Question

What is the equation of the line normal to `f(x)=sec4x-cot2x` at `x=pi/3`?

Answer 1

`"Normal"=>y=-(3x)/(8-24sqrt3)+(152sqrt3-120+3pi)/(24-72sqrt2)=>y~~0.089x-1.52`

Explanation:

The normal is the perpendicular line to the tangent.

`f(x)=sec(4x)-cot(2x)`
`f'(x)=4sec(4x)tan(3x)+2csc^2(2x)`

`f'(pi/3)=4sec((4pi)/3)tan((4pi)/3)+2csc^2((2pi)/3)=(8-24sqrt3)/3`

For normal, `m=-1/(f'(pi/3))=-3/(8-24sqrt3)`

`f(pi/3)=sec((4pi)/3)-cot((2pi)/3)=(sqrt3-6)/3`

`(sqrt3-6)/3=-3/(8-24sqrt3)(pi/3)+c`

`c=(sqrt3-6)/3+pi/(8-24sqrt3)=(152sqrt3-120+3pi)/(24-72sqrt2)`

`"Normal":y=-(3x)/(8-24sqrt3)+(152sqrt3-120+3pi)/(24-72sqrt2);y=0.089x-1.52`