Question

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

Answer 1

`y-(sqrt(2)+1)=(2-sqrt(2))/4(x-pi/8)`

Explanation:

To find the normal line, first find the slope of the tangent line of `f(x)=cotx` at `x=pi/8`.

This can be found by finding the derivative of `f(x)=cotx`, which is `f'(x)=-(cscx)^2`

proof

`f'(pi/8)` is the slope of the tangent line of `f(x)=cotx` at `x=pi/8`.
`f'(pi/8)=-(csc(pi/8))^2`
`f'(pi/8)=-(2/sqrt(2-sqrt(2)))^2` csc(pi/8)
`f'(pi/8)=-4/(2-sqrt(2))`

the slope of the normal like is the opposite reciprocal of that, which would be:
`-1/(-4/(2-sqrt(2)))`

`=1/(4/(2-sqrt(2)))`

`=(2-sqrt(2))/4`

the point the normal line passes through is `(pi/8,f(pi/8))`
`f(pi/8)=cot(pi/8)`
`=sqrt(2)+1` (cot(pi/8)

use point-slope form:
`y-(sqrt(2)+1)=(2-sqrt(2))/4(x-pi/8)`