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

Dec 28, 2016

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

#### Explanation:

The equation of the line normal to the curve $y = f \left(x\right)$ at $x = \overline{x}$, provided that $f \left(x\right)$ is differentiable is:

$y \left(x\right) = f \left(\overline{x}\right) - \frac{1}{f ' \left(\overline{x}\right)} \left(x - \overline{x}\right)$

In our case:

$f \left(x\right) = \ln \left({\sin}^{2} x\right)$
$f ' \left(x\right) = \frac{2 \sin x \cos x}{{\sin}^{2} x} = 2 \cot x$

and

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

$f ' \left(\frac{4 \pi}{3}\right) = 2 \cot \left(\frac{4 \pi}{3}\right) = \frac{2}{\sqrt{3}}$

So the normal line is:

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