Question

What is the equation of the line that is normal to `f(x)= cosx-sin^2x` at `x=(4pi)/3`?

Answer 1

vertical line `x = (4pi)/3`

Explanation:

Let's determine the slope of your normal line first.

To do so, you need to compute the derivative:

`f(x) = cos x - sin^2 x`

`f'(x) = - sin x - 2 sin x cos x`

Evaluate `f'(x)` for `x = (4 pi)/3`:

`f'((4 pi)/3) = - sin((4 pi) /3) - 2 sin((4 pi)/ 3) cos ((4 pi) / 3)`

To compute this, you can use

`sin ((4pi)/3) = - sin(pi/3) = - sqrt(3)/2`

and

`cos((4pi)/3) = cos(pi + pi/3) = - cos(pi/3) = - 1 /2`

Thus, we have

`f'((4pi)/3) = sqrt(3) / 2 - 2 * (- sqrt(3)/2) * (-1/2) = sqrt(3)/2 - 2 * sqrt(3) / 4 = 0`

This means that the slope of the tanget line at `x = (4pi)/3` is `0` and thus, the tangent line at `x = (4pi)/3` is a horizontal line.

However, as the the normal line is perpendicular to the tangent line, this also means that the normal line at `x = (4pi)/3` has to be a vertical line with the equation

`x = (4pi)/3`