# What is the slope of the line normal to the tangent line of f(x) = cscx-secx  at  x= (17pi)/12 ?

Dec 20, 2016

First off, you may want to refer to this answer as additional practice.

Furthermore, we will be using a similar method to the one in the answer linked above, and we'll also reference information that we've already figured out in that answer.

I got the line normal to the tangent line as:

f_N((17pi)/12) = -[1/(2(2 - sqrt3)^"3/2" + 2(2 + sqrt3)^"3/2")]x + 1/6 [12sqrt(2 + sqrt3) - 12sqrt(2 - sqrt3) - 17(2 - sqrt3)^"3/2"pi - 17(2 + sqrt3)^"3/2"pi]

The slope of this normal line is therefore $- \left[\frac{1}{2 {\left(2 - \sqrt{3}\right)}^{\text{3/2" + 2(2 + sqrt3)^"3/2}}}\right]$.

Or the approximate decimal normal line is:

${f}_{N} \left(\frac{17 \pi}{12}\right) \approx - 0.0680 x - 62.5816$

You can see the tangent line itself here:

graph{(cscx - secx - y)(14.6969x - 62.5816 - y) = 0 [2.392, 6.833, -0.24, 5.306]}

Linearization Equation

$\boldsymbol{{f}_{T} \left(a\right) = f \left(a\right) + f \text{'} \left(a\right) \left(x - a\right)}$

where $a = \frac{17 \pi}{12}$.

Since this isn't half of a convenient angle, we'll end up using the addition formulas later for $\sin$ and $\cos$, in addition to the half-angle relations we already figured out in a previous answer.

For now:

$f ' \left(x\right) = - \left(\csc x \cot x + \sec x \tan x\right)$
$f \left(x\right) = \csc x - \sec x$

so that

$f ' \left(\frac{17 \pi}{12}\right) = - \left[\csc \left(\frac{17 \pi}{12}\right) \cot \left(\frac{17 \pi}{12}\right) + \sec \left(\frac{17 \pi}{12}\right) \tan \left(\frac{17 \pi}{12}\right)\right]$

$f \left(\frac{17 \pi}{12}\right) = \csc \left(\frac{17 \pi}{12}\right) - \sec \left(\frac{17 \pi}{12}\right)$

Since $\csc x = \frac{1}{\sin} x$ and $\sec x = \frac{1}{\cos} x$, we can use the addition formulas to evaluate these trig functions directly. Notice how $\frac{17 \pi}{12}$ radians is also $\frac{3 \pi}{2} - \frac{\pi}{12}$.

Now, recall from here that:

$\sin \left(\frac{\pi}{12}\right) = \sqrt{\frac{1 - \cos \left(\pi \text{/} 6\right)}{2}} = \left[\ldots\right] = \frac{\sqrt{2 - \sqrt{3}}}{2}$
$\cos \left(\frac{\pi}{12}\right) = \sqrt{\frac{1 + \cos \left(\pi \text{/} 6\right)}{2}} = \left[\ldots\right] = \frac{\sqrt{2 + \sqrt{3}}}{2}$

$\sin \left(u \pm v\right) = \sin u \cos v \pm \cos u \sin v$
cos(u pm v) = cosucosv ∓ sinusinv

Thus:

$\textcolor{g r e e n}{\sin \left(\frac{17 \pi}{12}\right)} = \sin \left(\frac{3 \pi}{2} - \frac{\pi}{12}\right)$

$= \sin \left(\frac{3 \pi}{2}\right) \cos \left(\frac{\pi}{12}\right) - {\cancel{\cos \left(\frac{3 \pi}{2}\right) \sin \left(\frac{\pi}{12}\right)}}^{0}$

$= \textcolor{g r e e n}{- \frac{\sqrt{2 + \sqrt{3}}}{2}}$

$\textcolor{g r e e n}{\cos \left(\frac{17 \pi}{12}\right)} = \cos \left(\frac{3 \pi}{2} - \frac{\pi}{12}\right)$

$= {\cancel{\cos \left(\frac{3 \pi}{2}\right) \cos \left(\frac{\pi}{12}\right)}}^{0} + \sin \left(\frac{3 \pi}{2}\right) \sin \left(\frac{\pi}{12}\right)$

$= \textcolor{g r e e n}{- \frac{\sqrt{2 - \sqrt{3}}}{2}}$

This allows us to evaluate the following functions:

• $\csc \left(\frac{17 \pi}{12}\right) = \frac{1}{\sin} \left(\frac{17 \pi}{12}\right) = - \frac{2}{\sqrt{2 + \sqrt{3}}}$
• $\cot \left(\frac{17 \pi}{12}\right) = \cos \frac{\frac{17 \pi}{12}}{\sin} \left(\frac{17 \pi}{12}\right) = \frac{\sqrt{2 - \sqrt{3}}}{\sqrt{2 + \sqrt{3}}}$
• $\sec \left(\frac{17 \pi}{12}\right) = \frac{1}{\cos} \left(\frac{17 \pi}{12}\right) = - \frac{2}{\sqrt{2 - \sqrt{3}}}$
• $\tan \left(\frac{17 \pi}{12}\right) = \frac{1}{\cot} \left(\frac{17 \pi}{12}\right) = \frac{\sqrt{2 + \sqrt{3}}}{\sqrt{2 - \sqrt{3}}}$

