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

Feb 19, 2016

$f \left(x\right) = - \frac{2 \sqrt{3}}{3} x + \frac{9 + 2 \pi \sqrt{3}}{18}$

#### Explanation:

The normal line of a point on a function is the line perpendicular to the tangent line at that point. This involves three basic steps: finding the slope of the tangent line, taking the opposite reciprocal of that slope (giving us the slope of the normal line), and finding the y-intercept.

Step 1: Find the Slope of the Tangent Line
All we do here is find the derivative of $\sin x$ and evaulate it at $x = \frac{\pi}{6}$ - because the slope of the tangent line at some point is the derivative at that point.
$f \left(x\right) = \sin x$
$f ' \left(x\right) = \cos x \to$the derivative of sine is cosine

Step 2: Opposite Reciprocal
Taking the opposite reciprocal of the tangent line slopes gives us the normal line slope. The reciprocal of $\frac{\sqrt{3}}{2}$ is $\frac{2}{\sqrt{3}}$. We'll have to rationalize this denominator - having a square root down there is a no-no:
$\frac{2}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{2 \sqrt{3}}{3}$

To find the opposite of this result, we simply make it negative:
$\frac{2 \sqrt{3}}{3} \cdot - 1 = - \frac{2 \sqrt{3}}{3}$

Step 3: Finding the Equation of the Line
We have the slope, but we need a $y$-intercept. Remember, a normal line has the form $y = m x + b$, where $x$ and $y$ are points on the line, $m$ is the slope, and $b$ is the $y$-intercept. We know the normal line passes through the tangent line - they're perpendicular, meaning they intersect at a ${90}^{o}$ angle. And since we're finding the normal line at $x = \frac{\pi}{6}$, we use that for our point:
$f \left(x\right) = \sin x$
$f \left(\frac{\pi}{6}\right) = \sin \left(\frac{\pi}{6}\right) = \frac{1}{2}$

We have now identified the point $\left(\frac{\pi}{6} , \frac{1}{2}\right)$. We will now use this point to find the equation of the line:
$y = m x + b$
$\frac{1}{2} = - \frac{2 \sqrt{3}}{3} \cdot \frac{\pi}{6} + b$
$\frac{1}{2} = - \frac{\pi \sqrt{3}}{9} + b$
$\frac{1}{2} + \frac{\pi \sqrt{3}}{9} = b \to b = \frac{9 + 2 \pi \sqrt{3}}{18}$

Bingo! The $y$-intercept is $\frac{9 + 2 \pi \sqrt{3}}{18}$. That means the equation of the normal line is $y = - \frac{2 \sqrt{3}}{3} x + \frac{9 + 2 \pi \sqrt{3}}{18}$. We could combine this into one whole fraction, but we don't have to, and I like avoiding unnecessary work :). If you prefer hard numbers, we can approximate this to $y = - 1.155 x + 1.105$.