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

The slope of the line normal to the tangent line
m=1/((1+sqrt(2)/2)sqrt(2+sqrt2)+((3sqrt2)/2+1)sqrt(2-sqrt2)
$m = 0.18039870004873$

#### Explanation:

From the given:
$y = \sec x + \sin \left(2 x - \frac{3 \pi}{8}\right)$ at $\text{ } x = \frac{11 \pi}{8}$

Take the first derivative $y '$

$y ' = \sec x \cdot \tan x \cdot \frac{\mathrm{dx}}{\mathrm{dx}} + \cos \left(2 x - \frac{3 \pi}{8}\right) \left(2\right) \frac{\mathrm{dx}}{\mathrm{dx}}$

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

Take note: that by $\textcolor{B l u e}{\text{Half-Angle formulas}}$, the following are obtained
$\sec \left(\frac{11 \pi}{8}\right) = - \sqrt{2 + \sqrt{2}} - \sqrt{2 - \sqrt{2}}$

$\tan \left(\frac{11 \pi}{8}\right) = \sqrt{2} + 1$

and

$2 \cdot \cos \left(2 x - \frac{3 \pi}{8}\right) = 2 \cdot \cos \left(\frac{19 \pi}{8}\right)$
$= 2 \cdot \left(\frac{\sqrt{2}}{4}\right) \left(\sqrt{2 + \sqrt{2}} - \sqrt{2 - \sqrt{2}}\right)$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
continuation

$y ' = \left(- \sqrt{2 + \sqrt{2}} - \sqrt{2 - \sqrt{2}}\right) \left(\sqrt{2} + 1\right)$
$+ 2 \cdot \left(\frac{\sqrt{2}}{4}\right) \left(\sqrt{2 + \sqrt{2}} - \sqrt{2 - \sqrt{2}}\right)$

$y ' = - \left(\sqrt{2} + 1\right) \sqrt{2 + \sqrt{2}} - \left(\sqrt{2} + 1\right) \sqrt{2 - \sqrt{2}}$
$+ \frac{\sqrt{2}}{2} \cdot \sqrt{2 + \sqrt{2}} - \frac{\sqrt{2}}{2} \cdot \sqrt{2 - \sqrt{2}}$

further simplification

$y ' = \left(- 1 - \frac{\sqrt{2}}{2}\right) \sqrt{2 + \sqrt{2}} + \left(\frac{- 3 \sqrt{2}}{2} - 1\right) \sqrt{2 - \sqrt{2}}$

For the normal line: $m = \frac{- 1}{y '}$

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

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

$m = 0.180398700048733$

God bless....I hope the explanation is useful.