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

1 Answer
Jun 7, 2018

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