Question

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

Answer 1

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.