What is the slope of the line normal to the tangent line of #f(x) = cosx*sin(2x-pi/6) # at # x= pi/3 #?

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Answer 1 · 2017-07-25T19:16:17

#m=2/sqrt3=2/3sqrt3#

Let's first find the derivative of the function using the product rule :

#f'(x)=-sinxsin(2x-pi/6)+2cosxcos(2x-pi/6)#

The slope of the tangent line is #f'(pi/3)#

#f'(pi/3)=-sin(pi/3)sin((2pi)/3-pi/6)+2cos(pi/3)cos((2pi)/3-pi/6)=#

#-sqrt3/2sin(pi/2)+2*1/2cos(pi/2)=-sqrt3/2#

So now the slope of the normal line to the tangent is :

#m=-1/(f'(pi/3))=2/sqrt3=2/3sqrt3#