What is the slope of the line normal to the tangent line of #f(x) = cosx-cos^2x +1 # at # x= (17pi)/12 #?

1 Answer

Slope of the normal to the tangent line #m_n#

#=(-1)/(1/4(sqrt(6+3sqrt3)+sqrt(2-sqrt3))+1/8(sqrt(21+12sqrt3)-sqrt(21-12sqrt3)-2))#

#m_n=-0.68216275480416#

Explanation:

From the given equation #f(x)=cos x-cos^2 x+1#
Obtain the first derivative

#f(x)=cos x-cos^2 x+1#

#f' (x)=-sin x-2cos x*(-sin x)+0#

#f' (x)=-sin x+2 sin x cos x#

Slope #m=f' ((17pi)/12)#

#m=f' ((17pi)/12)=-sin ((17pi)/12)+2 sin ((17pi)/12) cos ((17pi)/12)#

Using trigonometric formulas like half-angle formulas and sum/difference formulas

#m=1/4(sqrt(6+3sqrt3)+sqrt(2-sqrt3))+1/8(sqrt(21+12sqrt3)-sqrt(21-12sqrt3)-2)#

Now, we have to solve for the slope of the normal line #m_n#

#m_n=(-1)/m#

#m_n=#

#=(-1)/(1/4(sqrt(6+3sqrt3)+sqrt(2-sqrt3))+1/8(sqrt(21+12sqrt3)-sqrt(21-12sqrt3)-2))#

#m_n=-0.68216275480416#

God bless.....I hope the explanation is useful.