cos(A - B) = cos(A) * cos(B) + sin(A) * sin(B)
We need to find cos(A), cos(B), sin(A), sin(B), and then plug these values into the formula.
Since A = arcsin(-3/5), we know sin(A) = -3/5, and we know that angle A is in Quadrant IV.
We use sin^2(A) + cos^2(A) = 1 to find cos(A).
(-3/5)^2 + cos^2(A) = 1 -----> 9/25 + cos^2(A) = 25/25 -----> cos^2(A) = 16/25 -----> cos(A) = 4/5
We pick the positive value for cos(A) because cosine is positive in quadrant IV.
Now we have to find sin(B) and cos(B).
The problem tells us that cos(B) = 5/6, and that angle B is also in quadrant IV (so sin(B) will be negative).
We use sin^2(B) + cos^2(B) = 1 to find sin(B).
sin^2(B) + (5/6)^2 = 1 -----> sin^2(B) + 25/36 = 36/36 -----> sin^2(B) = 11/36 -----> sin(B) = -[sqrt(11)]/6.
We now have all four values:
sin(A) = -3/5
cos(A) = 4/5
sin(B) = -[sqrt(11)]/6
cos(B) = 5/6
Plug these values into the formula:
cos(A - B) = cos(A) * cos(B) + sin(A) * sin(B)
cos(A - B) = (4/5) * (5/6) + (-3/5) * (-[sqrt(11)]/6)
cos(A - B) = 20/30 + [3sqrt(11)]/30
cos(A - B) = [20 + 3sqrt(11)]/30
Hope this helps!
