Please help ASAP?

π/2 < θ < π
π < 2θ < 2π

cos2θ = -5/12
2θ = π + cos⁻¹(5/12)
θ = [π + cos⁻¹(5/12)] / 2
θ = 4.28

Do you know this identity?

cos(a + b) = cos(a).cos(b) - sin(a).sin(b) → suppose that: a = b = θ

cos(θ + θ) = cos(θ).cos(θ) - sin(θ).sin(θ)

cos(2θ) = cos²(θ) - sin²(θ) → given: cos(2θ) = - 5/12

cos²(θ) - sin²(θ) = - 5/12



cos²(θ) - sin²(θ) = - 5/12 → recall: cos²(x) + sin²(x) = 1 → sin²(x) = 1 - cos²(x)

cos²(θ) - [1 - cos²(θ)] = - 5/12

cos²(θ) - 1 + cos²(θ) = - 5/12

2.cos²(θ) = - (5/12) + 1

2.cos²(θ) = - (5/12) + (12/12)

2.cos²(θ) = 7/12

cos²(θ) = 7/24

cos²(θ) = [± √(7/24)]²

cos(θ) = ± √(7/24) → given: π/2 < θ < θ → cos(θ) < 0

cos(θ) = - √(7/24)

cos(θ) = - (√7)/(√24)

cos(θ) = - (√7)/(2√6)

→ cos(θ) = - (√42)/12


cos²(θ) - sin²(θ) = - 5/12 → recall: cos²(x) + sin²(x) = 1 → cos²(x) = 1 - sin²(x)

[1 - sin²(θ)] - sin²(θ) = - 5/12

1 - sin²(θ) - sin²(θ) = - 5/12

- 2.sin²(θ) = - (5/12) - 1

- 2.sin²(θ) = - (5/12) - (12/12)

- 2.sin²(θ) = - 17/12

sin²(θ) = 17/24

sin²(θ) = [± √(17/24)]²

sin(θ) = ± √(17/24) → given: π/2 < θ < θ → sin(θ) > 0

sin(θ) = √(17/24)

sin(θ) = (√17)/(√24)

sin(θ) = (√17)/(2√6)

→ sin(θ) = (√102)/12

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