Question

What is the equation of the normal line of `f(x)= cscx` at `x = pi/8`?

Answer 1

Note that `f(pi/8)=csc(pi/8)`, so the normal line will pass through the point `(pi/8,csc(pi/8))`.

The slope of the normal line will be the inverse reciprocal of the slope of the tangent line at `x=pi/8`, since the lines are perpendicular.

To find the slope of the tangent line there, find the derivative of `f` at `x=pi/8`.

The derivative of `csc(x)`, if you don't have it memorized, is easiest found using the chain rule on `csc(x)=1/sin(x)=(sin(x))^-1`. (Note this isn't inverse sine— rather sine to the negative first power.)

`f(x)=(sin(x))^-1`

`=>f'(x)=-(sin(x))^-2*cos(x)=-cos(x)/sin^2(x)`

So the slope of the tangent line is `-cos(pi/8)/sin^2(pi/8)` and the slope of the normal line is `sin^2(pi/8)/cos(pi/8)`.

The line passing through `(pi/8,csc(pi/8))` with slope `sin^2(pi/8)/cos(pi/8)` is:

`y-1/sin(pi/8)=sin^2(pi/8)/cos(pi/8)(x-pi/8)`

We could find these decimals, but we can also find the exact values.

I like to use the double angle formulas:

`cos(2x)=2cos^2(x)-1`

So:

`cos(pi/4)=2cos^2(pi/8)-1`

`cos^2(pi/8)=1/2(cos(pi/4)+1)=1/2(1/sqrt2+1)=(1+sqrt2)/(2sqrt2)=(2+sqrt2)/4`

Then:

`cos(pi/8)=sqrt(2+sqrt2)/2`

The same process can be done using `cos(2x)=1-2sin^2(x)` to find that

`sin^2(pi/8)=(2-sqrt2)/4`

`sin(pi/8)=sqrt(2-sqrt2)/2`

Then the normal line becomes:

`y-2/sqrt(2-sqrt2)=(2-sqrt2)/4(2/sqrt(2+sqrt2))(x-pi/8)`

`y-2/sqrt(2-sqrt2)=(2-sqrt2)/(2sqrt(2+sqrt2))(x-pi/8)`