# What is the slope of the line normal to the tangent line of f(x) = cscx+cos^2(5x-pi)  at  x= (11pi)/8 ?

Jun 7, 2018

$m = - 0.2510$ (rounded)

#### Explanation:

This may be a wildly constructed problem, but here's my attempt to answer it:

To determine the slope of any tangent line, we need the derivative of the curve. For this problem:
$f ' \left(x\right) = - \csc x \cot x + 2 \cos \left(5 x - \pi\right) \cdot \left(- \sin \left(5 x - \pi\right)\right) \cdot 5$

$f ' \left(x\right) = - \cos \frac{x}{{\sin}^{2} x} - 10 \cos \left(5 x - \pi\right) \cdot \sin \left(5 x - \pi\right)$

When $x = \frac{11 \pi}{8}$ ,
the slope of the tangent line at that point is:
${m}_{T} = f ' \left(\frac{11 \pi}{8}\right) = 3.9839$ (rounded)

Since the lines are perpendicular, the slope of the normal line is the negative reciprocal of the slope of the tangent:
${m}_{N} = - \frac{1}{m} _ T$

So, ${m}_{N} = - 0.2510$