Recall the angle sum identity for cosine,
cos(α + β) = cos(α) cos(β) - sin(α) sin(β)
Then given sin(α) = -1/3, we have
cos(α + β) = cos(α) cos(β) + 1/3 sin(β)
α is in quadrant IV, so 3π/2 < α < 2π, and so sin(α) is negative and cos(α) is positive. Then from the Pythagorean identity, we get
sin²(α) + cos²(α) = 1
⇒ cos(α) = √(1 - sin²(α)) = 2√2/3
β is in quadrant III, so π < β < 3π/2, so both sin(β) and cos(β) are negative. If we multiply both sides of the previous identity by 1/cos²(α) and replace α with β, we get
sin²(β)/cos²(β) + cos²(β)/cos²(β) = 1/cos²(β)
⇒ tan²(β) + 1 = 1/cos²(β)
⇒ 1/cos(β) = -√(tan²(β) + 1) = -√58/3
⇒ cos(β) = -3/√58
Also recall that by definition,
tan(β) = sin(β)/cos(β)
⇒ sin(β) = tan(β) cos(β) = 7/3 • (-3/√58) = -7/√58
Putting everything together, we find that
cos(α + β) = 2√2/3 • (-3/√58) + 1/3 • (-7/√58) = -(6√2 + 7)/(3√58)
or more clearly,
[tex]\cos(\alpha + \beta) = -\dfrac{6\sqrt2 + 7}{3\sqrt{58}}[/tex]
