f(x) = cot^2 x + sin(x-pi/2) f(x)=cot2x+sin(x−π2)
First, find the slope of the tangent line of f(x)f(x) at x=(5pi)/6x=5π6 by finding the derivative of f(x)f(x):
f'(x) = 2(cotx)(-csc^2 x) + cos(x-pi/2)
f'(x) = frac{-2cosx}{sin^3 x} + cos(x-pi/2)
f'((5pi)/6) = frac{-2cos((5pi)/6)}{sin^3 ((5pi)/6)} + cos(frac{5pi-3pi}{6})
= frac{-2(-sqrt3/2)}{(1/2)^3} + cos(pi/3)
= frac{sqrt3}{1/8} + 1/2
= 8sqrt3 + 1/2
Use common denominators to combine the two terms into one:
f'((5pi)/6)= frac{16sqrt3 + 1}{2}
Now that we have the slope of the tangent line at the point where x=(5pi)/6, find the slope of the normal line at this same point by taking the opposite reciprocal of this value:
"Normal line slope" = frac{-1}{frac{16sqrt3 + 1}{2}}
"Normal line slope" = frac{-2}{16sqrt3 + 1}
