# What is the equation of the normal line of f(x)= (1-x)sinx at x = pi/8?

Sep 10, 2016

Equation of normal is

$\left(y - 1 + \frac{\pi}{8} \sin \left(\frac{\pi}{8}\right)\right) = - \frac{1}{\cos \left(\frac{\pi}{8}\right) - \sin \left(\frac{\pi}{8}\right) - \frac{\pi}{8} \cos \left(\frac{\pi}{8}\right)} \left(x - \frac{\pi}{8}\right)$

#### Explanation:

At $x = \frac{\pi}{8}$, $f \left(x = \frac{\pi}{8}\right) = \left(1 - \frac{\pi}{8}\right) \times \sin \left(\frac{\pi}{8}\right)$

Note that $\sin \left(\frac{\pi}{8}\right) = \frac{\sqrt{\sqrt{2} - 1}}{2}$ and $\cos \left(\frac{\pi}{8}\right) = \frac{\sqrt{\sqrt{2} + 1}}{2}$, but we leave it as it is to avoid complications.

Hence we are seeking a normal at $\left(\frac{\pi}{8} , \left(1 - \frac{\pi}{8}\right) \sin \left(\frac{\pi}{8}\right)\right)$

Now as normal is perpendicular to tangent and slope of tangent is given by $f ' \left(\frac{\pi}{8}\right)$

as $f ' \left(x\right) = - 1 \times \sin x + \left(1 - x\right) \times \cos x = \cos x - \sin x - x \cos x$

hence slope of tangent is $\cos \left(\frac{\pi}{8}\right) - \sin \left(\frac{\pi}{8}\right) - \frac{\pi}{8} \cos \left(\frac{\pi}{8}\right)$

and slope of normal is $- \frac{1}{\cos \left(\frac{\pi}{8}\right) - \sin \left(\frac{\pi}{8}\right) - \frac{\pi}{8} \cos \left(\frac{\pi}{8}\right)}$

and equation of normal is

$\left(y - 1 + \frac{\pi}{8} \sin \left(\frac{\pi}{8}\right)\right) = - \frac{1}{\cos \left(\frac{\pi}{8}\right) - \sin \left(\frac{\pi}{8}\right) - \frac{\pi}{8} \cos \left(\frac{\pi}{8}\right)} \left(x - \frac{\pi}{8}\right)$