# What is the slope of the line normal to the tangent line of f(x) = xtanx-xsin(x-pi/3)  at  x= (15pi)/8 ?

Jul 14, 2018

Slope of the normal is $- 0.15$

#### Explanation:

$f \left(x\right) = x \tan x - x \sin \left(x - \frac{\pi}{3}\right)$ at $x = \frac{15 \pi}{8} \approx 5.89 \left(2 \mathrm{dp}\right)$

Slope of the tangent is

${f}^{'} \left(x\right) = x {\sec}^{2} x + \tan x - \left\{x \cos \left(x - \frac{\pi}{3}\right) + \sin \left(x - \frac{\pi}{3}\right)\right\}$

${f}^{'} \left(5.89\right) = 5.89 \cdot {\sec}^{2} 5.89 + \tan 5.89 - \left\{5.89 \cdot \cos \left(5.89 - \frac{\pi}{3}\right) + \sin \left(5.89 - \frac{\pi}{3}\right)\right\}$ or

${f}^{'} \left(5.89\right) = 5.89 \cdot {\sec}^{2} 5.89 + \tan 5.89 - \left\{5.89 \cdot \cos 4.84 + \sin 4.84\right\}$ or

${f}^{'} \left(5.89\right) \approx 6.71$, slope of the tangent is ${m}_{t} \approx 6.71$

Let slope of the normal be ${m}_{n} \therefore {m}_{t} \cdot {m}_{n} = - 1$

or ${m}_{n} = - \frac{1}{m} _ t = - \frac{1}{6.71} \approx - 0.15$

Slope of the normal is $- 0.15$ [Ans]