# What is the equation of the line normal to  f(x)=(xsinx)/tanx at  x=pi/3?

Jun 20, 2016

Equation of normal is $\left(y - \frac{\pi}{6}\right) = \frac{2 \sqrt{3}}{\pi - \sqrt{3}} \times \left(x - \frac{\pi}{3}\right)$

#### Explanation:

As $f \left(x\right) = \frac{x \sin x}{\tan} x$, at $x = \frac{\pi}{3}$, $f \left(x\right) = \frac{\frac{\pi}{3} \times \frac{\sqrt{3}}{2}}{\sqrt{3}} = \frac{\pi}{6}$

Hence,m we are seeking equation of normal at $\left(\frac{\pi}{3} , \frac{\pi}{6}\right)$

As slope of tangent is given by $f ' \left(x\right)$, as $f \left(x\right) = \frac{x \sin x}{\sin \frac{x}{\cos} x} = x \cos x$

Hence $f ' \left(x\right) = \cos x - x \sin x$ and at $x = \frac{\pi}{3}$, $f ' \left(\frac{\pi}{3}\right) = \cos \left(\frac{\pi}{3}\right) - \frac{\pi}{3} \times \sin \left(\frac{\pi}{3}\right) = \frac{1}{2} - \frac{\pi}{3} \times \frac{\sqrt{3}}{2}$

= $\frac{1}{2} - \frac{\pi}{2 \sqrt{3}} = \frac{\sqrt{3} - \pi}{2 \sqrt{3}}$

Hence slope of tangent is $\frac{\sqrt{3} - \pi}{2 \sqrt{3}}$

and that of normal would be $\frac{- 1}{\frac{\sqrt{3} - \pi}{2 \sqrt{3}}} = \frac{2 \sqrt{3}}{\pi - \sqrt{3}}$

and equation of normal would be

$\left(y - \frac{\pi}{6}\right) = \frac{2 \sqrt{3}}{\pi - \sqrt{3}} \times \left(x - \frac{\pi}{3}\right)$