What is the slope of the line normal to the tangent line of #f(x) = cosx+sin(2x-pi/12) # at # x= pi/3 #?

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Answer 1 · 2016-08-20T19:15:27

The Reqd. Slope#=2/(sqrt6+sqrt3-sqrt2)#.

Slope of tgt. to the curve # C : f(x)=cosx+sin(2x-pi/12)# at #x=pi/3# is #f'(pi/3)#.

Hence, the slope of the normal at that pt. is #-1/(f'(pi/3))#.

Now, #f(x)=cosx+sin(2x-pi/12)#

#rArr f'(x)=-sinx+cos(2x-pi/12)*d/dx(2x-pi/12)#

#=-sinx+2cos(2x-pi/12)#

#rArr f'(pi/3)=-sin(pi/3)+2cos{2(pi/3)-pi/12}#

#=-sqrt3/2+2cos(7pi/12)#

Here, #2cos(7pi/12)=2cos(pi/3+pi/4)#

#=2{cos(pi/3)cos(pi/4)-sin(pi/3)sin(pi/4)}#

#=2{1/2*1/sqrt2-sqrt3/2*1/sqrt2}=2{(1-sqrt3)/(2sqrt2)}=(1-sqrt3)/sqrt2#

Hence, #f'(pi/3)=-sqrt3/2+(1-sqrt3)/sqrt2=-sqrt3/2+(sqrt2-sqrt6)/2#,

#=(sqrt2-sqrt3-sqrt6)/2#

Therefore, the Reqd. Slope#=2/(sqrt6+sqrt3-sqrt2)#.