What is the equation of the normal line of $f (x) = cos (2 x - \frac{π}{2})$ $f(x)=cos(2x-pi/2)$ at $x = \frac{π}{6}$ $x=pi/6$ ?

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Answer 1 · 2018-04-21T12:36:33

Equation of normal is $x + y = 1.39$ $x +y= 1.39$

$f (x) = cos (2 x - \frac{π}{2}); x = \frac{π}{6} \approx 0.524$ $f(x)= cos (2 x-pi/2) ; x =pi/6 ~~ 0.524$

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

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

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

$f(x)= cos (2 x-pi/2):.f'(x)= -sin (2 x-pi/2)*2$ or

$f'(x)= -2 sin (2 x-pi/2)$

$:. f'(pi/6)= -2 sin (2* pi/6-pi/2)$ or

$:. f'(pi/6)= -2 sin (-pi/6)= -2 *(-1/2)=1$

The slope of tangent at the point is $m_t=1$ , so

slope of normal at the point is $m_n=-1:.$ Equation of

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

$y- 0.866= -1(x-0.524) [y-y_1=m(x-x_1)]$ or

$x+y= 0.524+0.866 or x +y= 1.39$

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

What is the equation of the normal line of f(x)=cos(2x-pi/2)f(x)=cos(2x−π2)f(x)=cos(2x-pi/2) at x=pi/6x=π6x=pi/6?