Question

What is the slope of the line normal to the tangent line of `f(x) = secx-cos3x` at `x= (pi)/12`?

Answer 1

Slope of normal: `-((sqrt2+sqrt6)/(11-sqrt3))`

Explanation:

`f(x) = secx - cos3x`

since we know that the derivative of `secx = tanxsecx`
`f'(x) = tanxsecx - (3)(-sin3x)`
`f'(x) = tanxsecx +3sin3x`

if `x = pi/12`, then
`f(pi/12) = sec(pi/12) - cos3(pi/12)`
`f(pi/12) = 4/(sqrt2 + sqrt6) - sqrt2/2`
therefore, `f(pi/12)= (6+sqrt3)/(sqrt2+sqrt6)`

therefore, the point at which the normal passes is `(pi/12, (6+sqrt3)/(sqrt2+sqrt6))`

`f'(pi/12) = tan(pi/12)sec(pi/12) + 3sin3(pi/12)`
`f'(pi/12) = (2-sqrt3)(4/(sqrt2+sqrt6)) + (3sqrt2)/2`
^ this simplifies to:
`f'(pi/12) = (11-sqrt3)/(sqrt2+sqrt6)`

therefore, the slope of the tangent is `(11-sqrt3)/(sqrt2+sqrt6)`
since the slope of the normal is the negative reciprocal to the slope of a line, the slope of the normal will be:
Slope of normal: `-((sqrt2+sqrt6)/(11-sqrt3))`

Side note:

• I simplified a few things and skipped those steps purposely so that this question doesn't run forever

• I would say the hardest part about this question would be the simplification such as figuring out what `tan(pi/12)` is or what `sec(pi/12)` is