Question

What is the slope of the line normal to the tangent line of `f(x) = secx-cos^2(x-pi)` at `x= (pi)/12`?

Answer 1

`3sqrt(6)-5sqrt(2)+1/2`

Explanation:

the answer is `f'(pi/12)` where `f'(x)` is the derivative of `f(x)`

first simplify the `-(cos(x-pi))^2` part:
`=-(-cos(x))^2` proof
`=-(cosx)^2`

so `f(x)=secx-(cosx)^2`
`f'(x)=d/dx(secx)-d/dx((cosx)^2)`
`f'(x)=secxtanx-(2cosx*d/dx(cosx))` (secx derivative and chain rule)
`f'(x)=secxtanx-2cosx(-sinx)` (cosx derivative)
`f'(x)=secxtanx+2sinxcosx`
`f'(x)=secxtanx+sin(2x)` (double angle formula for sin(2x))

now evaluate at `x=pi/12`
`f'(pi/12)=sec(pi/12)tan(pi/12)+sin(2*(pi/12))`

use a calculator, memorize these, or use this table to evaluate:

`f'(pi/12)=(sqrt(6)-sqrt(2))*(2-sqrt(3))+sin(pi/6)`
`f'(pi/12)=2sqrt(6)-2sqrt(2)-sqrt(18)+sqrt(6)+1/2`
`f'(pi/12)=3sqrt(6)-2sqrt(2)-3sqrt(2)+1/2`
`f'(pi/12)=3sqrt(6)-5sqrt(2)+1/2`