if cosA = 1/3 in I, then sinA = √8/3 using Pythagoras
if sinB= -1/2 in IV, the cosB = √3/2
sin (A-B) = sinAcosB - cosAsinB
= (√8/3)(√3/2) - (1/3)(-1/2)
= (√24 + 1)/6
find the exact value given that cosA=(1/3) in quad I, sin B=(-1/2) in quad 4, and sin C=(1/4) in quad 2
sin(A-B)?
1 answer