If sinA = 3/5 and cosB = 5/13, where A and B are both acute, then sin(A + B) equals:

The 3-4-5 and 5-12-13 triangles give cosA = 4/5 and sinB = 12/13. Then sin(A + B) = sinA·cosB + cosA·sinB = 15/65 + 48/65 = 63/65 ≈ 0.969. Subtracting instead of adding produces sin(A − B) = −33/65, and 16/65 is the size of cos(A + B).