# What is the slope of the line normal to the tangent line of f(x) = secx-cos3x  at  x= (pi)/12 ?

Mar 6, 2018

Slope of normal: $- \left(\frac{\sqrt{2} + \sqrt{6}}{11 - \sqrt{3}}\right)$

#### Explanation:

$f \left(x\right) = \sec x - \cos 3 x$

since we know that the derivative of $\sec x = \tan x \sec x$
$f ' \left(x\right) = \tan x \sec x - \left(3\right) \left(- \sin 3 x\right)$
$f ' \left(x\right) = \tan x \sec x + 3 \sin 3 x$

if $x = \frac{\pi}{12}$, then
$f \left(\frac{\pi}{12}\right) = \sec \left(\frac{\pi}{12}\right) - \cos 3 \left(\frac{\pi}{12}\right)$
$f \left(\frac{\pi}{12}\right) = \frac{4}{\sqrt{2} + \sqrt{6}} - \frac{\sqrt{2}}{2}$
therefore, $f \left(\frac{\pi}{12}\right) = \frac{6 + \sqrt{3}}{\sqrt{2} + \sqrt{6}}$

therefore, the point at which the normal passes is $\left(\frac{\pi}{12} , \frac{6 + \sqrt{3}}{\sqrt{2} + \sqrt{6}}\right)$

$f ' \left(\frac{\pi}{12}\right) = \tan \left(\frac{\pi}{12}\right) \sec \left(\frac{\pi}{12}\right) + 3 \sin 3 \left(\frac{\pi}{12}\right)$
$f ' \left(\frac{\pi}{12}\right) = \left(2 - \sqrt{3}\right) \left(\frac{4}{\sqrt{2} + \sqrt{6}}\right) + \frac{3 \sqrt{2}}{2}$
^ this simplifies to:
$f ' \left(\frac{\pi}{12}\right) = \frac{11 - \sqrt{3}}{\sqrt{2} + \sqrt{6}}$

therefore, the slope of the tangent is $\frac{11 - \sqrt{3}}{\sqrt{2} + \sqrt{6}}$
since the slope of the normal is the negative reciprocal to the slope of a line, the slope of the normal will be:
Slope of normal: $- \left(\frac{\sqrt{2} + \sqrt{6}}{11 - \sqrt{3}}\right)$

Side note:

• I simplified a few things and skipped those steps purposely so that this question doesn't run forever

• I would say the hardest part about this question would be the simplification such as figuring out what $\tan \left(\frac{\pi}{12}\right)$ is or what $\sec \left(\frac{\pi}{12}\right)$ is