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

Feb 16, 2017

1.082x+y-1.009, nearly.. See the tangent-inclusive Socratic graph.

#### Explanation:

f at $x = \frac{\pi}{8}$ is sin(pi/8)=sqrt(1/2(1-cos(pi/4))

$= \frac{1}{2} \sqrt{1 - \frac{1}{\sqrt{2}}} = 0.5837$, nearly.

So, the foot of the normal P is $\left(\sin \left(\frac{\pi}{8}\right) , \frac{\pi}{8}\right) = \left(0.3927 , 0.5837\right)$

$f ' = \cos x = \cos \left(\frac{\pi}{8}\right) = \sqrt{\frac{1}{2} \left(1 + \cos \left(\frac{\pi}{4}\right)\right)} = \sqrt{\frac{1}{2} \left(1 + \frac{1}{\sqrt{2}}\right)} = 0.9239$,

nearly. at P.

The slope of the normal at P is $- \frac{1}{f '} = - \frac{1}{0.9239} = - 1.082$, nearly.

Now, the equation to the normal, at P( (0.5837, 0.3927 ), is

$y - 0 , 5837 = - 1.082 \left(x - 0.3927\right)$, giving

1.082x+y-1.009, nearly.

graph{(sinx-y)(1.082x+y-1.01)=0 [-10, 10, -5, 5]}