Given sin(α)=−5/6 and α is in quadrant III, find the exact value of sin(α/2).

Answer 1

sinA =-5/6

A is in quadrant III

cosA = -sqr(36-25)/6 = -sqr11/6

sinA/2 = [sqr(1-cosA)/2]

= exactly sqr[(1+sqr(11/6))/2]

check the answer

A = about 56.4+180 = 236.4

A/2 = 118.2

sin118.2 = about 0.88

sqr[(1+sqr11/6)]/2 = sqr[(1+.55)/2] =sqr(.78) = about 0.88