why does supplementary angles always have the same sin?

Answer 1

use the difference formula:

sin(π-x) = sinπ cosx - cosπ sinx

sinπ=0 and cosπ = -1, so you wind up with

0 - (-1)(sinx) = sinx

Also, if you draw the angles in standard position, recall that

sinθ = y/r
cosθ = x/r

since y is positive in QI and QII, the two angles have the same sine value.

This also shows why cos(π-x) = -cosx -- x is negative in QII.