posted by nik .
GIVEN THE ANGEL ALFA WITH COS ALFA = -1/a. IF ALFA IN QUADRANT TWO, THE VALUE OF SIN ALFA IS....
since sin^2 + cos^2 = 1,
sin^2 = 1 - cos^2 = 1-(1/a^2)
so sin = √(1 - 1/a^2) = 1/a √(a^2-1)
to see this, draw your triangle and label the adjacent leg=1 and the hypotenuse=a. Then the other leg is √(a^2-1) and the sin is just √(a^2-1)/a.
Since sin > 0 in QII, that's the value; no - sign needed. in fact, if you draw the angle in QII, you will see that's so, since sin=y/r and y>0.