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

1 Answer
Mar 6, 2018

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