# What is the slope of the line normal to the tangent line of f(x) = sec^3x+sin(2x-(3pi)/8)  at  x= (11pi)/8 ?

Jul 10, 2018

Slope of the normal is 1/0.8104=1.2339

#### Explanation:

$f \left(x\right) = {\sec}^{3} x + \sin \left(2 x - \frac{3 \pi}{8}\right)$

$x = \frac{11 \pi}{8}$

$f ' \left(x\right) = 3 {\sec}^{2} x \sec x \tan x + \cos \left(2 x - \frac{3 \pi}{8}\right) \times 2$

$f ' \left(x\right) = 3 {\sec}^{3} x \sec x + 2 \cos \left(2 x - \frac{3 \pi}{8}\right)$

$f ' \left(\frac{11 \pi}{8}\right) = 3 {\sec}^{3} \left(\frac{11 \pi}{8}\right) \tan \left(\frac{11 \pi}{8}\right) + 2 \cos \left(2 \left(\frac{11 \pi}{8}\right) - \frac{3 \pi}{8}\right)$

$= 3 {\sec}^{3} \left(\frac{\pi}{8}\right) \left(- \tan \left(\frac{\pi}{8}\right) + 2 \cos \left(\frac{19 \pi}{8}\right)\right)$

=-3sec^3(pi/8)tan(pi/8)+2cos((3pi)/8))

$= - 0.8104$
Slope of the tangent is -0.8104

Slope of the normal is 1/0.8104=1.2339