# What is the slope of the line normal to the tangent line of f(x) = tanx+sin(x-pi/4)  at  x= (5pi)/8 ?

Dec 29, 2016

$- 0.1290$, nearly.

#### Explanation:

At x=5/8pi,

y= f(5/8pi)=tan(5/8pi)+sin(5/8pi-pi/4)

$= \tan \left(\frac{\pi}{2} + \frac{\pi}{8}\right) + \sin \left(- \frac{\pi}{8}\right)$

$= - \cot \left(\frac{\pi}{8}\right) - \sin \left(\frac{\pi}{8}\right) = - 2.797$, nearly

The slope of the normal at this point

$P \left(\frac{5}{8} \pi , 2.0315\right) = P \left(1.9635 , 2.0315\right)$

is -1/y/ at P

$= - \frac{1}{\tan x + \sin \left(x - \frac{\pi}{4}\right)} '$ at $x = \frac{5}{8} \pi$

=-1/(sec^2(5/8pi)+cos(pi/8)

$= - \frac{1}{{\csc}^{2} \left(\frac{\pi}{8}\right) + \cos \left(\frac{\pi}{8}\right)}$

$= - 0.1290$, nearly.

The equation to the normal at P(1.9635,2.0315) is

y-2.0315=-0.1290(x-1.9635). Simplifying,

$y = - 0.1290 x + 2.057$