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

Jul 23, 2018

$y = \left(\frac{3}{2} + \sqrt{3} - \frac{2}{3} \cdot \pi - \frac{\pi}{\sqrt{3}}\right) x + \frac{1}{72} \left(- 36 + 18 \cdot \sqrt{3} + 3 \pi + 4 {\pi}^{2} + 2 \sqrt{3} \cdot {\pi}^{2}\right)$

#### Explanation:

given is

$f \left(x\right) = x \cot \left(x\right) - {\sin}^{2} \left(x - \pi\right)$ then we get by the sum , the chain and the product rule:

$f ' \left(x\right) = \cot \left(x\right) - x {\csc}^{2} \left(x\right) - 2 \sin \left(x\right) \cos \left(x\right)$

now we compute

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

and

$f \left(\frac{\pi}{12}\right) = \frac{1}{12} \left(2 + \sqrt{3}\right) \pi - \frac{1}{8} \cdot {\left(\sqrt{3} - 1\right)}^{2}$

The Tangent line is given by the equation

$y = m x + n$

$m = f ' \left(\frac{\pi}{12}\right)$

plugging $x = \frac{\pi}{12}$ and $f \left(\frac{\pi}{12}\right)$ in the given equation we get

$n = \frac{1}{72} \left(- 36 + 18 \sqrt{3} + 3 \pi + 4 {\pi}^{2} + 2 \sqrt{3} {\pi}^{2}\right)$