posted by kath
why does supplementary angles always have the same sin?
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.