Equation:
cosθ= -12/13 for π < θ< 3π /2
prompt:
find sin 2θ, cos 2θ, and tan 2θ

Cosθ = -12/13.   
For π <θ<3π /2 means  180° <θ< 270°. That is the third quadrant.

Let us just have the positive value of Cosθ = 12/13

Cosθ = Adjacent / Hypotenuse = 12 / 13

So we imagine a right angled triangle with adjacent side = 12, and Hypotenuse = 13.

To get the opposite side we apply Pythagoras' Theorem. Let the opposite side be x.

x² + 12² = 13²
x² + 144 = 169
x²  = 169 - 144
x² = 25
x = √25
x = 5.

Sinθ = Opposite / Hypotenuse = 5 / 13

Tanθ = Opposite / Adjacent = 5 / 12

Recall the angle is in the third quadrant, and in the third quadrant, only Tangent is positive, Cosine and Sine are both negative.

Therefore 
Cosθ = -12/13  Sinθ = -5/13  Tanθ = 5/12

Solving:

i) Sin2θ = 2SinθCosθ.            By Trigonometric Identity.
 
            =  2*(-5/13)*(-12/13)

            = 120/169

ii) Cos2θ = 2Cos²θ - 1

              = 2*(-12/13)(-12/13) - 1

              = 288/169 - 1

              = (288 - 169) / 288

              =  119/288

Tan2θ = 2Tanθ /(1 - Tan²θ)

         = 2*(5/12) / ( 1- (5/12)²)

         =   (5/6) / ( 1 - 25/144)

         = (5/6) / ( (144 -25)/144)

         = (5/6) / (169/25)

         = (5/6) * (25/169)

         = 125/1014    

I hope this helps.

Idon't know that language