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

$y = - 1.2033332887561 x + 0.63106054525314$ OR
$y = - \frac{\left(14 + 9 \sqrt{2}\right) \sqrt{2 - \sqrt{2}}}{17} \cdot x + \frac{\pi \left(14 + 9 \sqrt{2}\right) \sqrt{2 - \sqrt{2}}}{136}$
$+ \frac{\left(\sqrt{2} - 1\right) \sqrt{2 - \sqrt{2}}}{2}$

#### Explanation:

Given $y = \tan x \cdot \sin x$
and $x = \frac{\pi}{8}$

Solve for the Ordinate ${y}_{1}$ when Abscissa ${x}_{1} = \frac{\pi}{8}$

${y}_{1} = f \left(\frac{\pi}{8}\right) = \tan \left(\frac{\pi}{8}\right) \sin \left(\frac{\pi}{8}\right)$

Take note from Trigonometry:

$S \in \left(\frac{A}{2}\right) = \sqrt{\frac{1 - \cos A}{2}}$ and $\cos \left(\frac{A}{2}\right) = \sqrt{\frac{1 + \cos A}{2}}$
$T a n \left(\frac{A}{2}\right) = \frac{1 - \cos A}{S} \in A$
Let $A = \frac{\pi}{4}$ So that
$S \in \left(\frac{\pi}{8}\right) = \frac{\sqrt{2 - \sqrt{2}}}{2}$
$C o s \left(\frac{\pi}{8}\right) = \frac{\sqrt{2 + \sqrt{2}}}{2}$
$T a n \left(\frac{\pi}{8}\right) = \sqrt{2} - 1$

${y}_{1} = f \left(\frac{\pi}{8}\right) = \tan \left(\frac{\pi}{8}\right) \sin \left(\frac{\pi}{8}\right)$
${y}_{1} = \frac{\left(\sqrt{2} - 1\right) \cdot \sqrt{2 - \sqrt{2}}}{2}$

The First derivative $f ' \left(x\right)$:
$f ' \left(x\right) = \tan x \cdot \cos x + \sin x \cdot {\sec}^{2} x$
Also
$f ' \left(x\right) = \sin x \left(1 + {\sec}^{2} x\right)$
And for the Normal Line
Slope $m = - \frac{1}{f ' \left(\frac{\pi}{8}\right)}$

Slope m=-1/(sin x(1+sec^2x)

Slope m=-1/((sqrt(2-sqrt2)/2)*(1+(2/(sqrt(2+sqrt2)))^2)

Slope $m = - \frac{\left(14 + 9 \sqrt{2}\right) \sqrt{2 - \sqrt{2}}}{17}$

At this point, use now $\left({x}_{1} , {y}_{1}\right)$ and slope $m$ to find the Normal Line.
Use Point-Slope Form $y - {y}_{1} = m \left(x - {x}_{1}\right)$

$y - \frac{\left(\sqrt{2} - 1\right) \cdot \sqrt{2 - \sqrt{2}}}{2} =$
$- \frac{\left(14 + 9 \sqrt{2}\right) \sqrt{2 - \sqrt{2}}}{17} \left(x - \frac{\pi}{8}\right)$

$y = - \frac{\left(14 + 9 \sqrt{2}\right) \sqrt{2 - \sqrt{2}}}{17} \cdot x + \frac{\pi \left(14 + 9 \sqrt{2}\right) \sqrt{2 - \sqrt{2}}}{136}$
$+ \frac{\left(\sqrt{2} - 1\right) \sqrt{2 - \sqrt{2}}}{2}$

OR using Decimal Values:

$y = - 1.2033332887561 x + 0.63106054525314$

graph{y=-1.2033332887561x+0.63106054525314 [-2.5, 2.5, -1.25, 1.25]}

graph{y=tan x sinx [-2.5,2.5,-1.25,1.25]}