# What is the equation of the normal line of f(x)= 3lnsin2x at x = pi/8?

Apr 5, 2017

Equation of normal is $x + 6 y - 36 - \frac{\pi}{8} = 0$

#### Explanation:

While slope of tangent at curve $f \left(x\right)$ at $x = a$ is given by $f ' \left(a\right)$, as lope of normal at $x = a$ is $\frac{- 1}{f ' \left(a\right)}$ as normal is perpendicular to the tangent.

Both pass through the point $\left(a , f \left(a\right)\right)$ on the curve.

Here we have $f \left(x\right) = 3 \ln \sin 2 x$ and hence

$f ' \left(x\right) = \frac{3}{\sin 2 x} \times 2 \cos 2 x = 6 \cot 2 x$

and hence we are seeking a normal at $\left(\frac{\pi}{8} , 6 \cot \left(\frac{\pi}{4}\right)\right)$ or $\left(\frac{\pi}{8} , 6\right)$

and as slope of normal is $\frac{- 1}{6 \cot \left(\frac{\pi}{4}\right)} = - \frac{1}{6}$

and as equation of line passing through $\left({x}_{1} , {y}_{1}\right)$ and having a slope $m$ is $y - {y}_{1} = m \left(x - {x}_{1}\right)$ and hence equation of normal is

$y - 6 = \frac{- 1}{6} \left(x - \frac{\pi}{8}\right)$

or $6 y - 36 = - x + \frac{\pi}{8}$

or $x + 6 y - 36 - \frac{\pi}{8} = 0$