# What is the slope of the line normal to the tangent line of f(x) = cscx+sin(2x-(3pi)/8)  at  x= (11pi)/8 ?

Jul 13, 2017

$- \frac{\left(\sqrt{2 + \sqrt{2}}\right) \left(3 \sqrt{2} + 2\right)}{14}$

#### Explanation:

First, find the derivative of $f \left(x\right)$:

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left(\csc x\right) + \frac{d}{\mathrm{dx}} \left(\sin \left(2 x - \frac{3 \pi}{8}\right)\right)$

$f ' \left(x\right) = - \csc x \cot x + \cos \left(2 x - \frac{3 \pi}{8}\right) \cdot \frac{d}{\mathrm{dx}} \left(2 x - \frac{3 \pi}{8}\right)$

$f ' \left(x\right) = - \csc x \cot x + 2 \cos \left(2 x - \frac{3 \pi}{8}\right)$

Now we need to plug in $x = \frac{11 \pi}{8}$

$f ' \left(\frac{11 \pi}{8}\right) = - \csc \left(\frac{11 \pi}{8}\right) \cot \left(\frac{11 \pi}{8}\right) + 2 \cos \left(2 \frac{11 \pi}{8} - \frac{3 \pi}{8}\right)$

$f ' \left(\frac{11 \pi}{8}\right) = \frac{- 1}{\sin \left(\frac{11 \pi}{8}\right) \tan \left(\frac{11 \pi}{8}\right)} + 2 \cos \left(\frac{19 \pi}{8}\right)$

To evaluate these trig functions, use the half angle rules:

$\sin \left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 - \cos \theta}{2}}$

$\cos \left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 + \cos \theta}{2}}$

$\tan \left(\frac{\theta}{2}\right) = \frac{1 - \cos \theta}{\sin} \theta$

Using these rules, we can evaluate these trig functions like this:

$\frac{11 \pi}{8}$ is in the third quadrant, so sine and cosine will be negative, and tangent will be positive.

$\frac{3 \pi}{8}$ is in the first quadrant, so all trig functions are positive.

$\sin \left(\frac{11 \pi}{8}\right) = - \sqrt{\frac{1 - \cos \left(\frac{11 \pi}{4}\right)}{2}} = - \sqrt{\frac{1 - \cos \left(\frac{3 \pi}{4}\right)}{2}} = - \sqrt{\frac{1 + \frac{\sqrt{2}}{2}}{2}} = - \sqrt{\frac{2 + \sqrt{2}}{4}} = - \frac{\sqrt{2 + \sqrt{2}}}{2}$

$\tan \left(\frac{11 \pi}{8}\right) = \frac{1 - \cos \left(\frac{11 \pi}{4}\right)}{\sin} \left(\frac{11 \pi}{4}\right) = \frac{1 + \frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = \frac{1}{\frac{\sqrt{2}}{2}} + \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = \sqrt{2} + 1$

$\cos \left(\frac{19 \pi}{8}\right) = \cos \left(\frac{3 \pi}{8}\right) = \sqrt{\frac{1 + \cos \left(\frac{3 \pi}{4}\right)}{2}} = \sqrt{\frac{1 - \frac{\sqrt{2}}{2}}{2}} = \sqrt{\frac{2 - \sqrt{2}}{4}} = \frac{\sqrt{2 - \sqrt{2}}}{2}$

Now we plug these back into our function and simplify.

$f ' \left(\frac{11 \pi}{8}\right) = \frac{- 1}{\left(- \frac{\sqrt{2 + \sqrt{2}}}{2}\right) \left(\sqrt{2} + 1\right)} + 2 \left(\frac{\sqrt{2 - \sqrt{2}}}{2}\right)$

$= \frac{2}{\left(\sqrt{2 + \sqrt{2}}\right) \left(\sqrt{2} + 1\right)} + \sqrt{2 - \sqrt{2}}$

$= \frac{2}{\left(\sqrt{2 + \sqrt{2}}\right) \left(\sqrt{2} + 1\right)} + \sqrt{2 - \sqrt{2}} \left(\frac{\left(\sqrt{2 + \sqrt{2}}\right) \left(\sqrt{2} + 1\right)}{\left(\sqrt{2 + \sqrt{2}}\right) \left(\sqrt{2} + 1\right)}\right)$

$= \frac{2 + \sqrt{\left(2 - \sqrt{2}\right) \left(2 + \sqrt{2}\right)} \left(\sqrt{2} + 1\right)}{\left(\sqrt{2 + \sqrt{2}}\right) \left(\sqrt{2} + 1\right)}$

$= \frac{2 + \sqrt{2} \left(\sqrt{2} + 1\right)}{\left(\sqrt{2 + \sqrt{2}}\right) \left(\sqrt{2} + 1\right)}$

$= \frac{4 + \sqrt{2}}{\left(\sqrt{2 + \sqrt{2}}\right) \left(\sqrt{2} + 1\right)}$

Finally, since we're looking for the slope of the normal line, we need to flip this fraction and make it negative.

$\text{norm } = - \frac{\left(\sqrt{2 + \sqrt{2}}\right) \left(\sqrt{2} + 1\right)}{4 + \sqrt{2}}$

And then rationalize the denominator by multiplying by the conjugate.

$= - \frac{\left(\sqrt{2 + \sqrt{2}}\right) \left(\sqrt{2} + 1\right) \left(4 - \sqrt{2}\right)}{16 - 2}$

$= - \frac{\left(\sqrt{2 + \sqrt{2}}\right) \left(3 \sqrt{2} + 2\right)}{14}$