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

Jan 25, 2016

vertical line $x = \frac{4 \pi}{3}$

#### Explanation:

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

To do so, you need to compute the derivative:

$f \left(x\right) = \cos x - {\sin}^{2} x$

$f ' \left(x\right) = - \sin x - 2 \sin x \cos x$

Evaluate $f ' \left(x\right)$ for $x = \frac{4 \pi}{3}$:

$f ' \left(\frac{4 \pi}{3}\right) = - \sin \left(\frac{4 \pi}{3}\right) - 2 \sin \left(\frac{4 \pi}{3}\right) \cos \left(\frac{4 \pi}{3}\right)$

To compute this, you can use

$\sin \left(\frac{4 \pi}{3}\right) = - \sin \left(\frac{\pi}{3}\right) = - \frac{\sqrt{3}}{2}$

and

$\cos \left(\frac{4 \pi}{3}\right) = \cos \left(\pi + \frac{\pi}{3}\right) = - \cos \left(\frac{\pi}{3}\right) = - \frac{1}{2}$

Thus, we have

$f ' \left(\frac{4 \pi}{3}\right) = \frac{\sqrt{3}}{2} - 2 \cdot \left(- \frac{\sqrt{3}}{2}\right) \cdot \left(- \frac{1}{2}\right) = \frac{\sqrt{3}}{2} - 2 \cdot \frac{\sqrt{3}}{4} = 0$

This means that the slope of the tanget line at $x = \frac{4 \pi}{3}$ is $0$ and thus, the tangent line at $x = \frac{4 \pi}{3}$ is a horizontal line.

However, as the the normal line is perpendicular to the tangent line, this also means that the normal line at $x = \frac{4 \pi}{3}$ has to be a vertical line with the equation

$x = \frac{4 \pi}{3}$