Question

What is the slope of the line normal to the tangent line of `f(x) = secx+sin(2x-(3pi)/8)` at `x= (11pi)/8`?

Answer 1

The slope of the normal line is given by `1/(3sqrt(2+sqrt(2)))`

Explanation:

The derivative is given by

`f'(x)=2sin(pi/8+2x)+sec(x)*tan(x)`

for `x=11/8pi` we get

`f'(pi/8)=sqrt(2-sqrt(2))-2sqrt(2+sqrt(2))/(2-sqrt(2))`
simplified to

`f'(pi/8)=-3sqrt(2+sqrt(2))`

so the slope of the normal line is given by

`f'(pi/8)=1/(3sqrt(2+sqrt(2)))`