sin(2 theta) = 2 sin(theta)cos(theta). You got a 5/12/13 right triangle since 12^2 + x^2 = 13^2, 144 + x^2 = 169, x^2 = 25, and x = 5 (forget -5). The adjacent leg must be 5, so cos(theta) = 5/13.

sin(2 theta) = 2(12/13)(5/13) = (2 * 12 * 5) / 13^2 = 120/169. You got this right, good job.

sinθ = 12/13

cos²θ = 1-sin²θ = 25/169

cosθ = 5/13

sin(2θ) = 2sinθcosθ = 120/169

sin Ө = 12/13

cos Ө = 5/13

sin 2Ө = 2 sin Ө cos Ө

sin 2Ө = 2 x 12/13 x 5/13

sin 2Ө = 120/169______as you suggest.

sinT = 12/13, y = 12 and r = 13

x = √(r^2 - y^2) = √(13^2 - 12^2) = 5

cosT = x/r = 5/13

sin(2T) = 2 sinT cos T

= 2(12/13)(5/13)

= 120/169, True

Yes.

A quick check using complex numbers:

sinθ = 12/13, so arg(5+12i) = θ,

so 2θ = arg((5+12i)²) = arg((25-144)+2(5)(12)i) = arg(-119+120i), and |-119 + 120i| = √((-119)²+120²) = 169,

so sin(2θ) = 120/169

sin(2θ) = 2sin(θ)cos(θ)

sin^2(θ) + cos^2(θ) = 1

so if θ is in 1st quadrant where cos is positive, cos(θ) = √(1-sin^2(θ)) = √(25/169) = 5/13

sin(2θ) = 2(5/13)(12/13) = 120/169