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

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Answer 1 · 2018-04-21T12:36:33

Equation of normal is # x +y= 1.39#

#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]