqrt3(x-5/4))(x^2+y^2+5y)=0[-20,20,-10,10]}

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