What is the slope of the line normal to the tangent line of #f(x) = cosx+sin(2x-pi/12) # at # x= (5pi)/8 #?

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Answer 1 · 2016-07-30T09:08:32

Slope #m_p=((sqrt(2+sqrt2)-2sqrt3)(sqrt2+10))/(-49)#

Slope #m_p=0.37651589912173#

#f(x)=cos x+sin(2x-pi/12)" "#at #x=(5pi)/8#

#f'(x)=-sin x+ 2*cos(2x-pi/12)#

#f'((5pi)/8)=-sin((5pi)/8)+2*cos(2*((5pi)/8)-pi/12)#

#f'((5pi)/8)=-cos (pi/8)+2*cos((7pi)/6)#

#f'((5pi)/8)=-1/2sqrt(2+sqrt2)+2((-sqrt3)/2)#

#f'((5pi)/8)=(-sqrt(2+sqrt2)-2sqrt3)/2#

For the slope of the normal line

#m_p=-1/m=-1/(f'((5pi)/8))=2/(sqrt(2+sqrt2)+2sqrt3)#

#m_p=(2(sqrt(2+sqrt2)-2sqrt3))/(sqrt2-10)#

#m_p=(2(sqrt(2+sqrt2)-2sqrt3)(sqrt2+10))/(-98)#

#m_p=((sqrt(2+sqrt2)-2sqrt3)(sqrt2+10))/(-49)#

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