Plugging all of these in to $f ' \left(a\right)$ and $f \left(a\right)$, we get:

$\textcolor{g r e e n}{f ' \left(\frac{17 \pi}{12}\right)} = - \left[- \frac{2}{\sqrt{2 + \sqrt{3}}} \frac{\sqrt{2 - \sqrt{3}}}{\sqrt{2 + \sqrt{3}}} - \frac{2}{\sqrt{2 - \sqrt{3}}} \frac{\sqrt{2 + \sqrt{3}}}{\sqrt{2 - \sqrt{3}}}\right]$

$= - \left[- \frac{2 \sqrt{2 - \sqrt{3}}}{2 + \sqrt{3}} - \frac{2 \sqrt{2 + \sqrt{3}}}{2 - \sqrt{3}}\right]$

$= \frac{2 {\left(2 - \sqrt{3}\right)}^{\text{3/2")/cancel((2 + sqrt3)(2 - sqrt3))^(1) + (2(2 + sqrt3)^"3/2}}}{\cancel{\left(2 - \sqrt{3}\right) \left(2 + \sqrt{3}\right)}} ^ \left(1\right)$

$= \textcolor{g r e e n}{2 {\left(2 - \sqrt{3}\right)}^{\text{3/2" + 2(2 + sqrt3)^"3/2}}}$

$\textcolor{g r e e n}{f \left(\frac{17 \pi}{12}\right)} = - \frac{2}{\sqrt{2 + \sqrt{3}}} + \frac{2}{\sqrt{2 - \sqrt{3}}}$

$= - \frac{2 \sqrt{2 - \sqrt{3}}}{\cancel{\sqrt{2 + \sqrt{3}} \sqrt{2 - \sqrt{3}}}} ^ \left(1\right) + \frac{2 \sqrt{2 + \sqrt{3}}}{\cancel{\sqrt{2 - \sqrt{3}} \sqrt{2 + \sqrt{3}}}} ^ \left(1\right)$

$= \textcolor{g r e e n}{2 \sqrt{2 + \sqrt{3}} - 2 \sqrt{2 - \sqrt{3}}}$

Plugging these results into the linearization equation, we can obtain the tangent line equation to simplify.

${f}_{T} \left(\frac{17 \pi}{12}\right) = \stackrel{f \left(a\right)}{\overbrace{2 \sqrt{2 + \sqrt{3}} - 2 \sqrt{2 - \sqrt{3}}}} + \stackrel{f ' \left(a\right)}{\overbrace{\left[2 {\left(2 - \sqrt{3}\right)}^{\text{3/2" + 2(2 + sqrt3)^"3/2}}\right]}} \left(x - \frac{17 \pi}{12}\right)$

Multiply through:

$= 2 \sqrt{2 + \sqrt{3}} - 2 \sqrt{2 - \sqrt{3}} + 2 {\left(2 - \sqrt{3}\right)}^{\text{3/2"x + 2(2 + sqrt3)^"3/2"x - 2(2 - sqrt3)^"3/2"(17pi)/12 - 2(2 + sqrt3)^"3/2}} \frac{17 \pi}{12}$

Get common denominators going on:

= [2(2 - sqrt3)^"3/2" + 2(2 + sqrt3)^"3/2"]x + (12sqrt(2 + sqrt3))/6 - (12sqrt(2 - sqrt3))/6 - (17pi)/6 (2 - sqrt3)^"3/2" - (17pi)/6 (2 + sqrt3)^"3/2"

Combine fractions with common denominators:

$= \left[2 {\left(2 - \sqrt{3}\right)}^{\text{3/2" + 2(2 + sqrt3)^"3/2"]x + (12sqrt(2 + sqrt3) - 12sqrt(2 - sqrt3))/6 - [(17pi)/6 (2 - sqrt3)^"3/2" + (17pi)/6 (2 + sqrt3)^"3/2}}\right]$

$\implies \textcolor{g r e e n}{{f}_{T} \left(\frac{17 \pi}{12}\right) = \left[2 {\left(2 - \sqrt{3}\right)}^{\text{3/2" + 2(2 + sqrt3)^"3/2"]x + 1/6 [12sqrt(2 + sqrt3) - 12sqrt(2 - sqrt3) - 17(2 - sqrt3)^"3/2"pi - 17(2 + sqrt3)^"3/2}} \pi\right]}$

So, the line normal to the tangent line will have the negative reciprocal slope:

color(blue)(f_N((17pi)/12) = -[1/(2(2 - sqrt3)^"3/2" + 2(2 + sqrt3)^"3/2")]x + 1/6 [12sqrt(2 + sqrt3) - 12sqrt(2 - sqrt3) - 17(2 - sqrt3)^"3/2"pi - 17(2 + sqrt3)^"3/2"pi])

Or the approximate decimal answer is:

$\textcolor{b l u e}{{f}_{N} \left(\frac{17 \pi}{12}\right) \approx - 0.0680 x - 62.5816}$