Question

What is the slope of the line normal to the tangent line of `f(x) = cscx+cos^2(5x-pi)` at `x= (11pi)/8`?

Answer 1

`m=-0.2510` (rounded)

Explanation:

This may be a wildly constructed problem, but here's my attempt to answer it:

To determine the slope of any tangent line, we need the derivative of the curve. For this problem:
`f'(x) =-cscxcotx + 2cos(5x-pi)*(-sin(5x-pi))*5`

`f'(x) = -cosx/(sin^2x) -10cos(5x-pi)*sin(5x-pi)`

When `x=(11pi)/8` ,
the slope of the tangent line at that point is:
`m_T=f'((11pi)/8)=3.9839` (rounded)

Since the lines are perpendicular, the slope of the normal line is the negative reciprocal of the slope of the tangent:
`m_N=-1/m_T`

So, `m_N=-0.2510`