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

Jul 25, 2017

$m = \frac{2}{\sqrt{3}} = \frac{2}{3} \sqrt{3}$

#### Explanation:

Let's first find the derivative of the function using the product rule :

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

The slope of the tangent line is $f ' \left(\frac{\pi}{3}\right)$

$f ' \left(\frac{\pi}{3}\right) = - \sin \left(\frac{\pi}{3}\right) \sin \left(\frac{2 \pi}{3} - \frac{\pi}{6}\right) + 2 \cos \left(\frac{\pi}{3}\right) \cos \left(\frac{2 \pi}{3} - \frac{\pi}{6}\right) =$

$- \frac{\sqrt{3}}{2} \sin \left(\frac{\pi}{2}\right) + 2 \cdot \frac{1}{2} \cos \left(\frac{\pi}{2}\right) = - \frac{\sqrt{3}}{2}$

So now the slope of the normal line to the tangent is :

$m = - \frac{1}{f ' \left(\frac{\pi}{3}\right)} = \frac{2}{\sqrt{3}} = \frac{2}{3} \sqrt{3}$