# What is the slope of the line normal to the tangent line of f(x) = sin(2x-pi/6)  at  x= pi/3 ?

May 18, 2017

No slope

#### Explanation:

Find the derivative:

$f \left(x\right) = \sin \left(2 x - \frac{\pi}{6}\right)$
$f ' \left(x\right) = \cos \left(2 x - \frac{\pi}{6}\right) x$

Substitute $x = \frac{\pi}{3}$ to find the slope
$f ' \left(\frac{\pi}{3}\right) = \cos \left(2 \left(\frac{\pi}{3}\right) - \frac{\pi}{6}\right) \left(\frac{\pi}{3}\right) = \cos \left(\frac{\pi}{2}\right) \left(\frac{\pi}{3}\right) = 0$

The slope of a normal line is negative reciprocal of the derivative.
$- \frac{1}{0}$ is undefined then the it must be a vertical line.