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

Mar 5, 2017

The equation is $y = - \frac{\sqrt{3}}{4} x + \frac{\sqrt{3} \pi - 6}{12}$

#### Explanation:

Start by finding the y-coordinate of tangency.

$f \left(\frac{\pi}{3}\right) = \sin \left(\frac{\frac{\pi}{3}}{2} + \pi\right)$

$f \left(\frac{\pi}{3}\right) = \sin \left(\frac{\pi}{6} + \pi\right)$

$f \left(\frac{\pi}{3}\right) = \sin \left(\frac{7 \pi}{6}\right)$

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

Now we find the derivative of $f \left(x\right)$, using the chain rule. We let $y = \sin u$ and $u = \frac{x}{2} + \pi$. Then $\frac{\mathrm{dy}}{\mathrm{du}} = \cos u$ and $\frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{2}$ (since π is only a constant).

The derivative therefore is

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \cos u \cdot \frac{1}{2}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} \cos \left(\frac{x}{2} + \pi\right)$

We now find the slope of the tangent.

$\frac{\mathrm{dy}}{\mathrm{dx}} {|}_{x = \frac{\pi}{3}} = \frac{1}{2} \cos \left(\frac{\frac{\pi}{3}}{2} + \pi\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} {|}_{x = \frac{\pi}{3}} = \frac{1}{2} \cos \left(\frac{7 \pi}{6}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} {|}_{x = \frac{\pi}{3}} = \frac{1}{2} \left(- \frac{\sqrt{3}}{2}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} {|}_{x = \frac{\pi}{3}} = - \frac{\sqrt{3}}{4}$

We now have all the information we need to find the equation of the line.

$y - {y}_{1} = m \left(x - {x}_{1}\right)$

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

$y + \frac{1}{2} = - \frac{\sqrt{3}}{4} x + \frac{\sqrt{3} \pi}{12}$

$y = - \frac{\sqrt{3}}{4} x + \frac{\sqrt{3} \pi - 6}{12}$

Hopefully this helps!