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

Nov 12, 2017

Equation of normal line is $x + 1.17 y = 0.87$

#### Explanation:

x=pi/8 ~~0.39 ; f(x)=tanx = or f(x)= tan (pi/8) ~~ 0.41

So at $\left(0.39 , 0.41\right)$ tangent and normal is drawn.

Slope of the tangent is $f ' \left(x\right) = {\sec}^{2} x$. at $x = \frac{\pi}{8}$

${\sec}^{2} x = \frac{1}{\cos} ^ 2 x = \frac{1}{\cos \left(\frac{\pi}{8}\right)} ^ 2 \approx 1.17$ . Slope of tangent

is ${m}_{1} = 1.17$ Normal line is perpendicular to tangent , so

slope of the normal line ${m}_{2} = - \frac{1}{m} _ 1 = - \frac{1}{1.17}$

Equation of normal line at point (0.39,0.41) ; m_2= -1/1.17

is $y - {y}_{1} = {m}_{2} \left(x - {x}_{1}\right) \mathmr{and} \left(y - 0.41\right) = - \frac{1}{1.17} \left(x - 0.39\right)$ or

$1.17 \left(y - 0.41\right) = - \left(x - 0.39\right)$ or

$x + 1.17 y = 0.39 + \left(1.17 \cdot 0.41\right)$ or

$x + 1.17 y = 0.39 + \left(1.17 \cdot 0.41\right) \mathmr{and} x + 1.17 y = 0.87$

Equation of normal line is $x + 1.17 y = 0.87$ [Ans]