Question

What is the equation of the line normal to `f(x)=cos(5x+pi/4)` at `x=pi/3`?

Answer 1

`color(red)(y-((sqrt2+sqrt6))/4=-((sqrt2+sqrt6))/5*(x-pi/3)`

Explanation:

Given `f(x)=cos (5x+pi/4)` at `x_1=pi/3`

Solve for the point `(x_1, y_1)`

`f(pi/3)=cos((5*pi)/3+pi/4)=(sqrt2+sqrt6)/4`

point `(x_1, y_1)=(pi/3, (sqrt2+sqrt6)/4)`

Solve for the slope m

`f' (x)=-5*sin (5x+pi/4)`

`m=-5*sin ((5pi)/3+pi/4)`

`m=(-5(sqrt2-sqrt6))/4`

for the normal line `m_n`

`m_n=-1/m=-1/((-5(sqrt2-sqrt6))/4)=4/(5(sqrt2-sqrt6))`
`m_n=-(sqrt2+sqrt6)/5`

Solve the normal line

`y-y_1=m_n(x-x_1)`

`color(red)(y-((sqrt2+sqrt6))/4=-((sqrt2+sqrt6))/5*(x-pi/3)`

Kindly see the graph of `y=cos (5x+pi/4)` and the normal line `y-((sqrt2+sqrt6))/4=-((sqrt2+sqrt6))/5*(x-pi/3)`

graph{(y-cos (5x+pi/4))(y-((sqrt2+sqrt6))/4+((sqrt2+sqrt6))/5*(x-pi/3))=0[-5,5,-2.5,2.5]}

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