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

##### 1 Answer

vertical line

#### 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'((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

However, as the the normal line is perpendicular to the tangent line, this also means that the normal line at **vertical line** with the equation

#x = (4pi)/3#