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

Jun 22, 2016

$y = - 1.178 x + 1.097$

#### Explanation:

$f \left(x\right) = \frac{x}{\sqrt{\sin}} x$

by Quotient Rule:
$f ' \left(x\right) = \frac{x ' \sqrt{\sin} x - x \left(\sqrt{\sin} x\right) '}{\sqrt{\sin} x} ^ 2 = \frac{1 \cdot \sqrt{\sin} x - x \left(\sqrt{\sin} x\right) '}{\sin x}$

the oddest bit:

$\left(\sqrt{\sin} x\right) ' = \frac{1}{2} \frac{1}{\sqrt{\sin} x} \cos x$

so we have

$f ' \left(x\right) = \frac{\sqrt{\sin} x - x \left(\frac{1}{2}\right) \frac{1}{\sqrt{\sin} x} \cos x}{\sin x}$

$= \frac{\sin x - \frac{1}{2} x \cos x}{\sin x} ^ \left\{\frac{3}{2}\right\}$

$f ' \left(\frac{\pi}{8}\right) = \frac{\sin \left(\frac{\pi}{8}\right) - \frac{1}{2} \left(\frac{\pi}{8}\right) \cos \left(\frac{\pi}{8}\right)}{\sin \left(\frac{\pi}{8}\right)} ^ \left\{\frac{3}{2}\right\} = 0.85024$

that is the slope, $s$, of the tangent. slope of the normal is $- \frac{1}{s} = - 1.178$

$f \left(\frac{\pi}{8}\right) = 0.634$

$y = m x + c \setminus \implies 0.634 = - 1.178 \left(\frac{\pi}{8}\right) + c \setminus \implies c = 1.097$

So the line is: $y = - 1.178 x + 1.097$ 