given that tan a=5/12, a is in QIII, find sin(2a) and cos(2a)?

Question

a=angle
a is in quadrant III

Answer 1

sin a = - 5 / 13
cos a = - 12 / 13

sin 2a = 2 sin a cos a
sin 2a = 2 (5 / 13) (12 / 13)
sin 2a = 120 / 169

cos 2a = cos ² a - sin ² a
cos 2a = 144 / 169 - 25 / 169
cos 2a = 119 / 169

Answer 2

Since tan a = 5/12 and a is in QIII,
and because 5,12,13 is a right triangle.
sin a = -5/13
cos a = -12/13

Using double angle formulas
sin(2a) = 2(sina)(cosa)
sin(2a) = 2(-5/13)(-12/13)
sin(2a) = 120/169

cos(2a) = cos^2(a)-sin^2(a)
cos(2a) = (-12/13)^2-(-5/13)^2
cos(2a) =144/169 - 25/169
cos(2a) = 119/169

Answer 3

I believe mathman is close except I think sin(2a) and cos(2a) must stay negative because 2a is still in QIII. Someone correct me if I'm wrong.