Question

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

Answer 1

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

Explanation:

f at `x = pi/8` is `sin(pi/8)=sqrt(1/2(1-cos(pi/4))`

`=1/2sqrt(1-1/sqrt2)=0.5837`, nearly.

So, the foot of the normal P is `(sin(pi/8), pi/8)=( 0.3927, 0.5837)`

`f'=cosx=cos(pi/8)=sqrt(1/2(1+cos(pi/4))) = sqrt(1/2(1+1/sqrt2))=0.9239`,

nearly. at P.

The slope of the normal at P is `-1/(f')=-1/0.9239=-1.082`, nearly.

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

`y-0,5837=-1.082(x-0.3927)`, giving

1.082x+y-1.009, nearly.

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