# What is the equation of the normal line of f(x)= cotx at x = pi/8?

Jan 10, 2018

$y - \left(\sqrt{2} + 1\right) = \frac{2 - \sqrt{2}}{4} \left(x - \frac{\pi}{8}\right)$

#### Explanation:

To find the normal line, first find the slope of the tangent line of $f \left(x\right) = \cot x$ at $x = \frac{\pi}{8}$.

This can be found by finding the derivative of $f \left(x\right) = \cot x$, which is $f ' \left(x\right) = - {\left(\csc x\right)}^{2}$

$f ' \left(\frac{\pi}{8}\right)$ is the slope of the tangent line of $f \left(x\right) = \cot x$ at $x = \frac{\pi}{8}$.
$f ' \left(\frac{\pi}{8}\right) = - {\left(\csc \left(\frac{\pi}{8}\right)\right)}^{2}$
$f ' \left(\frac{\pi}{8}\right) = - {\left(\frac{2}{\sqrt{2 - \sqrt{2}}}\right)}^{2}$ csc(pi/8)
$f ' \left(\frac{\pi}{8}\right) = - \frac{4}{2 - \sqrt{2}}$

the slope of the normal like is the opposite reciprocal of that, which would be:
$- \frac{1}{- \frac{4}{2 - \sqrt{2}}}$

$= \frac{1}{\frac{4}{2 - \sqrt{2}}}$

$= \frac{2 - \sqrt{2}}{4}$

the point the normal line passes through is $\left(\frac{\pi}{8} , f \left(\frac{\pi}{8}\right)\right)$
$f \left(\frac{\pi}{8}\right) = \cot \left(\frac{\pi}{8}\right)$
$= \sqrt{2} + 1$ (cot(pi/8)

use point-slope form:
$y - \left(\sqrt{2} + 1\right) = \frac{2 - \sqrt{2}}{4} \left(x - \frac{\pi}{8}\right)$