Question

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

Answer 1

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

Slope `m_p=0.37651589912173`

Explanation:

`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.