# What is the slope of the line normal to the tangent line of f(x) = secx-cos^2(x-pi)  at  x= (pi)/12 ?

Dec 21, 2017

$3 \sqrt{6} - 5 \sqrt{2} + \frac{1}{2}$

#### Explanation:

the answer is $f ' \left(\frac{\pi}{12}\right)$ where $f ' \left(x\right)$ is the derivative of $f \left(x\right)$

first simplify the $- {\left(\cos \left(x - \pi\right)\right)}^{2}$ part:
$= - {\left(- \cos \left(x\right)\right)}^{2}$ proof
$= - {\left(\cos x\right)}^{2}$

so $f \left(x\right) = \sec x - {\left(\cos x\right)}^{2}$
$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left(\sec x\right) - \frac{d}{\mathrm{dx}} \left({\left(\cos x\right)}^{2}\right)$
$f ' \left(x\right) = \sec x \tan x - \left(2 \cos x \cdot \frac{d}{\mathrm{dx}} \left(\cos x\right)\right)$ (secx derivative and chain rule)
$f ' \left(x\right) = \sec x \tan x - 2 \cos x \left(- \sin x\right)$ (cosx derivative)
$f ' \left(x\right) = \sec x \tan x + 2 \sin x \cos x$
$f ' \left(x\right) = \sec x \tan x + \sin \left(2 x\right)$ (double angle formula for sin(2x))

now evaluate at $x = \frac{\pi}{12}$
$f ' \left(\frac{\pi}{12}\right) = \sec \left(\frac{\pi}{12}\right) \tan \left(\frac{\pi}{12}\right) + \sin \left(2 \cdot \left(\frac{\pi}{12}\right)\right)$

use a calculator, memorize these, or use this table to evaluate:

$f ' \left(\frac{\pi}{12}\right) = \left(\sqrt{6} - \sqrt{2}\right) \cdot \left(2 - \sqrt{3}\right) + \sin \left(\frac{\pi}{6}\right)$
$f ' \left(\frac{\pi}{12}\right) = 2 \sqrt{6} - 2 \sqrt{2} - \sqrt{18} + \sqrt{6} + \frac{1}{2}$
$f ' \left(\frac{\pi}{12}\right) = 3 \sqrt{6} - 2 \sqrt{2} - 3 \sqrt{2} + \frac{1}{2}$
$f ' \left(\frac{\pi}{12}\right) = 3 \sqrt{6} - 5 \sqrt{2} + \frac{1}{2}$