Recall the angle sum identity for cosine,
cos(α + β) = cos(α) cos(β) - sin(α) sin(β)
We're given sin(α) = -1/3 and tan(β) = 7/3. Since α terminates in quadrant IV, we know that cos(α) is positive, and since β terminates in quadrant III, both sin(β) and cos(β) are negative.
Recall the Pythagorean identities,
cos²(x) + sin²(x) = 1
and
1 + tan²(x) = sec²(x)
Using the first identity, we find
cos²(α) = 1 - sin²(α)
cos(α) = + √(1 - (-1/3)²)
cos(α) = + √(8/9)
cos(α) = √8/3
Using the second identity,
sec²(β) = 1 + tan²(β)
sec(β) = - √(1 + tan²(β))
sec(β) = - √(1 + (7/3)²)
sec(β) = - √(58/9)
sec(β) = -√58/3
cos(β) = -3/√58
Then by definition of tangent,
tan(β) = sin(β) / cos(β)
sin(β) = tan(β) cos(β)
sin(β) = (7/3) (-3/√58)
sin(β) = -7/√58
Putting everything together,
cos(α + β) = cos(α) cos(β) - sin(α) sin(β)
cos(α + β) = (√8/3) (-3/√58) - (-1/3) (-7/√58)
cos(α + β) = -(7 + 3√8)/(3√58)
