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

Jul 6, 2018

$- \frac{2 - \sqrt{2}}{2 \left[2 - \sqrt{2} - \sqrt{2 + \sqrt{2}}\right]} = 0.232$, to three significant figures

#### Explanation:

We get the slope (${m}_{t}$) of the tangent line from the value of the first derivative $\frac{\mathrm{df}}{\mathrm{dx}}$ at the given $x$ value. The slope of the line normal to that we get from noting that the slopes of two perpendicular lines multiply to -1 - assuming that neither line is vertical.

Firstly note that we may immediately simplify the given function - $\cos x$ has period $2 \pi$, so $\cos 2 x$ has period $\pi$, so $\cos \left(2 x - \pi\right) = \cos 2 x$ and $f \left(x\right) = \sec x + {\cos}^{2} 2 x$.

Find the function derivative

Use the quotient rule for differentiation for the first term of $f$ and the chain rule twice for the second term:

$\frac{\mathrm{df}}{\mathrm{dx}} = \sec x \tan x + 2 \cos 2 x \cdot \left(- \sin 2 x\right) \cdot 2$
$\frac{\mathrm{df}}{\mathrm{dx}} = \sec x \tan x - 4 \sin 2 x \cos 2 x$

Recall the identity $\sin 2 x = 2 \sin x \cos x$

$\frac{\mathrm{df}}{\mathrm{dx}} = \sec x \tan x - 2 \sin 4 x$

Evalute the derivative

Now evaluate $\frac{\mathrm{df}}{\mathrm{dx}}$ at the given $x$:

$\frac{\mathrm{df}}{\mathrm{dx}} \left(\frac{11 \pi}{8}\right) = \sec \left(\frac{11 \pi}{8}\right) \tan \left(\frac{11 \pi}{8}\right) - 2 \sin \left(\frac{11 \pi}{2}\right)$

Recall that $\cos \left(x + \pi\right) = - \cos x$ (so $\sec \left(x + \pi\right) = - \sec x$). Then $\sec \left(\frac{11 \pi}{8}\right) = - \sec \left(\frac{3 \pi}{8}\right)$
Similarly, recall that $\tan \left(x + \pi\right) = \tan x$, so $\tan \left(\frac{11 \pi}{8}\right) = \tan \left(\frac{3 \pi}{8}\right)$. If these aren't easy to recall, think of the graphs of the functions.

As $\sin \left(x + \pi\right) = - \sin x$, $\sin \left(\frac{11 \pi}{2}\right) = - \sin \left(\frac{\pi}{2}\right) = - 1$

So
$\frac{\mathrm{df}}{\mathrm{dx}} \left(\frac{11 \pi}{8}\right) = - \sec \left(\frac{3 \pi}{8}\right) \tan \left(\frac{3 \pi}{8}\right) + 2$

Now the trig values of the angle $\frac{3 \pi}{8}$ are not standard ones, but they can be calculated.

$\sin \left(\frac{3 \pi}{8}\right) = \frac{\sqrt{2 + \sqrt{2}}}{2}$, as derived here:
https://socratic.org/questions/how-do-you-use-the-half-angle-identity-to-find-exact-value-of-sin-3pi-8

$\cos \left(\frac{3 \pi}{8}\right) = \frac{\sqrt{2 - \sqrt{2}}}{2}$, as derived here:
https://socratic.org/questions/how-do-you-find-the-exact-values-of-cos-3pi-8-using-the-half-angle-formula

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

So
$\frac{\mathrm{df}}{\mathrm{dx}} \left(\frac{11 \pi}{8}\right) = - \sec \left(\frac{3 \pi}{8}\right) \tan \left(\frac{3 \pi}{8}\right) + 2 = - \frac{2}{\sqrt{2 - \sqrt{2}}} \sqrt{\frac{2 + \sqrt{2}}{2 - \sqrt{2}}} + 2$

$\frac{\mathrm{df}}{\mathrm{dx}} \left(\frac{11 \pi}{8}\right) = 2 - \frac{2 \sqrt{2 + \sqrt{2}}}{2 - \sqrt{2}} = \frac{2}{2 - \sqrt{2}} \left(2 - \sqrt{2} - \sqrt{2 + \sqrt{2}}\right)$

This (complicated) value is the slope of the tangent line at the given point.

Calculate the normal slope

We now calculate the normal slope ${m}_{n}$via ${m}_{n} {m}_{t} = - 1$.

${m}_{t} = 2 - \frac{2 \sqrt{2 + \sqrt{2}}}{2 - \sqrt{2}} = \frac{2}{2 - \sqrt{2}} \left(2 - \sqrt{2} - \sqrt{2 + \sqrt{2}}\right)$

so

${m}_{n} = - \frac{1}{m} _ t = - \frac{2 - \sqrt{2}}{2 \left[2 - \sqrt{2} - \sqrt{2 + \sqrt{2}}\right]}$

Numerically, we calculate to three significant figures that ${m}_{t} = - 4.31$ and ${m}_{n} = 0.232$.

Sanity check these answers by plotting the graphs of the function and these lines for the given $x$ value
graph{(y-secx-(cos(2x))^2)(y+4.308644x-16.49888)(y-0.232x+3.115689)=0 [2, 6, -5, 0]}