Answer: sin2θ = 240/289
cos2θ = -161/289
tan2θ = -240/161
2θ is in quadrant 2
Step-by-step explanation: tanθ = 15/8
sinθ/cosθ = 15/8
sinθ = 15cosθ/8
sin²θ + cos²θ = 1
(15cosθ/8)² + cos²θ = 1
225cos²θ/64 + cos²θ = 1
225cos²θ + 64cos²θ = 64
289cos²θ = 64
cos²θ = 64/289
cosθ = ±√64/289
cosθ = ±8/17
As tanθ in quad 3 ⇒ tanθ is +, cosθ is - and sinθ is -
So, cosθ = -8/17
as sin θ = 15cosθ/8 = -15/17
For sin2θ = 2sinθcosθ = 2.(-15/17).(-8/17) = 240/289
For cos2θ = cos²θ - sin²θ = (-8/17)² - (-15/17)² = -161/289
For tan2θ = sin2θ/cos2θ = (240/289) / (-161/289) = -240/161
As sin2θ + and cos2θ - ⇒ 2θ is in quadrant 2
