Question

How to solve for sec(-pi/9)sin(-23pi/6)cos(pi/9) using the values on the unit circle and not decimal values??

Answer 1

sec(-x) = sec(x). True for cosine too. cos(-x) = cos(x).
sec(-pi/9) = sec(pi/9) = 1 / cos(pi/9)

sec(-pi/9) * sin(-23pi/6) * cos(pi/9) = ( 1 / cos(pi/9) ) * sin(-23pi/6) * cos(pi/9) =
sin(-23pi/6)

Keep adding as many multiples of 2pi and sine value will not change.
Therefore, -23pi/6 + 2pi = pi( -23/6 + 2) = pi( (-23+12)/6 ) = pi(-11/6)
Add another 2pi
pi(-11/6) + pi(2) = pi(-11/6 + 2) = pi( (-11 + 12) / 6 ) = pi/6

sin(-23pi/6) = sin(-11pi/6) = sin(pi/6) = 1/2

Answer 2

@aum what happens to the cos(pi/9) though?

Answer 3

cos(pi/9) cancels out sec(pi/9).
cos(pi/9) times sec(pi/9) is 1. They are reciprocals of each other.

Since sec(pi/9) = 1 / cos(pi/9),
cos(pi/9) x sec(pi/9) = cos(pi/9) x 1 / cos(pi/9) = 1