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

Apr 13, 2017

Note that $f \left(\frac{\pi}{8}\right) = \csc \left(\frac{\pi}{8}\right)$, so the normal line will pass through the point $\left(\frac{\pi}{8} , \csc \left(\frac{\pi}{8}\right)\right)$.

The slope of the normal line will be the inverse reciprocal of the slope of the tangent line at $x = \frac{\pi}{8}$, since the lines are perpendicular.

To find the slope of the tangent line there, find the derivative of $f$ at $x = \frac{\pi}{8}$.

The derivative of $\csc \left(x\right)$, if you don't have it memorized, is easiest found using the chain rule on $\csc \left(x\right) = \frac{1}{\sin} \left(x\right) = {\left(\sin \left(x\right)\right)}^{-} 1$. (Note this isn't inverse sine— rather sine to the negative first power.)

$f \left(x\right) = {\left(\sin \left(x\right)\right)}^{-} 1$

$\implies f ' \left(x\right) = - {\left(\sin \left(x\right)\right)}^{-} 2 \cdot \cos \left(x\right) = - \cos \frac{x}{\sin} ^ 2 \left(x\right)$

So the slope of the tangent line is $- \cos \frac{\frac{\pi}{8}}{\sin} ^ 2 \left(\frac{\pi}{8}\right)$ and the slope of the normal line is ${\sin}^{2} \frac{\frac{\pi}{8}}{\cos} \left(\frac{\pi}{8}\right)$.

The line passing through $\left(\frac{\pi}{8} , \csc \left(\frac{\pi}{8}\right)\right)$ with slope ${\sin}^{2} \frac{\frac{\pi}{8}}{\cos} \left(\frac{\pi}{8}\right)$ is:

$y - \frac{1}{\sin} \left(\frac{\pi}{8}\right) = {\sin}^{2} \frac{\frac{\pi}{8}}{\cos} \left(\frac{\pi}{8}\right) \left(x - \frac{\pi}{8}\right)$

We could find these decimals, but we can also find the exact values.

I like to use the double angle formulas:

$\cos \left(2 x\right) = 2 {\cos}^{2} \left(x\right) - 1$

So:

$\cos \left(\frac{\pi}{4}\right) = 2 {\cos}^{2} \left(\frac{\pi}{8}\right) - 1$

${\cos}^{2} \left(\frac{\pi}{8}\right) = \frac{1}{2} \left(\cos \left(\frac{\pi}{4}\right) + 1\right) = \frac{1}{2} \left(\frac{1}{\sqrt{2}} + 1\right) = \frac{1 + \sqrt{2}}{2 \sqrt{2}} = \frac{2 + \sqrt{2}}{4}$

Then:

$\cos \left(\frac{\pi}{8}\right) = \frac{\sqrt{2 + \sqrt{2}}}{2}$

The same process can be done using $\cos \left(2 x\right) = 1 - 2 {\sin}^{2} \left(x\right)$ to find that

${\sin}^{2} \left(\frac{\pi}{8}\right) = \frac{2 - \sqrt{2}}{4}$

$\sin \left(\frac{\pi}{8}\right) = \frac{\sqrt{2 - \sqrt{2}}}{2}$

Then the normal line becomes:

$y - \frac{2}{\sqrt{2 - \sqrt{2}}} = \frac{2 - \sqrt{2}}{4} \left(\frac{2}{\sqrt{2 + \sqrt{2}}}\right) \left(x - \frac{\pi}{8}\right)$

$y - \frac{2}{\sqrt{2 - \sqrt{2}}} = \frac{2 - \sqrt{2}}{2 \sqrt{2 + \sqrt{2}}} \left(x - \frac{\pi}{8}\right)$