Question

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

Answer 1

Equation of normal is

`(y-1+pi/8sin(pi/8))=-1/(cos(pi/8)-sin(pi/8)-pi/8cos(pi/8))(x-pi/8)`

Explanation:

At `x=pi/8`, `f(x=pi/8)=(1-pi/8)xxsin(pi/8)`

Note that `sin(pi/8)=sqrt(sqrt2-1)/2` and `cos(pi/8)=sqrt(sqrt2+1)/2`, but we leave it as it is to avoid complications.

Hence we are seeking a normal at `(pi/8,(1-pi/8)sin(pi/8))`

Now as normal is perpendicular to tangent and slope of tangent is given by `f'(pi/8)`

as `f'(x)=-1xxsinx+(1-x)xxcosx=cosx-sinx-xcosx`

hence slope of tangent is `cos(pi/8)-sin(pi/8)-pi/8cos(pi/8)`

and slope of normal is `-1/(cos(pi/8)-sin(pi/8)-pi/8cos(pi/8))`

and equation of normal is

`(y-1+pi/8sin(pi/8))=-1/(cos(pi/8)-sin(pi/8)-pi/8cos(pi/8))(x-pi/8)`