# What is the equation of the normal line of f(x)=cos(2x-pi/2) at x=pi/6?

Apr 21, 2018

Equation of normal is $x + y = 1.39$

#### Explanation:

f(x)= cos (2 x-pi/2) ; x =pi/6 ~~ 0.524

 :.f(pi/6)= cos (2*pi/6-pi/2) ; or

 :f(pi/6)= cos (pi/3-pi/2)= cos (-pi/6) ~~ 0.866 ;

So the point is $\left(0.524 , 0.866\right)$ at which normal is drawn.

$f \left(x\right) = \cos \left(2 x - \frac{\pi}{2}\right) \therefore f ' \left(x\right) = - \sin \left(2 x - \frac{\pi}{2}\right) \cdot 2$ or

$f ' \left(x\right) = - 2 \sin \left(2 x - \frac{\pi}{2}\right)$

$\therefore f ' \left(\frac{\pi}{6}\right) = - 2 \sin \left(2 \cdot \frac{\pi}{6} - \frac{\pi}{2}\right)$ or

$\therefore f ' \left(\frac{\pi}{6}\right) = - 2 \sin \left(- \frac{\pi}{6}\right) = - 2 \cdot \left(- \frac{1}{2}\right) = 1$

The slope of tangent at the point is ${m}_{t} = 1$ , so

slope of normal at the point is ${m}_{n} = - 1 \therefore$ Equation of

normal having slope $- 1$ at point$\left(0.524 , 0.866\right)$ is

$y - 0.866 = - 1 \left(x - 0.524\right) \left[y - {y}_{1} = m \left(x - {x}_{1}\right)\right]$ or

$x + y = 0.524 + 0.866 \mathmr{and} x + y = 1.39$

Equation of normal is $x + y = 1.39$ [Ans]