COS(α+β); SINα=- 1/3, α IN QUADRANT IV, TANβ=7/3, β IN QUADRANT III PLEASE, I HAVE BEEN TRYING TO FIND THE ANSWER FOR 3 HOURS, AND I HAVE SPENT LIKE 1000 POINTS

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,

the first one is x^2+2x+2

the second one is x^2+4

the last one is x^2-4x+8