# What is the equation of the normal line of f(x)=x-sinx at x=pi/6?

Sep 19, 2017

$y = \left(- 4 - 2 \sqrt{3}\right) x + \frac{\sqrt{3} \pi}{3} + \frac{5 \pi}{6} - \frac{1}{2}$

#### Explanation:

The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point. (If needed, then the normal is perpendicular to the tangent so the product of their gradients is $- 1$).

We have:

$f \left(x\right) = x - \sin x$

Differentiating wrt $x$ we have:

$f ' \left(x\right) = 1 - \cos x$

So when $x = \frac{\pi}{6}$ we have;

$\setminus f \left(\frac{\pi}{6}\right) = \frac{\pi}{6} - \frac{1}{2}$
$f ' \left(\frac{\pi}{6}\right) = 1 - \frac{\sqrt{3}}{2} = \frac{2 - \sqrt{3}}{2}$

So, the gradient of the normal is:

${m}_{N} = - \frac{2}{2 - \sqrt{3}}$
$\setminus \setminus \setminus \setminus \setminus \setminus = - \frac{2}{2 - \sqrt{3}} \frac{2 + \sqrt{3}}{2 + \sqrt{3}}$
$\setminus \setminus \setminus \setminus \setminus \setminus = - \frac{2 \left(2 + \sqrt{3}\right)}{4 - 3}$
$\setminus \setminus \setminus \setminus \setminus \setminus = - \left(4 + 2 \sqrt{3}\right)$

So the tangent passes through $\left(\frac{\pi}{6} , \frac{\pi}{6} - \frac{1}{2}\right)$ and has gradient ${m}_{N} = - \left(4 + 2 \sqrt{3}\right)$, and using the point/slope form $y - {y}_{1} = m \left(x - {x}_{1}\right)$ the normal equation we seek is;

$y - \left(\frac{\pi}{6} - \frac{1}{2}\right) = - \left(4 + 2 \sqrt{3}\right) \left(x - \frac{\pi}{6}\right)$

$\therefore y - \frac{\pi}{6} + \frac{1}{2} = - 4 x + \frac{4 \pi}{6} - 2 \sqrt{3} x + \frac{2 \sqrt{3} \pi}{6}$

$\therefore y = - 4 x - 2 \sqrt{3} x + \frac{\sqrt{3} \pi}{3} + \frac{2 \pi}{3} + \frac{\pi}{6} - \frac{1}{2}$

$\therefore y = \left(- 4 - 2 \sqrt{3}\right) x + \frac{\sqrt{3} \pi}{3} + \frac{5 \pi}{6} - \frac{1}{2}$

We can verify this graphically: