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)?
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
To find the value of sin(A - B), we'll need to determine the values of A and B first.
Since cosA = 1/3 in quadrant I, we can use the Pythagorean identity to find sinA:
sinA = √(1 - cos^2(A))
= √(1 - (1/3)^2)
= √(1 - 1/9)
= √(8/9)
= √8 / 3
Since sinB = -1/2 in quadrant IV, we can use the Pythagorean identity to find cosB:
cosB = √(1 - sin^2(B))
= √(1 - (-1/2)^2)
= √(1 - 1/4)
= √(3/4)
= √3 / 2
Since sinC = 1/4 in quadrant II, we can use the Pythagorean identity to find cosC:
cosC = -√(1 - sin^2(C))
= -√(1 - (1/4)^2)
= -√(1 - 1/16)
= -√(15/16)
= -√15 / 4
Now that we have the values of sinA, cosB, and cosC, we can find sin(A - B) using the trigonometric identity:
sin(A - B) = sinA * cosB - cosA * sinB
Substituting the values we found:
sin(A - B) = (√8 / 3) * (√3 / 2) - (1/3) * (-1/2)
= (√24 / 6) - (1/6)
= (√24 - 1) / 6
Therefore, sin(A - B) = (√24 - 1) / 6.